@@sparkyfire2443 because essentially so if you go up the 1 mile in 15 mph, there's no time left to go down, if you tried to average 30 mph over 2 miles, took me a bit to understand too. really had to think there, i thought it'd be 45. here let me try and re phrase it for you. if you're trying to run 200 meters in an hour, and your top speed is able to run 100 meters per hour, then how fast will you have to run the other 100 meters if you're trying to average 200 meters per hour? well it's impossible since your top speed is 100 meters per hour. either that or you take 0 minutes for the latter 100 meters
That was my initial thought but I knew mph seemed suspicious so I needed a way to check my logic. I converted it to time actually spent and got the right answer.
For those who are wondering why it isn't 45 mph: Average speed is calculated distance/time. To show this, an example of a unit of average speed is mph = m/h = miles (unit of distance)/hour (unit of time). If you travel first mile at 15 mph and second mile at 45 mph, your average speed is not the average of the two speeds because you spend different amounts of time on each mile. Imagine it this way: if you travelled the same amount of time, let's say 4 minutes from the question, first at 15 mph, then at 45 mph (so 8 minutes travel time total), indeed your average speed would be 30 mph. However, you would also travel 4 miles! But the question said we have to travel only 2 miles exactly. Thus, the restriction on distance also creates a restriction on time.
I just keep confusing myself. "If you travel first mile at 15 mph and second mile at 45 mph, your average speed is not the average of the two speeds because you spend different amounts of time on each mile." If I travel with 15 mph, I take 1/15th hour to travel 1 mile, which is 4 minutes, which is 240 seconds If I travel with 45 mph, I take 1/45th hour to travek 1 mile, which is 1.33 minutes, which is 79.99 seconds (80 seconds for sanity's sake) Why is it incorrect to do (240+80)/2, which is 160 seconds, which is 2,66 minutes, which is 0.0444 hour, which is 22.5 mph? Is it incorrect to say "over teh course of this [2 mile] journey, on average I spent 2.66 minutes on each mile"?
So the only reason this problem doesn't work is the assumption that the average is distance/time. But why would it be calculated that way? MPH is a defined set amount of motion(or speed) not an average. Going at 1MPH for 1 mile will take 1 hour. This is a measurement for motion. So the average for this should be motion/distance. So 15mph+45mph=60mph 60mph/2miles = 30mph average.
This was a frustrating one, because I spent a significant amount of time trying to figure out where I went wrong. Turns out I wasn't wrong, it is indeed impossible.
@@vincenttommasini4048 Sure, let's work it out. The first half of the trip takes 1/15 of an hour. With your speed, the second half of the trip takes 1/30720 of an hour. So the time for the entire journey is 1/15 + 1/30720, which equals 2049/30720 or 683/10240 of an hour. That means the average speed over the entire 2 miles is: 2/(683/10240) mph ≈ 29.985 mph. Not 30 mph.
Step 1- Get a flux capacitor. Step 2- Attach it to your car. Step 3- Travel 88 mph. Step 4- Travel back to time and kill Max Wartheimer. Step 5- Live peacefully.
Simple: Pause your clock or watch for the time you come down the hill, t=0 Although it's a artificial process, but the main thing is you got yourself a life
if you think at it in a more absurd way, a fool can create a nonsense and there's no genius that can math it out, also a sensated jibberish is way easier to come up with than to solve... it's kinda the legit reverse of "to destruct" vs "to build", one is hard the other is simple
For anyone wondering then what's the answer , I will tell you. I am going to assume that distance travelled is given(2miles) and the watch through which time is measured is in car . First we need to understand the relation of time which is inside and time which is outside Let's say time outside is t And time inside is t' t'=t√(1-v²/c²) For calculation purposes I am going to use si units So total time vehicle need is 2/30 hrs = 240 seconds Now we will be calculating time that is flowing outside vehicle 1/15hrs = 240 seconds But hang on time inside vehicle is less then that is 240√(1-v²/c²) = t' v=15mph=6.7056 m/s Sadly no calculator was able to calculate such small value but fear not we will use calculas my friends !! Remaining time left for 2nd part of journey is 240-240√(1-v²/c²) = t"( time left for other trip) 240(1-√(1-v²/c²))=t" Now using expansion of root using mclauren series we can neglect other terms(they are very small) and keep the first the term 240(1-(1-0.5v²/c²))=t" 240×0.5×v²/c²=t" 120×v²/c²=t" Now calculator gave the answer thanks to calculas 0.000000000000060036467002990743346703130254940792991950836435617448761427151923217429171811094289749944538966505597887983802322401518797793890192189656149242982301575455283765168756435935707823136503331530207014184 ( it's t") Now let's assume speed of car going downwards is v' So time taken in going down ( this time is from outside perspective ) is 1609.344/v' But hang on this time is from outside perspective we need to calculate from inside 1609.344/v'×√(1-v'²/c²) = t" We have to calculate for v' we can't use mclauren series to estimate because this time v' is bigger v'=c×1609.344/ √ (x²+(c.t")²) v'=99.999999999999993746201353855131488035053196294571173429056176961242868216566225763338544466330421185462845993920903672772712136295199625718519695883561631704644951046866851689889849005332679328390186 % of speed of light Hahahahaa so car needs to travel this much speed So car has to travel v'= 299792457.99999998125158332035157644711226187877905984138241039911293970197614465141374188531703495207365180468193000769641759046666368909394835035750345423363226239891629886666833531582557539043583883 m/s Let's transform this into mph lest Americans will be angry v'=670616628.7685007780609863492531369132546843006244598512699575894491186279322226850367990279586936394919624747170011414560958111092380413595302843003339505755661462749383256073362706618588409009968468 mph This took lots of the time and time in my country is 4:29 am what the hell I am doing with my life !!!
Chit c ctrl Ism the problem lies within the question: are they asking average speed over distance or average speed over time? In the real world these are two different things. That’s why they are separate variables in the equation. Most people understand speed = distance/time but the question was intentionally misleading.
thats not 100 percent accurate because the time at top of hill is also flowing fast than the ground...so calculate that also buts its little hard...yo have to use calculus cuz each rise in hills height will make tha time fast wrt ground..so could you calculate also those😅😅😂
of course according to general relativity....because more the gravity more slow the time will pass wrt observer at low gravity region and as you move up the hill height increases and unnoticeable amount of gravitational force will decrease........and bro what about the length contraction....you should also use it..😅😂
Got it right away. Usually I fail miserably at these kinds of things, but calculating speed-time-distance was something I learned as a radar plotter in the navy. This was in the pre-computer days where we used paper and pencil to develop a picture using a plotting table. We got very good at doing relative velocity calculations on the fly in our heads. As soon as I saw the numbers, I was twigged. 1/15, bay-beh.
Ah yes, during my time as a massivity neutronian outlander, we surveyed plots of gravitonian hemispheres on Jupiter’s moons. Elementary my dear watson. By the time my coworkers slid the cover off their Ti-83s, I already had the answer jotted down on papyrus.
At first I didn't get it, but after I grabbed a piece of paper and started thinking about the units, I quickly realized that I needed to start by computing the trip and leg times.
@@stefanluginger3682 Yeah, my mistake was probably not writing anything down, could have been a bit quicker if I did 😅 And true, we aren't. One can dream though
Great riddle. Solved it before watching the solution. The error we tend to make (when we think 45 mph is the answer) is that we try to calculate average speed over the distance, while average speed is calculate over time.
No, the error we make is to assume that because of the low speed at the start, you obviously don't need to run that fast just to reach this average speed
@@Melpheos1er I think Dirk is right. Usually you think that since it is 2 miles, the other mile just needs to be 45 so (45+15)/2 = 30. However, that is wrong because as Dirk said, average speed is divided by time and not distance.
@@JJviniciuss interesting point but let’s say we do travel the second mike at 45mph wouldn’t we still get an average speed al together of 30 miles per hour.
@@JJviniciuss ah I got confused because I didn’t think there was an actual time limit and heard it again and he said “average 30 miles in just one hour” so the time would be spent in the first half of the trip due to lack of faster speed.
Part of the reason I got this one quickly is because I used to do distance running, and runners are used to thinking in terms of minutes per mile. When you're used to working with that unit, the answer becomes obvious pretty quickly.
I am a machinist, and I deal with velocity a lot and conversions a lot, although averages aren't really my job description in this context. In any case, just converting units to see the amount of total time you have shows that both sides of the question are 4 minutes, and therefor unless you can quantum tunnel, teleport, gain omnipresence, or time travel, you cannot physically be at the middle and end point at the exact same time and even if you could, the only way for the speed to actually average correctly would be to reverse or freeze time to make up the discrepancy. The other methods jsut replicate the same results, but don't actually solve the problem with speed.
@@mountainrun I've just posted this, but I'd say too many people are over thinking this question... It's not actually a trick question.... You never specified whether the average was over time, or distance. If you consider the average being over time, then you have to calculate this as you did. But if the average is over distance, then the answer is 45mph. If half of a journey was completed at 15 mph, and we know that the average speed over the distance of the whole journey was 30mph, then logically the remaining speed for the second half of the journeys distance was 45mph. Think of it this way, if we split the distance into 1% chunks, and after each 1% of distance travelled, we drop 15 sweets for the first half, and "X" number of sweets for each remaining percent after 50%, for us to reach and average of 30 sweets per 1% travelled, we need to drop 45 sweets per percentage travelled. When time is not a factor, then the speed is irrelevant, only the value needed to be assigned to each percentage, no matter how much faster we may have travelled it.
@@Ste-The-Leo I mean the video says it was a trick question, even Einstein himself almost got tricked by it. Tbf I also got it wrong on the 1st place, but the explanation made it clear. Many ppl also hv the same answer like the video since they're accustomized with that unit (runner, cyclist, etc.) Meanwhile, I think it simple like u since I'm only a freshman student. Imo, the problem with ur solution (which was also the 1st solution that I thought after seeing the question) is the difference in time passed on each speed. Imo, The easiest way to disprove something is to take the extreme case. Lets say ur solution is right. When the car goes uphill, it takes 1/15 hour for the car to reach the top of the hill (1m/15mph=1h). Meanwhile, the time it takes for the car to downhill is 1/45 hour (1m/45mph=1h). That means, in that 2 mile trip, the car takes 3/4 times of the trip going 15mph, while it goes fast (45mph) for only a quarter of the "trip time". If we take it to the extreme case like I said before, then taking another example, what is the speed needed for the car in the last 0.1 mile if it already goes 15mph for 1.9 mile, to get the same 30mph average like in the 1st question? To get 45mph as the solution (speed of downhill) like u said, then the car should get to the top of the hill on its first 1/4 trip times traveled by going 15mph, then now the car CAN ride the rest trip times by going 45mph, like u said. That means, the hill shape is different than the question in this video (the top of the hill is on 1/2 mile, then it downhill for 3/4 mile). After thinking about it.. I just remember it's a highschool physics problem (static movement)😅. hope u got enlightened😊
@@felixfong3834 To travel 1 mile at 15 mph takes the same time as to travel 2 miles at 30 mph. If you double both distance and speed the time will remain constant.
@@michaelbrannon8452 That doesn't work. To average 30mph in a fixed course of 2 miles, you have to complete the trip in 4 minutes. At 15mph, the car trundled along, and finished the first mile in 4 minutes. In order for that car to average 30mph, it can no longer spend any time traveling at all, so there is no speed the car can travel at to reach the stated goal. Even Star Trek-esque warp travel couldn't make up the difference. It is a trick question.
@@michaelbrannon8452 That only works if the car stays at 15mph and 45mph for equal amounts of time, in which case the trip would take eight minutes across four miles.
what's the frame of reference for the trip? technically, if you consider the trip from the point of view of the car and not outside, there is a way to make the downhill trip last 0 minutes: travel at the speed of light.
@@Jack-ts1jq Well - it didn't seem that Einstein was stumped by the riddle for days on end. In his answering letter he stated 'I had to calculate that one' - which probably took a pencil and a sheet of paper and maximal as long as Presh here needed to explain it here. That's the fun with trick questions. One assumes to get a 'real' riddle with a solution that makes sense. The problem is figuring out, someone is messing with you, and the counter-intuitive answer is indeed correct.
@@robertnett9793 I doubt Einstein had to take pencil to paper, I'm sure he could calculate it in his head, as I did. I'm no Einstein, I went at first the '60 mph speed for the descent' route before I figured in my head how much time it took for ascent compared to time it would take for 2 miles at 30 mph. btw, My first thinking was 15 is half of desired product for first half, added to 60 for second half, which is double the desired result.
Yup. I was planning on going back to college to finish my degree next semester.. When a simple-looking problem like this goes SO FAR over my head, I've come to realize that I needn't bother. Lol.. I'm kidding, of course, but not really. I mean, c'mon, if the answer isn't 45, then I'm lost beyond all redemption. Haha
It’s actually pretty easy. If it weren’t for the bit about Einstein being fooled, it would not be intimidating at all. Of course, you might fall into the trap which is assuming you can just find the number whose average with 15 is 30.
They were mass producing automobiles in 1900 in France and the USA. My grandfather had a 1921 vehicle a Ford we believe. So, in 1935 that would be considered old.
I was tempted to do the math. Instead I wondered how much time was left to complete the trip averaging 30mph. Since the car had travelled half the distance at half the desired speed, there was zero time left. Good video!
So would there be an answer if the question remained exactly the same but the first mile was not driven at 15 mph but rather 20 mph? Or something faster than 15?
The way I thought about it was to imagine a second car, leaving at the same time, and travelling at the desired 30 mph. So at the exact time the first car completes their one mile at 15 mph, the second car completes the their two miles at 30 mph. Which means that there is no time left for the first car to catch up with the second car.
Got this wrong with the 45mph assumption. I rejoice in getting things like this wrong though as I always learn something new from it. Thanks for the video.
It's divided into 1 time frame because you are averaging. One overall rate (30 mph), one total distance (two miles), and one overall time (not given). If knowing that d=rt is having a wild imagination, then I'm proud to have one. You can (and have to) assume time because without time, rate problems (and life) fall apart. There are many things that are implied in math (and life, but I digress) that are too obvious to state as that would take away most of or all of the problem. But GaryLifo is right, it doesn't matter if you got it wrong, so long as you learn.
@@thewhinjaninja3610 Because it is impossible to reach an average of 30mph over 2 miles because you've already gone 15mph over 1 mile....so you've spent 0.06666 hr (4 mins) going up hill. In order to average 30mph over 2 miles, you would need to travel over the 2 miles in 0.06666hr, or 4 minutes. You've already spent the time necessary to travel an average of 2 miles at 30mph by the mid point of the 2 miles. As someone pointed out in another comment, Speed / Velocity is averaged over the time domain, not the spatial domain.
What if someone actually travelled 1 mile with 15 mph speed, and another mile with the 45 mph, what would be the average speed over those 2 miles in that case? Obviously not 30 mph
@@meso_p Yes Jup Yo Jo Exactly I agree with you Jep That's what 'Jup' means for me, maybe it's odd for you because i'm from germany and you'd more likely understand what 'Jup' means there, but i thought english speakers would understand it too
Given his friend knew of Einstein’s special relativity and E=mC^2 accomplishments, this problem was clearly meant to be amusing. Travelling down the hill at the speed of light results in an average speed of 29.99996 mph which is essentially 30 mph.
@@goodtoGoNow1956 Yeah we all understand the car runs out of time (4 min.) at one mile. But if it somehow travelled at C for the second mile the speed would be very very close to 30 mph average (don't forget - time slows down as we approach the speed of light). BTW, it's impossible for the average to be >30 mph as the total travel time is greater than 4 minutes.
@@goodtoGoNow1956 But to cheat, we can use significant digits. you can say a tailwind helped him get to 15.4 mph (approx 15). this will give him time to get down the hill at C.
@@goodtoGoNow1956 From the drivers perspective of time its possible. Due to driver's motion when he reaches the end of the 1st mile his elapsed time would be under 4min so he would have an almost infinitely small amount of time to traverse the next mile.
People's minds can easily go to "I think they're just asking me to average 15 with some other number to get 30". And the answer to that question is clearly 45. But that is NOT the question, and that's where people go wrong. If you give the correct answer to the wrong question, then you're probably going to be giving the wrong answer to the actual question. People who point out that the literal rules of the question state that you have 4 minutes to go 2 miles, and you take 4 minutes to go the first mile, leaving no time for the 2nd mile, are obviously correct. But I like to look at it from the other direction: why can't it be 45 mph? Why indeed? Okay, so if you think you can go 15 mph for a mile and then 45 mph for a mile, then what you're ALSO saying is that you're going to go 15 mph for 4 minutes and 45 mph for 1 1/3 minutes. That's the literal arithmetic. Now, if the question were asked that way-- "If you spend 3x as long going 15 mph as you do going 45 mph, is your average speed 30 mph?"-- then it's likely no one would be fooled. Anyone who has ever driven cross country realizes that the actual amount of time you spend at each speed matters. You can also literally do the math: 4 minutes on the 1st mile (15 mph) and 1 1/3 minutes on the 2nd mile (45 mph) = 5 1/3 minutes to cover 2 miles. But if you spend 5 1/3 minutes to cover 2 miles, you have NOT averaged 30 mph: you've averaged 22.5 mph.
Great little puzzle! Common sense and intuition will lead you halfway into the forest. Then leave you there. Should definitely be the first lesson for all project managers: It is very easy to fool yourself into thinking "we can make up the lost time" long after it is mathematically impossible to do so.
@@klovell6793 I've typed this for someone else and I'm copying and pasting it.The average it's trying to meet is 30 miles per hour, right? Traveling 2 miles at 30 miles per hour would take 4 minutes, which means you have a total of four minutes to meet the average. Here's the thing: traveling one mile at 15 miles per hour also takes four minutes. So once you clear the first mile, you've already expended the time you needed to reach the average. Say you don't factor time in though, and go with 45 miles per hour as the answer. The second mile at 45 miles per hour would go by faster than the first mile. That means more of the trip would be spent at 15 miles per hour rather than 45 miles per hour, and you wouldn't get your average. Hope that makes sense.
Person a: I've got a riddle... how do you add 2 and 2 and get 5? Person b: (spends a week thinking about it) a: You can't, it's impossible. b: Cool riddle.
60mph = 1 mile per minute 30mph = 1 mile per 2 minutes 15mph = 1 mile per 4 minutes. To average 30mph for 2 miles requires 4 minutes. If you've already spent that 4 minutes in the first mile, there's no time left for the 2nd mile.
Thank you. I literally didn't know why or how there was this "magic 4 minute time limit" until I read this comment. But even now it doesn't make sense. "Mph" is a rate of speed. It was never stated anywhere that it couldn't take longer than 4 minutes
@@AggroBen To "average" 30 mph is equal to saying average 2 miles in 4 minutes. The fraction 30mile/60min. is equal to 2mile/4 min. The problem states the travel distance must be 2 miles at that rate and the rate involves 4 min. Since the first mile spent 4 minutes, it's impossible to meet that rate.
@@KenPaulsenArchitect thank you for the reply. I must've lost a few brain cells over the years because even after this reply I still thought it was wrong lol it took me a while to finally comprehend it all. For some reason I thought one could still get an average 30mph even if it took longer than 4 minutes total. But then it would no longer be 30mph average. Thanks again
@@AggroBen You're welcome! Many of these puzzles stump me, but I got lucky on this one. I'm missing brain cells too! You definitely got it - yeah, it would no longer be 30mph average if it took longer than 4 minutes. Good way to phrase it.
At 15 mph it would take you 4 min to travel 1 mile. At 30 mph it would take 2 min to travel 1 mile and at 30 mph for 2 miles it would take 4 min. If you have traveled for 4 min at 15 m/h, you already run out of time to average out at 30 mph.There is no way that you can make up the time to make up for the last mile that is not traveled.
@@martinpattison8916 i literally don’t care lol. That wasn’t even the point of my comment 🤦♂️ ppl like u rly just wanna start arguments. As i said, it works the same for km, so it doesn’t matter
Why did I read the comments? MPH includes both distance and time - it miles per hour. If you are given a distance and a mph you can calculate how long it would take. If you are given a time and a mph you can work out how far you can go. Thats the basics. 30mph for 2 miles takes 4 minutes. 15mph for 1 mile is also 4 minutes. If it was 16mph up then you would get to the 1 mile in 3 minutes 45s . So that would need 240mph down. 15.1 mph takes 3m 58.4s to go a mile up - that would take 5,760mph to average 30mph. 15 mph takes 4 minutes - no time left. Time is implicit in mph.
Who calculated and got 0 as the answer, knew it was impossible, assuming they were wrong watched the video, realized they were right and felt like a genius?
I initially thought 45 mph immediately, but had to rethink due to there being no reason to the vid if it was that simple. Broke it down to time and realized he’d travel 1 mile in 4 mins @ 15 mph. In order to average 30 mph over a 2 mile stretch, you need to make the trip in 4 mins. Which means he’d need to instantly teleport from the midpoint to the end.
It's impossible. My physics teacher gave my class the exact same problem (with different numbers) & the class laughed at me when I took longer than they did. My initial answer was correct, it couldn't be done but at 17 I didn't trust that so reworked it & got an immensely higher answer to which they again laughed until the teacher said I was a lot closer than they were.
bro I read a comment that said "raise your hand if you thought it was 45mph and I stopped reading and was like "alright im wrong... how" and I started to really think until I hit "wait it would take 4 min to reach the halfway point ok so how frickin fast do I have to go to adverage that out,... and I was stumped again until I was lik ok how long does it take to go 2 miles at 30mph? 4 min... oh so he has to go the speed of light? then i watched the vid and no its just "impossible" xD
45 would never make sense. You're trying to reach 30 average. 15 is half of 30. To balance out something that is HALF you need to DOUBLE the other side... not a 1.5x. 45 is literally the worst answer you can come up with. Even though this is impossible, your mental thought patterns were still ridiculous to witness. You don't just add 15 and 45 together to get 60, then divide by 2. It doesn't work that way with speed and time. It's just crazy how so many people came to the most ridiculous conclusion that makes the LEAST sense, and THAT is the most popular answer? Yikes you generic people just need to hide and be quiet...and just LISTEN...until you know how to think (but still don't talk at this point, you need to just shut up and pay attention, still).
My math was flawed. Happy Craig 11 iq. However Einstein didn't even read the riddle. It's speed not time. It's over 45 due to the decrease in travel time down hill. 1.33 minutes and this will effect the avg. My mistake was thinking of travel for four minutes each mile.
I really enjoyed solving this, my first thought was 45mph but then wrote the problem out and saw that 15mph is 4min/mile and 30mph is 2min/mile I could see that the car had to travel infinitely fast and arrive instantly at the bottom. What a fun puzzle!
He made a mistake here when he said "Velocity needs to be infinite" We have: t(trip) = 2/30 = 4 min = 240 secs As he mentioned. However, there would exist a dilation of time (this is Einstein we're talking about) so t(ascending) would not be 4 mins. From t' = t x root(1 - v²/c² ), (where t' is the dilated time) We get t' = (4x60) x root (1- 6.7056² / 299792458²) t' = 240 x root(1 - 5x10^-16) t' = 240root(0.9999999999999995) Therefore, t(ascending) = 240 - t' Thus we have: t(descending) = 240-(240-t') = t' Now there is a finite (albeit very small) time that we have to descend. Let V = velocity needed. V = d/t V = 1 / t' As t' > 0, V is not infinite! Of course, this would be much greater than the speed of light, so it is impossible, but certainly not infinite.
Time dilation only counts if you are an observer to the thing moving at that speed. Relative to yourself, as this riddle implies, you don't experience any dilation. But yes, if you were someone with a stop watch observing, you could get there theoretically.
I instantly knew it took 4 minutes to go up the hill. 15mph is a 4 minute mile. This has never left my brain since high school cross country. I easily calculated double the speed half the time; times 2 for 2 miles. It would take 2 min per mile at an average 30mph. You'd have to have an MC Escher hill to end up at the bottom at the same time you end up at the top. Infinite speed at infinite acceleration.
Basically in order to average 30 mph for 2 miles you have to travel the 2 miles in 4 minutes. But since you went 15 mph for the first mile, that mile took you 4 minutes, so it is impossible to travel the whole 2 miles in that 4 minute time restraint that would allow you to average the 30 mph.
To me this answer doesn’t even answer the question let’s put this on a flat surface ! I’m sorry if tell you , to do 1 mile your drive at 15mph so how many mph do you need to do another mile and for the total being 30 ! You’ll answer 15 !! WE DONT CARE ABOUT THE TIME !! ITS THE NOT THE PROBLEM ! if the guy take (a hypothetical) day driving at mph he will do 24 Mph ! The fact that he is on a hill will just made him slow down or accelerate in order to stabilize on that 15 mph Why wanting to say he took 4 minutes , it’s not the problem here because at the middle of the problem it’s like everything will restart , we talk about speed not time ! I understood the answer but to me it’s like saying “how effective is to drink this specific water and you answer by saying there is no water in your cup right now”
@@d.o.a9236 it took me a while but i understood everyone is wrong, because, yeah, he took 4 minutes to reach the top... but that is not what we're looking for, we just need to know how many speed he need going down to make those 2 miles into a 30 mph average, and well, that would be 45 mph right? just a regular calculation of average
@@d.o.a9236 jlnsadas xD but why i´m wrong, my GF explained me but i still dont get it, but told me something that i didn´t know, that you can´t do that kind of calculation to miles, but i dont know if that apllies only to miles or when you're doing a distance average speed, becayse others told me that the average speed of 15 + 45 would be 22,5 i dont get it why Tw T
@@marcelinebeckerrosas9023 basically I had a whole discussion under this video with someone so try to search my other comment (I precise that I first post it so people answered to me afterward and it’s one of the most recent comment under this video post 5 hours ago and sometimes I said thing that I thought was clear but wasn’t)
I think this is easily solved without algebra. Suppose you average 15 mph over a 1 mile distance. You can immediately see that if you happened to be at the 2 mile post by this time instead of the 1 mile post then you would have averaged 30 mph already. Therefore there is no time left to reach the second milepost and you would have to travel at infinite speed to get there.
I solved it mentally... But i bet its because i was careful unlike Einstein that time who didnt know it could be hard. The thumbnail probably let me solve it carefully
The “trick part” to this problem is from the definition of AVERAGE VELOCITY. We have been taught in grade school that to get the average you just add up the numbers then divide the result by the number of numbers. In physics, this is not the case for Average Velocity. Definition of Average Velocity: (Average Velocity) = (total distance)/(total time) PERIOD You can’t use (15+45)/2=30
@@tushar9655 harmonic average, c, of a and b satisfies 1/c = (1/a + 1/b)/2 linear average is c = (a+b)/2, geometric average is c^2 = ab quadrTIC mean is RMS: c^2 = (a^2+b^2)/2
Well, yes and no. The wording of the question is critical. Specifically, the "for the entire 2-mile trip" part. While it's true that the speed for the two-mile trip _as a unit_ can never reach 30mph, the average speed _per one-mile leg_ of the trip is unbounded. This is because we calculate the former as (total) distance over (total) time, but we would calculate the latter as total speed over total legs. (To be pedantic, distance-over-time is just _speed,_ regardless of whether it's for an entire trip or just a segment. "Average speed" is either redundant or something different.)
@@Stubbari It's what we do sometimes when referring to racers' performance. Admittedly, we say things like "trial", "heat", or "race" instead of "leg" in such cases, but the operation remains the same.
I like this. My answer to thumbnail question (which i only thought twice about because it said something about fooling people) is that the car cant do it unless it completes the second mile in an instant, cause the time elapsed after 1 mile is the target time for the whole trip. Definitely unintuitive though
The key to solving this problem is to know the definition of average velocity. Average velocity is the displacement of the object divided by the time it takes the object to travel that displacement.
yes thats how I solved it too. I nearly thought I made a mistake when calculating the time elepsed on the ascent (1/15 hour) and the full trip (1/15 hour), so speed had to be infinite. luckily it was a trick question.
All you did here is define what mph is measuring. That's what it measures. Miles traveled over 1 hour, or mph = miles/hour ( velocity = distance/time). Your statement is still incorrect however. You said "average velocity", but then gave the definition for "velocity". Average velocity is essentially and average of averages. In angular acceleration the collection of times and distances are represented by delta, or av = Δd/Δt.
@@vvhitevvabbit6479 Wrong! I gave you the physics textbook definition of average velocity. Lookup every physics textbook out there. MPH is a unit. I didn't define any units! Miles and hours are units that need to be defined operationally. You could also solve this problem using the definition of a weighted average which when simplified ends up being the definition of average velocity that I just gave you.
I can relate to that. When I have to go to an appointment, but I leave the house late, it is very difficult to make up the lost time. Even when I run I often barely make it on time.
I was dead on the answer being 45. I even replied to several asserting the answer was definitely 45mph. Then it hit me. The time/velocity/time relationship is not linear. This question is a perfect example.
1 hour at 15 and 1 hour at 45 would be 30, but here its 1 mile at 15, and since its scaled on distances, the car doesn't spend half the time at 15 mph, but half the distance
Just looking at the thumbnail before clicking the video I instinctively thought 45 mph, of course, but knowing it would be a trick I decided to calculate it and came up with 'infinite', for the reasons laid out in this video. But it got me thinking, why is there this difference? So I thought, if I were to travel at 15 mph for 1 hour and then 45 mph for 1 hour I would travel a total of 60 miles in 2 hours, which averages out to 30 mph. So this works over long times, why not over short times? So I thought of breaking it up into 15mph + 45 mph segments to see why it didn't work and then it hit me. It's because we're not measuring in units of time. We're measuring in units of distance. If you spend half the _time_ travelling at 15 mph and the other half the _time_ travelling at 45 mph you're going to average a speed of 30 mph. But if you spend half the _distance_ travelling at 15 mph then no matter how fast you go you'll never get up to even twice that speed on average across the whole distance. Because you travelled for that pace for that distance, not for a specific time.
Actually, I think you have missed out a trick here too. I've just posted this, but there is another way to consider what the average speed is over the DISTANCE. It's not actually a trick question.... You never specified whether the average was over time, or distance. If you consider the average being over time, then you have to calculate this as you did. But if the average is over distance, then the answer is 45mph. If half of a journey was completed at 15 mph, and we know that the average speed over the distance of the whole journey was 30mph, then logically the remaining speed for the second half of the journeys distance was 45mph. Think of it this way, if we split the distance into 1% chunks, and after each 1% of distance travelled, we drop 15 sweets for the first half, and "X" number of sweets for each remaining percent after 50%, for us to reach and average of 30 sweets per 1% travelled, we need to drop 45 sweets per percentage travelled. When time is not a factor, then the speed is irrelevant, only the value needed to be assigned to each percentage, no matter how much faster we may have travelled it.
Spot on! Because averages are weighted - in this case by time. So when you have a set of speeds, in order to find the overall average, you have to multiply them by the times travelled at those speeds (to find the total distance travelled) before you can divide by the total time. When the time intervals are all the same, the weighting is equal so they all cancel out. If the distance intervals are the same, but the speeds are not, the times cannot be the same, thus they must be weighted. S = D / T = (d1 + d2 + ... + dn) / (t1 + t2 + ... + tn) when t1 = t2 = ... = tn = t: S = (d1 + d2 + ... + dn) / (n * t) = (1/n) * (d1 / t + d2 / t + ... + dn / t) but s1 = d1 / t, s2 = d2 / t .... sn = dn / t => S = (1/n) * (s1 + s2 + ... + sn) = (s1 + s2 + ... + sn) / n But, when d1 = d2 = ... = dn = d: S = D / T = (n * d) / (t1 + t2 + ... + tn) t1 = d / s1, t2 = d / s2, etc => S = D / T = (n * d) / (d / s1 + d / s2 + ... + d / sn) It got messy after this, so I'll stop there. I hope it points you in the direction as to why it's different though. Cheers.
@@diegomarxweiller1814 No, it says what speed would you need to go at to achieve a 30 mph _speed_ across the 2 miles. Speed is distance over time. So to average 30 mph across the two miles the second mile needs to be traversed instantaneously.
The issue, for those who still don't understand, is that the distance travelled is limited to 2 miles, total. If there was no limit specified, then an answer of 45 mph for the same time as the 15 mph travel time would be acceptable. In the 45 mph travel speed, the car would travel 3x further than the 15 mph travel speed time. The problem here is that distance for the 15 mph section was defined as one mile AND the unknown speed section ALSO defined as one mile. Now the answer is constrained. The answer must account for a total distance of 2 miles, with one mile being 15 mph. As you can intuit, since the car must travel the same distance as the 15 mph section, but faster, time is also a limiting factor. Which is why figuring out the travel time for the first 15 mph section is important. And so there is no value that will meet the multiple constraints, other than instantaneous teleportation over the second mile.
Actually there’s multiple answers to this question because the question states the car averaged 15/mph during that one mile time it didn’t say how long it took to travel that mile so it’s necessarily 4 min
@@hotshot2101 D = V x T D is known, as is V (over the entire D as an aggregate). Therefore you simply solve for T. There is only one answer. For the first section, D is one mile, while V is 15mph, therefore T is one fifteenth of an hour, or 4 minutes. Now you figure for an average speed of 30 over two miles. D = 2 miles, while V = 30 V divided by D is 15, therefore the car travelling 2 miles at an average of 30 mph takes one fifteenth of an hour, or 4 minutes... If it took 4 minutes to travel one mile already, it cannot take ANY time to travel the second mile to meet the 4 minute constraint imposed by the constraints of the distance and the required average V over 2 miles when a V of 15mph over the first mile is defined by the question. It requires instantaneous travel of the second mile, after taking 4 minutes to travel the first.
For those of you who are confused... averages and ratios (like mph) are essentially the same thing. There is no way to express velocity without using distance as a ratio relative to time. Think of it this way: Imagine that there are TWO cars racing from point A to point B which is 2 miles away. Car#1 has to drive at 30mph for the entire 2 miles. Car #2 can only drive at 15mph for the first mile but can then go as fast as it wants to catch up to Car # 1. How long does it take Car #1 to reach the finish line? - going 30mph for 2 miles takes 4 minutes. How long does it take Car #2 to reach the 1-mile point, and then the finish line? - going half as fast as Car #1 for half the distance as Car #1 takes the same amount of time, which is 4 minutes. - so, after 4 minutes, Car #2 has gone half the distance, and now has to cover another mile with NO time left on the clock. So, if the restriction exists on the distance in which the average must be calculated, then there is also a restriction on the amount of time it can take.
Car 1 doesn't have to drive 30mph for the entire 2 miles. They just have to average 30mph over two miles, so that when they reach 2 miles their average speed was 30 mph. That is where the problem in the question lies. Saying the "entire" 2 mile trip may give it away to some but not to most imo. I get that it's a trick to fool people but i suppose it appears the writer of the question was trying to trick and therefore was purposefully not very clear in what was wanted. Just my opinion.
@@joeshawcroft7121 Well, there's not really a problem in the question because the unit for velocity is an equation that behaves like an averagee if any of the terma of the equation are given. So, regardless of how it is phrased, the point is this: if Car 1 takes 4 mins to travel 2 miles (30mph) and Car 2 takes 4 mins to travel 1 mile (15mph), it is impossible for Car 2 to travel the distance necessary in the amount of time required to also average 30 mph like Car 1. There's no time left in order for Car 2 to travel that 2nd mile, which means the equation/ratio for velocity has a denominator of zero. The original question posed to Einstein just uses one car. Same result.
@@robjames4160 the question is about speed. And your answer is time ? That is like asking for the weigth of a given object and give the answer in feet. Nowhere in the question is there any limitation in time spent.
@@briankristiansen821 You have to calculate that time limitation - average speed of 30mph is 1 mile every 2 minutes, so 2 miles have to be done in 4 minutes to average that speed.
@@KasparsRumba8 you are not adressing my critizism. The question calls for an average speed V(velocity) and your answer is four minuts T(time). It is a fallecy of wrong units.
i can use a smartpone better than him... and i know that feel, i pretend to be a techie and sometimes ppl or owners of stuff teach me how to access some things
I got the answer right away in my head but since you said Einstein figured it out I assumed I was wrong somehow. Did the calculations 2 more times and said "it can't be done" Then watched the rest of the video. ya got me.
I did actually get the answer on this one, though I have to admit when I first came up with infinite velocity for the trip down I dismissed this as impossible and had to keep working the problem for a while. Once I was sure that the total time for a one mile trip was actually equal to the time of the two miles trip did I come back to the conclusion and watch the rest of the video. Thanks Presh, it has been several decades since I'd encountered this particular puzzle. It was fun to work it out again.
Same here. Found the answer pretty quickly using mental calculus. I was almost certain I had fucked up and was curious how and why but... no i was correct lol!
You’re all wrong. The driver only needs to travel the speed of light as time stops from the driver’s perspective at the speed of light. No frame of reference was stated the in riddle.
... just spend 10 minutes on figuring out why my awnserd was 1800/v2 = 0. As in the car must have infinit speed. After admitting defeat i checked and i was correct 😅
The reason 45mph is wrong is because we cannot find the average of speed using the conventional average formula . It should be calculated using total distance / total time . For example , a car moves from 2 miles in 3 hrs and the next 17 miles in 6 hrs . Using the conventional formula we get the the average of speed as - ((2/3)+(17/6))/2 = 1.75mph . But if you use the right formula , you will get (2+17)/(3+6) = 2.1mph. Hence you cannot average the speed using the conventional method. If you follow the same logic in the given puzzle , you can see why there's no time left downway.
At first glance you think "oh, (15 + 45) / 2 = 30, easy!" And then you remember(maybe, some people aren't familiar with this concept) that speed, distance, and time are factors! haha I do like even the "simple" problems. I've seen similar questions posed in the form of "If I drive for 1 mile at 15 miles per hour, how long (or far) do I have to drive at 45 miles per hour in order to average 30 miles per hour?" This question helps people understand the relationship of speed, distance, and time. This question is really only a riddle if you aren't familiar with this relationship or don't use it often so it doesn't pop into your head immediately. I guess that applies to a lot of riddles though.
The assumption here is sampling w.r.t. time which is how average speed is usually calculated (it is literally distance over time) but you can technically sample w.r.t. distance, something you might do if you wanted speed to vary as a function of distance
This is an evil problem: before even watching the video I noticed it wasn’t 45 because the car would spend less time on the 2nd mile, pulling the average closer to 15 than 45.
My thought process was: Oh Einstein almost messed this up gotta pay attention. 1/15 hours to travel the first mile because 1 mile 15 mph. Avg trip is 30mph so 2/30hours spent for the whole trip, which is already all spent on the first mile
If you got 45 mph for the 2nd mile, you are correct. The solution presented here, comparing v=d/t resulting in 4 minutes and rendering the riddle unsolvable is not the correct formula given the information we have about the entire trip. We already know the formula doesn't address the riddle because we've been given a speed for the first mile and the average speed for the trip. We're not solving for the average speed because we already have it. If you think I'm lying, or must be incorrect because Einstein or whatever, take it up with the PhDs and experts at Omnicalculator. They too will say if you have an initial speed and an average speed, you find the final speed by doubling the average and subtracting the initial, directly answering the riddles question. Obviously, to verify the average speed, you add your initial and final speeds together, and divide by 2, just as you've likely assumed. Your total trip takes 5.33 minutes, 4 minutes uphill at 15 mph, 1 1/3 minutes downhill at 45 mph. You only get it done in 4 minutes if your speed is 30mph both miles. Either this amusing Einstein anecdote is misleading, or the experts are. Check it out at their average speed page and feel free to prove them wrong.
@@underscoredfrisk but the question doesn't say that the trip has to be defined by an amount of time, it could just be an amount of distance, and if the trip is just a 2 mile stretch, then 45 is the correct answer
If anyone's looking for an idiots explanation to this (I could have used one while struggling through comments from people who know a lot more maths/physics than me!) : If you travel for 1 minute at 15mph then 1 minute at 45mph you'll average 30mph, but you'll have traveled a lot further during your minute at 45mph. But here we're limited by distance as well as time, so we can't travel further to compensate. To do 2 miles at 30mph will always take 4 mins. However, you've already traveled for 4 mins doing 1 mile at 15mph. So it doesn't matter how fast you go, you've already used all your time! To reach an average speed of 30mph after doing 1 mile at 15mph (which takes 4mins) you'd need to travel for another 4 mins at 45mph, which would mean you traveling 3 more miles instead of the 1 allowed in the question.
but... why you can apply 15 + 45 = 60/2 = 30 when the unit is a minute, but you can´t when is applied to miles? i think i almost got it, but that is the thing that keep me from solving it, also, someone tell me what the average is 22.5, but i don´t understand the formula and why is planted the way it is. could you tell why is wrong?
I love this, despite the fact that I was clearly wrong in my initial assumption! It's not until I locked into the parts of the puzzle which were constrained that I realized that the answer was an impossibility. It's so simple, but my brain still wanted to ignore the obvious. (Solve for x... but x must equal zero) As a musician, I translated the terms of reference to musical tempos, which made no more sense, but proved the futility of the endeavour. Thanks so much.
The 'trick' here is that the information given to you is using an average...which messes things up. So you have to get back to the numbers that allow you to calculate. This is similar to the question: a car gets 10 mpg's going up the one mile hill...then 100 mpg's going back down again....what's the average for the round trip? You can't use either mpg number to find the answer other than getting back to actual amount of fuel consumed...which then applied over the distance traveled ends up being in the high 18's. Good stuff!
That explains why I was able to do it first try 😂 my brain went I feel like I need to calculate the total time and time already taken, but idk why.... thanks Derek!
Got it - I assumed averaging velocities would not be reasonable, so I look at time taken and discovered we'd need to travel 1 mile in 0 seconds. Cool puzzle.
The way I solved it was that the total time is 1/15 + 1/v, where v is the downhill velocity. Average speed is therefore 2/(1/15+1/v) = 30 ==> (1/15+1/v) = 1/15 ==> 1/v = 0 which is impossible!
I've seen a total of 3 videos from this channel. 1 had an obvious answer that wasn't even a challenge, two had no answer and were time wasters. Sometimes there is such a thing as a bad question.
Well, you could have just not commented! Commenting on this video fuels the algorithm, and disliking the video (which you probably did), ALSO fuels the algorithm.
60 mph = 1 minute for each mile 30 mph = 2 minutes for each mile 15 mph = 4 minutes for each mile At 30 mph the entire 2 mile trip = 4 minutes At 15 miles per hour during the first mile, the entire 4 minutes is used up, making it impossible to achieve a 30 mph average over the 2 mile trip. If you went down the hill faster than the speed of light, it would still take you longer than 4 minutes to make the trip.
I did this in my head. Well first I came up with 45 mph after seeing the thumb but after he laid the problem out I used exactly this process to realize the answer is infinity. Pretty cool. Because of the nice round numbers very easy to work out in your head. You could force someone to calculate using non-round numbers but I like the simplicity of it.
I did the same thing, at first glance you think "oh, (15 + 45) / 2 = 30, easy!" And then you remember that speed, distance, and time are factors! haha I do like even the "simple" problems. I've seen similar questions posed in the form of "If I drive for 1 mile at 15 miles per hour, how long (or far) do I have to drive at 45 miles per hour in order to average 30 miles per hour?" This question helps people understand the relationship of speed, distance, and time. This question is really only a riddle if you aren't familiar with this relationship or don't use it often so it doesn't pop into your head immediately.
I didn't pick the answer being "infinite" but I figured out very quickly in my head that the trip down would need to be excessively fast... after reaching what I thought was an answer, and immediately saying to myself "no, that's still not fast enough" I clicked play and discovered that yes, infinite speed is required, which is indeed, very fast :)
@@soldjahboy Tell me where it exists ? There is always a number to everything. For example you can say there is infinite water on earth or there are infinite number of atoms in the universe but it never is. It’s just too large of a number but still it’s not infinite.
@@zarifshoeb How do you expect me to explain something to you, that you claim doesn't exist? What's the biggest number you can think of? What comes after that? What about after that, and after that.... and, after that? Where do they end? They do not. Ergo, infinite.
This feels fundamentally flawed. I am pretty sure its misdirecting the flow into separate pools that have nothing to do with the vehicles overall average over a given range. Regardless of direction the car traveled (up or down), the average speed over two miles, even if it took more time than the 4 minutes, would still be able to be reached at 30mph per traveling at 45mph. I get how its a psuedo riddle, but I think it was answered in a flawed way. If a mile at 15mph, then travel a mile at 45mph, then my average speed over those two miles is 30mph. It has nothing to do with time here, but everything to do with distance and speed.
Speed is measured in units of distance divided by units of time (miles per hour) and if you need 4 minutes to travel 2 miles at an average speed of 30mph, but take 4 minutes for the first mile, even if you could travel down that hill at light speed, you would get only get very close to the 30mph average, but not quite, even if the difference would be miniscule
No. If you want to add two speeds together and divide by 2, it is only valid as an average speed if you traveled for the same interval of time at each speed. It is not a pseudo riddle in anyway but, respectfully, your 'average' is a pseudo average.
Speed is never "nothing to do with time" - speed a measurement of distance travelled and time taken. Without time, speed as a concept doesn't really exist. If you drive a mile at 15mph, it's impossible to do it in anything other than 4 minutes.
First i assume it's 45 mph, then i solved it to find average speed of 22.5mph Then i took the value of 60mpg, average speed became 24mph I assume 120mph, average speed became 26.66mph I assume 240mph, average speed became 28.25 mph I assumed 2400mph, average speed became 29.998mph I understood. The car came down at warp speed.
Oh, you solved the problem! The trick is to plan this drive for when Daylight Savings takes effect, then drive 1mph down the backside of the hill, taking one hour, then set your clock back when you reach the bottom, thus arriving at the same time you left!
Great riddle. Solved it before going through the solution. People tend to calculate average speed as the arithmetic mean of all the speeds, however average speed is actually calculated as the total distance covered in the trip divided by the total time taken.
If you got 45 mph for the 2nd mile, you are correct. The solution presented here, comparing v=d/t resulting in 4 minutes and rendering the riddle unsolvable is not the correct formula given the information we have about the entire trip. We already know the formula doesn't address the riddle because we've been given a speed for the first mile and the average speed for the trip. We're not solving for the average speed because we already have it. If you think I'm lying, or must be incorrect because Einstein or whatever, take it up with the PhDs and experts at Omnicalculator. They too will say if you have an initial speed and an average speed, you find the final speed by doubling the average and subtracting the initial, directly answering the riddles question. Obviously, to verify the average speed, you add your initial and final speeds together, and divide by 2, just as you've likely assumed. Your total trip takes 5.33 minutes, 4 minutes uphill at 15 mph, 1 1/3 minutes downhill at 45 mph. You only get it done in 4 minutes if your speed is 30mph both miles. Either this amusing Einstein anecdote is misleading, or the experts are. Check it out at their average speed page and feel free to prove them wrong.
@@JL-fg2vk I see. However, I think you might be confusing velocity with speed. While calculating velocity, since the direction of displacement (shortest distance) is also considered, we may calculate the average velocity as the arithmetic mean of initial and final velocities, just like you have explained. This can be proved using v=u+at and s=ut+1/2at^2. As far as speed is concerned, doing that won't be right as speed is a scalar quantity as distance does not have direction. So, when we calculate the avg speed as the arithmetic mean, we completely rule out the possibility of speed in opposite direction.
I feel like some people still don’t understand because of the iffy explanation in the video. Here’s mine: The average of 15 and 45 is 30, so from that math alone, 45mph. However, if the car traveled 2 miles at 30mph, it would take 4 minutes. But because the car has already gone for 4 minutes, the 45mph solution would theoretically work, but in our universe you have to factor in the laws of time, meaning you could not go 1 mile instantly. If the car were to descend the hill at 45mph, it would actually work out so that the average speed of the car is 22.5mph because speed equals distance over time, and distance is 2 miles, and it would take 4 minutes to go up the hill and 4/3 minutes to descend, which means that speed equals 2/(16/3) which is 0.375 miles per minute, or 22.5 miles per hour
Hey, I was among those telling 45;) It seems simple: 1st minute 15mph, 2nd minute 45 mph, gives on average 30mph UNTIL you notice going uphill takes 4 minutes, the entire time you have to go up&down. Nice puzzle:)
Except the 4 minute time constraint wasn’t included in puzzle as presented at the beginning. It was only added after, during the presentation showing everyone how foolish they were thinking this was just a math problem of average speed, where the correct answer for the puzzle, as presented, was 45, as most people assumed. This would have been a better presentation if the full puzzle constraints were made available at the beginning. That way people wouldn’t be puzzled at the confusion of a renowned physicist over an elementary arithmetic problem.
@@tcorris If you travel one mile at a speed of 15 mph, it will take 4 minutes. That is a calculated fact, not an added constraint. If it took you any time OTHER than 4 minutes to get to the top of the hill, you couldn't have driven exactly one mile, or you must have been going a different speed.
I remember the different wording "you are driving to your friends house 30km away, you drive the first half of the trip at 15km per hour, how fast would you have to drive for the second half to average 30km per hour." Same concept but slightly more obvious.
It's actually the exact same concept and wording; only tiny distinctions are the removal of the hill (the framing device that explains why the car might be able to go faster in specifically the second half) and the change of the units.
-"For -_-the second half-_- to average 30 kph." That's a big difference. This problem asks for the average of -_-the entire trip.-_ I see my mistake now
@@mordekaihorowitz It's a 30km trip, the first half, naturally would be 15km. If you travel for the first half at 15km per hour, how long have you been driving? The answer to that question demonstrates why it is impossible to average 30km/h for the entire trip, regardless of how fast the second half is.
@@mordekaihorowitz No it still asks for the average of the entire trip it’s just poorly worded. It’s not “How fast would you have to drive for: The second half to average 30km/h”, it’s “How fast would you have to drive for the second half: To average 30km/h (For the entire trip)”.
I spend several minutes trying to understand why my answer was 0 minute. There is another way to see that it's impossible. If you traveled x mile with a velocity v in average in t hours, and what to double the average velocity v'=2v of the whole travel in the next x miles, then v'=2x/t' 2v=2x/t' v=x/t' therefore t=t', so it's impossible to that because there is no time left to make the second part.
I remember having to do a problem very similar to this in my physics class. The professor gave this to us as a group assignment. We were all stumped. If I remember correctly one of us did get zero but it was rejected thinking it must be wrong.
@@VIShesh_Agrawal_happy how do you know he didn't get zero though? i think the original comment meant one of them got zero, but they thought it was wrong so they didn't submit it, it didn't mean the *professor* thought it was wrong
I was on the right track thinking that the second half will take a shorter amount of time at a higher speed and to find the average speed one has to take that into account. I wrote a whole equation where I tried to solve for x and got a nonsensical answer. Only then I checked how long a trip of 2 miles at 30 mph would actually take. When I found it was also 4 minutes, the same as the first half, I realized what was going on.
Doable in your head if you know the key to these problems is often total time driven, and at 60 mph you cover 1m/min. Total distance is 2 miles - at 60 mph average that's 2 min, at 30 mph its 4 min. 1 mile at 15 mph = 4 min - so you've no time left Though travelling the second mile at an appreciable portion of the speed of light would probably "allow" You to do it via rounding. There isn't much difference between 240 seconds and 240.0000053 seconds
Intuitively, I guessed 60 for the second leg. Then did the math and chuckled at myself. Does anyone have a good formulation to generalize this puzzle? "You cannot double your average trip speed if you are past the half-way point" or something like that? Not sure it's very practical, but an interesting thing to consider.
Your max average speed is your current average speed times one over the part of the journey you've already completed. I.e. 10mph times 1-over-1/3(=3) makes a max average speed of 30moh
Average speed works "normally" over equal time, not equal distance. Therefore, distance needs to be proportional to speed. If you go 10 mph for 10 miles and then 40 mph for 40 miles, average will be 25 mph over 50 miles. But if you reduce the second part of the distance and make it equal to the first one, you will reduce the time spent on it which will reduce the impact that second part has on the average speed: (10+10)/(10/10+10/40) = 16 mph over 20 miles. The formula is (X1+X2) / (X1/Y1+X2/Y2), where X1, X2 - first and second part of the way (might not necessarily be equal halfs); Y1, Y2 - first and second speed. Which creates the "limit". Even if you assume that second part of the way is just teleported over, it just means that X2/Y2=0. You are still left with (X1+X2)/(X1/Y1). This is the limit. In simpler words, limit is equal to the entire way divided by the time that was already wasted. If you go 10 mph for 10 miles and teleport another 10 miles, the average speed is 20. If you stand still for an hour, then teleport through all the 20 miles instead, the average is still 20 mph.
@@LazyOtaku So, we know the first mile takes 4 minutes. at 60mph the second mile takes 1 minute. so the mph for the whole trip is 2 x 60/(4+1) = 24 mph
I got it right! At first I thought it would be 45mph, until I realized that the real question was how long it would take to do that first mile. At half the speed (15mph) it takes the same amount of time to do the first half of the trip as you are allowed to use for the full trip at twice the speed (30mph). 45mph would be the right answer if the question were a two minute full trip, rather than a two mile trip.
wat? nowere is talks about time other than in the distance/time. of course it takes less time, but you want the average speed in set distance. it is 45
In my head I said "oh, easy. 15 + 45 = 60, 60/2=30" But somehow that extra mile changed things about how I thought about it because there was a D involved. Then time became important. I really didn't see it at first and was convinced 45 was the right answer.
@@diegomarxweiller1814 Time is sorta built in to the problem. MILES PER HOUR. If it takes too long to go the first mile, there is no way to make it up in the second mile. @30 MPH, you have to be doing 2 minute miles. It took 4 minutes to do the first mile...that is the total time you would have had to complete the whole trip. That you have another mile doesn't matter. It is definitely counter-intuitive. But my guess is you shouldn't be arguing with Einstein.
@@TR1CK I don't think that's right either. The point of the problem is that you are already out of time. it took 4 minutes to do the first mile. you have to average 2 minute miles to do 30 MPH. You already used all the time you have. Speed has to be infinite.
Also highlights the importance of double-checking your solutions! Inputting 45 mph as the second speed yields an average speed of 22.5 mph. AND, this seems like a good way to exercise determination of limits---in this case, there IS a limit to the possible average velocity.
Average speed, not the average speed over the distance. You guys are all over thinking your answer. I've just posted this, but you're welcome to check it and give me your thoughts. It's not actually a trick question.... You never specified whether the average was over time, or distance. If you consider the average being over time, then you have to calculate this as you did. But if the average is over distance, then the answer is 45mph. If half of a journey was completed at 15 mph, and we know that the average speed over the distance of the whole journey was 30mph, then logically the remaining speed for the second half of the journeys distance was 45mph. Think of it this way, if we split the distance into 1% chunks, and after each 1% of distance travelled, we drop 15 sweets for the first half, and "X" number of sweets for each remaining percent after 50%, for us to reach and average of 30 sweets per 1% travelled, we need to drop 45 sweets per percentage travelled. When time is not a factor, then the speed is irrelevant, only the value needed to be assigned to each percentage, no matter how much faster we may have travelled it.
@@Ste-The-Leo speed is always distance / time, so your second answer doesn't make any sense. And speed cannot be "irrelevant" because speed is in the question - what (average) *speed" does the car need to travel.... to make the overall average *speed*...
@@iandavidson471 Think of it another way then. A simpler way. I travelled 2 miles. For the first mile I travelled at 15mph, and for the second half of the distance I travelled at 45mph. You created fixed distance, or even infinite, data collection points across this distance to measure the speed I was going. What would the average speed I travelled across all of these data points come to?
@@Ste-The-Leo That would work... if you were measuring based on time. Sure, spend half the time going 15mph and the other half going 45mph, and you get a 30mph average. The problem is that in physics, time is always related to speed and distance. When you take distance into consideration, even if you take half the distance (a mile) at 15mph, there is no way you can travel 45mph for the other mile of the distance unless your car is traveling at light speed, since you've already spent all your time going uphill and you've got 0 seconds to complete that other half...
@@seisosimp Not quite, time is not something being measured in the example I provided. Actually, time is completely irrelevant. The only unit being measured is speed, and the points it is being measured is across a distance. How the speed has been calculated is arbitrary and for the purpose of providing a figure that can be considered "an average speed" (not necessarily the true average speed the vehicle would travel), then averaging the speed over the distance is the factor. Your calculation of the true average speed is, of course correct, but there is another way to interpret what an average can be considered to be, one of which is taking the known results, adding them together and dividing by the number of results. Think of it this way, a row of speed cameras across the distance measure the speed. 10 cameras across the first half measure 15, and ten cameras across the second half measure 45. Of all the results taken that day, the average speed measured by those cameras would be 30. Heck, if the first mile was fraction of a centimetre longer than the second mile, I could just as logically say the average speed was 15, if considering a median average across the distance.
Him: *impossible*
Me who thought 45mph descent: laughs nervously
i thought the same but it should have been logic since 15 + 45 : 2 = 30 mph
It should be 45mph, i dont understand how 4 mins is at all useful.
Average speed is not average of speeds magnitude it is total distance/total time.
@@sparkyfire2443 because essentially so if you go up the 1 mile in 15 mph, there's no time left to go down, if you tried to average 30 mph over 2 miles, took me a bit to understand too. really had to think there, i thought it'd be 45.
here let me try and re phrase it for you. if you're trying to run 200 meters in an hour, and your top speed is able to run 100 meters per hour, then how fast will you have to run the other 100 meters if you're trying to average 200 meters per hour?
well it's impossible since your top speed is 100 meters per hour. either that or you take 0 minutes for the latter 100 meters
@@sparkyfire2443 essentially the 4 minutes is useful because you find out that there's no time left to go down hill.
Raise your hand if you thought the answer to be 45 mph after listening to question.😃😉
same 😂😂
That was my initial thought but I knew mph seemed suspicious so I needed a way to check my logic. I converted it to time actually spent and got the right answer.
45 divided by g times mg sin x
Me:-P
I got tricked because of the use of the word 'average'
For those who are wondering why it isn't 45 mph:
Average speed is calculated distance/time. To show this, an example of a unit of average speed is mph = m/h = miles (unit of distance)/hour (unit of time). If you travel first mile at 15 mph and second mile at 45 mph, your average speed is not the average of the two speeds because you spend different amounts of time on each mile.
Imagine it this way: if you travelled the same amount of time, let's say 4 minutes from the question, first at 15 mph, then at 45 mph (so 8 minutes travel time total), indeed your average speed would be 30 mph. However, you would also travel 4 miles! But the question said we have to travel only 2 miles exactly. Thus, the restriction on distance also creates a restriction on time.
Thank you for explaining this correctly.
The fact that this doesn't have more upvotes baffles me. Thank you for the clean and concise explanation.
Yes, this is exactly the point to notice. 👍
I just keep confusing myself.
"If you travel first mile at 15 mph and second mile at 45 mph, your average speed is not the average of the two speeds because you spend different amounts of time on each mile."
If I travel with 15 mph, I take 1/15th hour to travel 1 mile, which is 4 minutes, which is 240 seconds
If I travel with 45 mph, I take 1/45th hour to travek 1 mile, which is 1.33 minutes, which is 79.99 seconds (80 seconds for sanity's sake)
Why is it incorrect to do (240+80)/2, which is 160 seconds, which is 2,66 minutes, which is 0.0444 hour, which is 22.5 mph?
Is it incorrect to say "over teh course of this [2 mile] journey, on average I spent 2.66 minutes on each mile"?
So the only reason this problem doesn't work is the assumption that the average is distance/time. But why would it be calculated that way? MPH is a defined set amount of motion(or speed) not an average. Going at 1MPH for 1 mile will take 1 hour. This is a measurement for motion.
So the average for this should be motion/distance.
So 15mph+45mph=60mph
60mph/2miles = 30mph average.
This was a frustrating one, because I spent a significant amount of time trying to figure out where I went wrong. Turns out I wasn't wrong, it is indeed impossible.
Unless you have a teleporter! Think outside the box! Or..........Einstein Rosen Bridge!
I don’t understand when i did this 30 720mph worked for me? Can someone please explain why my answer is wrong?
@@vincenttommasini4048 Sure, let's work it out. The first half of the trip takes 1/15 of an hour. With your speed, the second half of the trip takes 1/30720 of an hour. So the time for the entire journey is 1/15 + 1/30720, which equals 2049/30720 or 683/10240 of an hour. That means the average speed over the entire 2 miles is: 2/(683/10240) mph ≈ 29.985 mph. Not 30 mph.
Me too! I was not happy and gave up. I watched the solution only to find I was right. :-)
@@vincenttommasini4048bro how 💀😭
Step 1- Get a flux capacitor.
Step 2- Attach it to your car.
Step 3- Travel 88 mph.
Step 4- Travel back to time and kill Max Wartheimer.
Step 5- Live peacefully.
Simple:
Pause your clock or watch for the time you come down the hill, t=0
Although it's a artificial process, but the main thing is you got yourself a life
But in my way... you get to time travel
And you get to kill someone 😈😈
I get that...but.....
*TIME TRAVEL*
Read it backwards.
Great scott
"any darn fool can make something complex, it takes a genius to make something simple"
👏
@@Emir-wd8bs that surgery is not possible yet... sorry
if you think at it in a more absurd way, a fool can create a nonsense and there's no genius that can math it out, also a sensated jibberish is way easier to come up with than to solve... it's kinda the legit reverse of "to destruct" vs "to build", one is hard the other is simple
Brilliant
@@Emir-wd8bs well technically you aren't wrong if the numbers we named were 1,3,2,4,5 instead of 1,2,3,4,5.
You do you Bo bo
For anyone wondering then what's the answer , I will tell you. I am going to assume that distance travelled is given(2miles) and the watch through which time is measured is in car .
First we need to understand the relation of time which is inside and time which is outside
Let's say time outside is t
And time inside is t'
t'=t√(1-v²/c²)
For calculation purposes I am going to use si units
So total time vehicle need is
2/30 hrs = 240 seconds
Now we will be calculating time that is flowing outside vehicle
1/15hrs = 240 seconds
But hang on time inside vehicle is less then that is
240√(1-v²/c²) = t'
v=15mph=6.7056 m/s
Sadly no calculator was able to calculate such small value but fear not we will use calculas my friends !!
Remaining time left for 2nd part of journey is
240-240√(1-v²/c²) = t"( time left for other trip)
240(1-√(1-v²/c²))=t"
Now using expansion of root using mclauren series we can neglect other terms(they are very small) and keep the first the term
240(1-(1-0.5v²/c²))=t"
240×0.5×v²/c²=t"
120×v²/c²=t"
Now calculator gave the answer thanks to calculas
0.000000000000060036467002990743346703130254940792991950836435617448761427151923217429171811094289749944538966505597887983802322401518797793890192189656149242982301575455283765168756435935707823136503331530207014184 ( it's t")
Now let's assume speed of car going downwards is v'
So time taken in going down ( this time is from outside perspective ) is
1609.344/v'
But hang on this time is from outside perspective we need to calculate from inside
1609.344/v'×√(1-v'²/c²) = t"
We have to calculate for v' we can't use mclauren series to estimate because this time v' is bigger
v'=c×1609.344/ √ (x²+(c.t")²)
v'=99.999999999999993746201353855131488035053196294571173429056176961242868216566225763338544466330421185462845993920903672772712136295199625718519695883561631704644951046866851689889849005332679328390186 % of speed of light
Hahahahaa so car needs to travel this much speed
So car has to travel
v'= 299792457.99999998125158332035157644711226187877905984138241039911293970197614465141374188531703495207365180468193000769641759046666368909394835035750345423363226239891629886666833531582557539043583883 m/s
Let's transform this into mph lest Americans will be angry
v'=670616628.7685007780609863492531369132546843006244598512699575894491186279322226850367990279586936394919624747170011414560958111092380413595302843003339505755661462749383256073362706618588409009968468 mph
This took lots of the time and time in my country is 4:29 am what the hell I am doing with my life !!!
Chit c ctrl Ism the problem lies within the question: are they asking average speed over distance or average speed over time? In the real world these are two different things. That’s why they are separate variables in the equation. Most people understand speed = distance/time but the question was intentionally misleading.
@@urbangangsta too bad my comment has only 2 likes where as 45 mph ones have 4000 likes
thats not 100 percent accurate because the time at top of hill is also flowing fast than the ground...so calculate that also buts its little hard...yo have to use calculus cuz each rise in hills height will make tha time fast wrt ground..so could you calculate also those😅😅😂
@@unknownnepali772
how will time at top of hill will flow fast than ground ?
Did i hear general relativity ?
of course according to general relativity....because more the gravity more slow the time will pass wrt observer at low gravity region and as you move up the hill height increases and unnoticeable amount of gravitational force will decrease........and bro what about the length contraction....you should also use it..😅😂
Got it right away. Usually I fail miserably at these kinds of things, but calculating speed-time-distance was something I learned as a radar plotter in the navy. This was in the pre-computer days where we used paper and pencil to develop a picture using a plotting table. We got very good at doing relative velocity calculations on the fly in our heads. As soon as I saw the numbers, I was twigged. 1/15, bay-beh.
Wow! Nice :)
Ah yes, during my time as a massivity neutronian outlander, we surveyed plots of gravitonian hemispheres on Jupiter’s moons. Elementary my dear watson. By the time my coworkers slid the cover off their Ti-83s, I already had the answer jotted down on papyrus.
@@tschantz Impressive incoherence there, Spiff. ;) You write Star Wars fanfic, mayhap? It's not quite Trek Drek yet, tho. Keep diggin'. . .
At first I didn't get it, but after I grabbed a piece of paper and started thinking about the units, I quickly realized that I needed to start by computing the trip and leg times.
@@tschantzCreative writing
This is not a puzzle to Einstein. This is a motivation for him to find teleportation!
yes
Yes
yes
Ye
Shush
Took 30 minutes with a calculator before I got 0 minutes and assumed I was just wrong
Same solution with a piece of paper. Then I thought, give up. You are not Einstein. 😂😂
@@stefanluginger3682 Yeah, my mistake was probably not writing anything down, could have been a bit quicker if I did 😅
And true, we aren't. One can dream though
@Cyber Cactus. Do you have some scientific background?
If you watch until the end that was the answer.
Spent literally 1 min and I came up with 45mph
Great riddle. Solved it before watching the solution. The error we tend to make (when we think 45 mph is the answer) is that we try to calculate average speed over the distance, while average speed is calculate over time.
No, the error we make is to assume that because of the low speed at the start, you obviously don't need to run that fast just to reach this average speed
@@Melpheos1er I think Dirk is right. Usually you think that since it is 2 miles, the other mile just needs to be 45 so (45+15)/2 = 30. However, that is wrong because as Dirk said, average speed is divided by time and not distance.
What's this "we"? I suspect most people do these two simple divisions and realise it's impossible.
@@JJviniciuss interesting point but let’s say we do travel the second mike at 45mph wouldn’t we still get an average speed al together of 30 miles per hour.
@@JJviniciuss ah I got confused because I didn’t think there was an actual time limit and heard it again and he said “average 30 miles in just one hour” so the time would be spent in the first half of the trip due to lack of faster speed.
Part of the reason I got this one quickly is because I used to do distance running, and runners are used to thinking in terms of minutes per mile. When you're used to working with that unit, the answer becomes obvious pretty quickly.
I got it almost instantly for a similar reason. As a cyclist I'm always calculating average seed in my head. Definitely not an Einstein though.
When my first half mile takes 4 minutes, I'm pretty sure I won't be doing a sub-4 minute mile.
I am a machinist, and I deal with velocity a lot and conversions a lot, although averages aren't really my job description in this context. In any case, just converting units to see the amount of total time you have shows that both sides of the question are 4 minutes, and therefor unless you can quantum tunnel, teleport, gain omnipresence, or time travel, you cannot physically be at the middle and end point at the exact same time and even if you could, the only way for the speed to actually average correctly would be to reverse or freeze time to make up the discrepancy. The other methods jsut replicate the same results, but don't actually solve the problem with speed.
@@mountainrun I've just posted this, but I'd say too many people are over thinking this question...
It's not actually a trick question.... You never specified whether the average was over time, or distance. If you consider the average being over time, then you have to calculate this as you did. But if the average is over distance, then the answer is 45mph.
If half of a journey was completed at 15 mph, and we know that the average speed over the distance of the whole journey was 30mph, then logically the remaining speed for the second half of the journeys distance was 45mph.
Think of it this way, if we split the distance into 1% chunks, and after each 1% of distance travelled, we drop 15 sweets for the first half, and "X" number of sweets for each remaining percent after 50%, for us to reach and average of 30 sweets per 1% travelled, we need to drop 45 sweets per percentage travelled. When time is not a factor, then the speed is irrelevant, only the value needed to be assigned to each percentage, no matter how much faster we may have travelled it.
@@Ste-The-Leo I mean the video says it was a trick question, even Einstein himself almost got tricked by it.
Tbf I also got it wrong on the 1st place, but the explanation made it clear. Many ppl also hv the same answer like the video since they're accustomized with that unit (runner, cyclist, etc.) Meanwhile, I think it simple like u since I'm only a freshman student.
Imo, the problem with ur solution (which was also the 1st solution that I thought after seeing the question) is the difference in time passed on each speed.
Imo, The easiest way to disprove something is to take the extreme case. Lets say ur solution is right. When the car goes uphill, it takes 1/15 hour for the car to reach the top of the hill (1m/15mph=1h). Meanwhile, the time it takes for the car to downhill is 1/45 hour (1m/45mph=1h). That means, in that 2 mile trip, the car takes 3/4 times of the trip going 15mph, while it goes fast (45mph) for only a quarter of the "trip time".
If we take it to the extreme case like I said before, then taking another example, what is the speed needed for the car in the last 0.1 mile if it already goes 15mph for 1.9 mile, to get the same 30mph average like in the 1st question?
To get 45mph as the solution (speed of downhill) like u said, then the car should get to the top of the hill on its first 1/4 trip times traveled by going 15mph, then now the car CAN ride the rest trip times by going 45mph, like u said. That means, the hill shape is different than the question in this video (the top of the hill is on 1/2 mile, then it downhill for 3/4 mile).
After thinking about it.. I just remember it's a highschool physics problem (static movement)😅. hope u got enlightened😊
1 mile - 15 miles per hour
2 miles - 30 miles per hour avg
Something was not clicking from the start
What was not clicking? I don’t understand 😔😔
@@felixfong3834 To travel 1 mile at 15 mph takes the same time as to travel 2 miles at 30 mph. If you double both distance and speed the time will remain constant.
You basterd
Felix Fong If u drive at 15 mph speed it will stay consant for 1 mile or2 or 737372 miles
Telhias I thought there wouldn't be enough time to get 30mph avg since it is all used up in the 15mph start
I would've used the road nearby
Eks Vay and I would have just walked lmao
Simple Solutions
You'r right😂😂
Nahhhh
Nahhhh
Everybody: Kilometre Metre Centimetre
USA: Miles Foot Inch
Ha ha ha
You forgot Millimeter, Micrometer and Nanometer
@@fetendo8291 Picometer 😀
EVERYBODY: SCHOOL
USA: SHOOTING ZONE
The UK also uses the metric system...
actually got it which is nice to say for once. this one plays brutally well off of how easy it is to take any question at face value
This points out a fundamental principle of time management. Once time is gone, you can't catch up or get it back.
No you can't however you can increase your speed with gravity on the downhill ( let's say to 45 mph).
15+45=60
60+2=
30 mph AVERAGE speed.
EASY!
@@michaelbrannon8452 I don't think you fully understood this problem.
@@michaelbrannon8452 That doesn't work. To average 30mph in a fixed course of 2 miles, you have to complete the trip in 4 minutes.
At 15mph, the car trundled along, and finished the first mile in 4 minutes. In order for that car to average 30mph, it can no longer spend any time traveling at all, so there is no speed the car can travel at to reach the stated goal. Even Star Trek-esque warp travel couldn't make up the difference. It is a trick question.
@@michaelbrannon8452 That only works if the car stays at 15mph and 45mph for equal amounts of time, in which case the trip would take eight minutes across four miles.
what's the frame of reference for the trip? technically, if you consider the trip from the point of view of the car and not outside, there is a way to make the downhill trip last 0 minutes: travel at the speed of light.
This taught me more about my lack of self-confidence than it did about math.
Lmao same. Sat here thinking surely I'm working out 1/15 * 60 wrong if Einstein was confused by this riddle
Yup
@@Jack-ts1jq Well - it didn't seem that Einstein was stumped by the riddle for days on end. In his answering letter he stated 'I had to calculate that one' - which probably took a pencil and a sheet of paper and maximal as long as Presh here needed to explain it here.
That's the fun with trick questions. One assumes to get a 'real' riddle with a solution that makes sense. The problem is figuring out, someone is messing with you, and the counter-intuitive answer is indeed correct.
@@robertnett9793 I doubt Einstein had to take pencil to paper, I'm sure he could calculate it in his head, as I did. I'm no Einstein, I went at first the '60 mph speed for the descent' route before I figured in my head how much time it took for ascent compared to time it would take for 2 miles at 30 mph.
btw, My first thinking was 15 is half of desired product for first half, added to 60 for second half, which is double the desired result.
Yup. I was planning on going back to college to finish my degree next semester.. When a simple-looking problem like this goes SO FAR over my head, I've come to realize that I needn't bother. Lol.. I'm kidding, of course, but not really. I mean, c'mon, if the answer isn't 45, then I'm lost beyond all redemption. Haha
"This riddle almost fooled Einstein."
And here I am, fooling myself I'll be able to solve it.
Well I solved it and...
same lol
It’s actually pretty easy. If it weren’t for the bit about Einstein being fooled, it would not be intimidating at all. Of course, you might fall into the trap which is assuming you can just find the number whose average with 15 is 30.
I did the entire thing in my friggin mind!
I don't get how you should know the answer of the top of your head. Of course you have to calculate.
the thing I love most about most riddles is the fact that somebody actually came up with them (as it is with jokes and original stories)
That's a suprise, "There were *old* cars in 1935" hmm...
I had the same thought :)
They were mass producing automobiles in 1900 in France and the USA. My grandfather had a 1921 vehicle a Ford we believe. So, in 1935 that would be considered old.
@@mchapman132 well depends how long “old” is but yeah
First car engine was invented in 1890s by Karl Benz, owner of Mercedes-Benz
@@jayeetasinha3181 yea
88 mph and he would have plenty of time, but this is something Einstein would not have known at the time.
What I didn't get it?
@@pranavsuren9489 have you seen back to the future?
what if einstein is a descendant of Doc though?
@William White what are you smoking?
@@Sooka_Phatwon Back To the Future(1985)?
"nearly fooled Einstein"
Me: *"Maybe I can solve it.."*
I was tempted to do the math. Instead I wondered how much time was left to complete the trip averaging 30mph. Since the car had travelled half the distance at half the desired speed, there was zero time left. Good video!
So would there be an answer if the question remained exactly the same but the first mile was not driven at 15 mph but rather 20 mph? Or something faster than 15?
@@valeriereneeharper Sure, if there's any time left and some distance to travel, the only limit is the speed of light!
The way I thought about it was to imagine a second car, leaving at the same time, and travelling at the desired 30 mph. So at the exact time the first car completes their one mile at 15 mph, the second car completes the their two miles at 30 mph. Which means that there is no time left for the first car to catch up with the second car.
Got this wrong with the 45mph assumption. I rejoice in getting things like this wrong though as I always learn something new from it. Thanks for the video.
Why is it wrong? A lot of people here are letting their imagination get too wild by assuming factors that are not even mentioned in the puzzle.
@@thewhinjaninja3610 yeah I’m so confused people talking like the whole course is divided into two equal time frames. But it’s not.
It's divided into 1 time frame because you are averaging. One overall rate (30 mph), one total distance (two miles), and one overall time (not given). If knowing that d=rt is having a wild imagination, then I'm proud to have one. You can (and have to) assume time because without time, rate problems (and life) fall apart. There are many things that are implied in math (and life, but I digress) that are too obvious to state as that would take away most of or all of the problem. But GaryLifo is right, it doesn't matter if you got it wrong, so long as you learn.
@@thewhinjaninja3610
Because it is impossible to reach an average of 30mph over 2 miles because you've already gone 15mph over 1 mile....so you've spent 0.06666 hr (4 mins) going up hill. In order to average 30mph over 2 miles, you would need to travel over the 2 miles in 0.06666hr, or 4 minutes.
You've already spent the time necessary to travel an average of 2 miles at 30mph by the mid point of the 2 miles. As someone pointed out in another comment, Speed / Velocity is averaged over the time domain, not the spatial domain.
What if someone actually travelled 1 mile with 15 mph speed, and another mile with the 45 mph, what would be the average speed over those 2 miles in that case? Obviously not 30 mph
difficulty level on MYD:
kindergarten
nobel prize
Jup
what the hell is a Jup
@@meso_p
Yes
Jup
Yo
Jo
Exactly
I agree with you
Jep
That's what 'Jup' means for me, maybe it's odd for you because i'm from germany and you'd more likely understand what 'Jup' means there, but i thought english speakers would understand it too
I know it's "Kindergarten"
I thought it was "Kindergarden" too..
@@derekmadonna1732 thank you for your correction
Given his friend knew of Einstein’s special relativity and E=mC^2 accomplishments, this problem was clearly meant to be amusing. Travelling down the hill at the speed of light results in an average speed of 29.99996 mph which is essentially 30 mph.
Not possible. You run out of time after the first mile. You have no time left even at the speed of light. You will actually be a bit above 30MPH
@@goodtoGoNow1956 Yeah we all understand the car runs out of time (4 min.) at one mile. But if it somehow travelled at C for the second mile the speed would be very very close to 30 mph average (don't forget - time slows down as we approach the speed of light). BTW, it's impossible for the average to be >30 mph as the total travel time is greater than 4 minutes.
@@toddarmstrong7038 Oh, yes. My mistake. You are right.
@@goodtoGoNow1956 But to cheat, we can use significant digits. you can say a tailwind helped him get to 15.4 mph (approx 15). this will give him time to get down the hill at C.
@@goodtoGoNow1956 From the drivers perspective of time its possible. Due to driver's motion when he reaches the end of the 1st mile his elapsed time would be under 4min so he would have an almost infinitely small amount of time to traverse the next mile.
People's minds can easily go to "I think they're just asking me to average 15 with some other number to get 30". And the answer to that question is clearly 45. But that is NOT the question, and that's where people go wrong. If you give the correct answer to the wrong question, then you're probably going to be giving the wrong answer to the actual question.
People who point out that the literal rules of the question state that you have 4 minutes to go 2 miles, and you take 4 minutes to go the first mile, leaving no time for the 2nd mile, are obviously correct. But I like to look at it from the other direction: why can't it be 45 mph?
Why indeed? Okay, so if you think you can go 15 mph for a mile and then 45 mph for a mile, then what you're ALSO saying is that you're going to go 15 mph for 4 minutes and 45 mph for 1 1/3 minutes. That's the literal arithmetic. Now, if the question were asked that way-- "If you spend 3x as long going 15 mph as you do going 45 mph, is your average speed 30 mph?"-- then it's likely no one would be fooled. Anyone who has ever driven cross country realizes that the actual amount of time you spend at each speed matters.
You can also literally do the math: 4 minutes on the 1st mile (15 mph) and 1 1/3 minutes on the 2nd mile (45 mph) = 5 1/3 minutes to cover 2 miles. But if you spend 5 1/3 minutes to cover 2 miles, you have NOT averaged 30 mph: you've averaged 22.5 mph.
Great little puzzle! Common sense and intuition will lead you halfway into the forest. Then leave you there.
Should definitely be the first lesson for all project managers: It is very easy to fool yourself into thinking "we can make up the lost time" long after it is mathematically impossible to do so.
Well, confronted my players with a similiar problem on our RPG-table.... unfortunately it's a sci-fi system with faster-than-light travel...
It’s 45 miles per hour. If Einstein were so smart he’d still be alive.
@@klovell6793 whatre you going on about dude
@@klovell6793 I've typed this for someone else and I'm copying and pasting it.The average it's trying to meet is 30 miles per hour, right? Traveling 2 miles at 30 miles per hour would take 4 minutes, which means you have a total of four minutes to meet the average. Here's the thing: traveling one mile at 15 miles per hour also takes four minutes. So once you clear the first mile, you've already expended the time you needed to reach the average. Say you don't factor time in though, and go with 45 miles per hour as the answer. The second mile at 45 miles per hour would go by faster than the first mile. That means more of the trip would be spent at 15 miles per hour rather than 45 miles per hour, and you wouldn't get your average. Hope that makes sense.
Person a: I've got a riddle... how do you add 2 and 2 and get 5?
Person b: (spends a week thinking about it)
a: You can't, it's impossible.
b: Cool riddle.
ITS LIKE TOM RIDDLE. SORRY IF U R NOT A POTTERHEAD .
This is literally it. My brain melted trying to calculate is and the answer is the riddle's broken. Thanks a lot.
Start with one... Add two and two and you get five.
Add 2 and 2 and get 5 by
..
.
Mistake 😂
The riddle technically is not solved by either twat because they did not reach there destination. This is why education breeds many fools. 🙄
60mph = 1 mile per minute
30mph = 1 mile per 2 minutes
15mph = 1 mile per 4 minutes.
To average 30mph for 2 miles requires 4 minutes.
If you've already spent that 4 minutes in the first mile, there's no time left for the 2nd mile.
Easiest and simple explanation compared to the other commentors
Thank you. I literally didn't know why or how there was this "magic 4 minute time limit" until I read this comment. But even now it doesn't make sense. "Mph" is a rate of speed. It was never stated anywhere that it couldn't take longer than 4 minutes
@@AggroBen To "average" 30 mph is equal to saying average 2 miles in 4 minutes. The fraction 30mile/60min. is equal to 2mile/4 min. The problem states the travel distance must be 2 miles at that rate and the rate involves 4 min. Since the first mile spent 4 minutes, it's impossible to meet that rate.
@@KenPaulsenArchitect thank you for the reply. I must've lost a few brain cells over the years because even after this reply I still thought it was wrong lol it took me a while to finally comprehend it all. For some reason I thought one could still get an average 30mph even if it took longer than 4 minutes total. But then it would no longer be 30mph average. Thanks again
@@AggroBen You're welcome! Many of these puzzles stump me, but I got lucky on this one. I'm missing brain cells too! You definitely got it - yeah, it would no longer be 30mph average if it took longer than 4 minutes. Good way to phrase it.
At 15 mph it would take you 4 min to travel 1 mile. At 30 mph it would take 2 min to travel 1 mile and at 30 mph for 2 miles it would take 4 min. If you have traveled for 4 min at 15 m/h, you already run out of time to average out at 30 mph.There is no way that you can make up the time to make up for the last mile that is not traveled.
,,you gotta travel infinitely fast to make this possible, so this is impossible''
Vsauce: hold my beer
*How to count past Infinity*
Sakib Hasan me too haha
Mine’s is the infinite hotel paradox
@@SakibHasan-ks2fe yeah. I got that recommendation too.
Ayaan M same me too
And‚ if you could travel infinitely fast, the travel would take you 0+ minutes so the average speed would be 29.999... mph so it's impossible.
0:01 turn on captions,"Hey this is precious tall walker"😁
If he's white then we're doomed
He is white.
@@tm92211 so that makes him a white walker
And he likes math.
I miss Paul Walker
0:19 okay now we are supposed to believe that a german scientist used imperial?
units are merely playthings
That was my first thought too.
The problem works the same for km. He probably switched it to miles since his audience is mostly from america
@@LLLLogicccc I live in the US and just about all the problems my children are given at school are in metric.
@@martinpattison8916 i literally don’t care lol. That wasn’t even the point of my comment 🤦♂️ ppl like u rly just wanna start arguments. As i said, it works the same for km, so it doesn’t matter
Why did I read the comments? MPH includes both distance and time - it miles per hour. If you are given a distance and a mph you can calculate how long it would take. If you are given a time and a mph you can work out how far you can go. Thats the basics. 30mph for 2 miles takes 4 minutes. 15mph for 1 mile is also 4 minutes.
If it was 16mph up then you would get to the 1 mile in 3 minutes 45s . So that would need 240mph down.
15.1 mph takes 3m 58.4s to go a mile up - that would take 5,760mph to average 30mph.
15 mph takes 4 minutes - no time left. Time is implicit in mph.
Who calculated and got 0 as the answer, knew it was impossible, assuming they were wrong watched the video, realized they were right and felt like a genius?
Not you
meeeeee, i did it in the shower so I couldn't be sure of myself.
Han Huo meeee
*raises hands*
I remember doing something like this problem in physics so I was sure that 0 is right😂
I initially thought 45 mph immediately, but had to rethink due to there being no reason to the vid if it was that simple. Broke it down to time and realized he’d travel 1 mile in 4 mins @ 15 mph. In order to average 30 mph over a 2 mile stretch, you need to make the trip in 4 mins. Which means he’d need to instantly teleport from the midpoint to the end.
So the 2nd part of the trip going downhill doesn't matter at all.
It's impossible. My physics teacher gave my class the exact same problem (with different numbers) & the class laughed at me when I took longer than they did. My initial answer was correct, it couldn't be done but at 17 I didn't trust that so reworked it & got an immensely higher answer to which they again laughed until the teacher said I was a lot closer than they were.
The fastest average speed of the car can only be 25mph
bro I read a comment that said "raise your hand if you thought it was 45mph and I stopped reading and was like "alright im wrong... how" and I started to really think until I hit "wait it would take 4 min to reach the halfway point ok so how frickin fast do I have to go to adverage that out,... and I was stumped again until I was lik ok how long does it take to go 2 miles at 30mph? 4 min... oh so he has to go the speed of light? then i watched the vid and no its just "impossible" xD
45 would never make sense. You're trying to reach 30 average. 15 is half of 30. To balance out something that is HALF you need to DOUBLE the other side... not a 1.5x.
45 is literally the worst answer you can come up with.
Even though this is impossible, your mental thought patterns were still ridiculous to witness.
You don't just add 15 and 45 together to get 60, then divide by 2. It doesn't work that way with speed and time.
It's just crazy how so many people came to the most ridiculous conclusion that makes the LEAST sense, and THAT is the most popular answer? Yikes you generic people just need to hide and be quiet...and just LISTEN...until you know how to think (but still don't talk at this point, you need to just shut up and pay attention, still).
Yes he can travel at an average of 30 mph,it is possible
But he only need to change his car
Edit:I never got so many likes
DeLorean, perhaps?
He needs to go at 45 miles per hour doesn't he?
@أحمد قاضي why
My math was flawed. Happy Craig 11 iq. However Einstein didn't even read the riddle. It's speed not time. It's over 45 due to the decrease in travel time down hill. 1.33 minutes and this will effect the avg. My mistake was thinking of travel for four minutes each mile.
@@redvelvet727 why tf did you divide;
I really enjoyed solving this, my first thought was 45mph but then wrote the problem out and saw that 15mph is 4min/mile and 30mph is 2min/mile I could see that the car had to travel infinitely fast and arrive instantly at the bottom. What a fun puzzle!
He made a mistake here when he said "Velocity needs to be infinite"
We have:
t(trip) = 2/30 = 4 min = 240 secs
As he mentioned.
However, there would exist a dilation of time (this is Einstein we're talking about) so t(ascending) would not be 4 mins.
From t' = t x root(1 - v²/c² ),
(where t' is the dilated time)
We get
t' = (4x60) x root (1- 6.7056² / 299792458²)
t' = 240 x root(1 - 5x10^-16)
t' = 240root(0.9999999999999995)
Therefore,
t(ascending) = 240 - t'
Thus we have:
t(descending) = 240-(240-t') = t'
Now there is a finite (albeit very small) time that we have to descend.
Let V = velocity needed.
V = d/t
V = 1 / t'
As t' > 0, V is not infinite!
Of course, this would be much greater than the speed of light, so it is impossible, but certainly not infinite.
Time dilation only counts if you are an observer to the thing moving at that speed. Relative to yourself, as this riddle implies, you don't experience any dilation.
But yes, if you were someone with a stop watch observing, you could get there theoretically.
I got this straight away. I said to myself, 'I'll never figure this out, it's impossible.'
And I was right!
I instantly knew it took 4 minutes to go up the hill. 15mph is a 4 minute mile. This has never left my brain since high school cross country. I easily calculated double the speed half the time; times 2 for 2 miles. It would take 2 min per mile at an average 30mph. You'd have to have an MC Escher hill to end up at the bottom at the same time you end up at the top. Infinite speed at infinite acceleration.
@@djdickey At the speed of light, where time stops.
Me who also didn't see the solution until I started calculating: "You know, I'm something of an Einstein myself."
German go figure,,,lol, I picked 45.
Basically in order to average 30 mph for 2 miles you have to travel the 2 miles in 4 minutes. But since you went 15 mph for the first mile, that mile took you 4 minutes, so it is impossible to travel the whole 2 miles in that 4 minute time restraint that would allow you to average the 30 mph.
To me this answer doesn’t even answer the question let’s put this on a flat surface !
I’m sorry if tell you , to do 1 mile your drive at 15mph so how many mph do you need to do another mile and for the total being 30 ! You’ll answer 15 !!
WE DONT CARE ABOUT THE TIME !! ITS THE NOT THE PROBLEM ! if the guy take (a hypothetical) day driving at mph he will do 24 Mph ! The fact that he is on a hill will just made him slow down or accelerate in order to stabilize on that 15 mph
Why wanting to say he took 4 minutes , it’s not the problem here because at the middle of the problem it’s like everything will restart , we talk about speed not time !
I understood the answer but to me it’s like saying “how effective is to drink this specific water and you answer by saying there is no water in your cup right now”
@@d.o.a9236 it took me a while but i understood
everyone is wrong, because, yeah, he took 4 minutes to reach the top... but that is not what we're looking for, we just need to know how many speed he need going down to make those 2 miles into a 30 mph average, and well, that would be 45 mph right? just a regular calculation of average
@@marcelinebeckerrosas9023 that would be 45 for a normal average idk why in my head 15+15/2 = 30 😂😂 but that’s not the answer
@@d.o.a9236 jlnsadas xD
but why i´m wrong, my GF explained me but i still dont get it, but told me something that i didn´t know, that you can´t do that kind of calculation to miles, but i dont know if that apllies only to miles or when you're doing a distance average speed, becayse others told me that the average speed of 15 + 45 would be 22,5 i dont get it why Tw T
@@marcelinebeckerrosas9023 basically I had a whole discussion under this video with someone so try to search my other comment
(I precise that I first post it so people answered to me afterward and it’s one of the most recent comment under this video post 5 hours ago and sometimes I said thing that I thought was clear but wasn’t)
Me a “genius” ; 45mph average d
You are not alone.xd
me too
the average velocity it is not equal to velocity average
Me too lol
Xd
I think this is easily solved without algebra. Suppose you average 15 mph over a 1 mile distance. You can immediately see that if you happened to be at the 2 mile post by this time instead of the 1 mile post then you would have averaged 30 mph already. Therefore there is no time left to reach the second milepost and you would have to travel at infinite speed to get there.
Nice thinking. Thanks
Thank you, out of all the comments yours was the one that made it click for me
James Lacey AVFX this is exactly what he said in the video though 😂
Or travel at the speed of light, where time stops.
Why do you guys don't use SI units in America? Use Km/hr for speed and Kg for mass instead of pounds.
@ShadowOfChemistry but they're in SI system just bigger units. Any unit with gram(g) like Kg, dg, g, mg are considered in SI system
ShadowOfChemistry 3600 not 360
@ShadowOfChemistry lol you destroyed manpreet singh with knowledge. I am proud of you my son.
@@imperialrecker7111 or with his half knowledge 😄
@@real4487 no he just use a little knowledge to prove that there are people in the world that aren't aware of standard SI units 😄
without watching the video, it's impossible. The first mile would take 4 mins, but to average 30mph for the whole 2 miles is 4 minutes.
I am as smart as Einstein 😎, because I needed to calculate too!
lmao
I solved it mentally...
But i bet its because i was careful unlike Einstein that time who didnt know it could be hard. The thumbnail probably let me solve it carefully
R/Woosh
Im smarter than Einstein. I skipped the vid until the answer comes up.
"Nothing can travel faster than speed of light."
Albus Dumbledore: "Hold my wine."
it's beer not wine
@@howardxing5885 wrong butterbeer
*Hold my nimbus2000
HOLD MY ELDER WAND XD.
When i am moving 1m forwards for you, I am actually travelling 8 times around the earth
The “trick part” to this problem is from the definition of AVERAGE VELOCITY. We have been taught in grade school that to get the average you just add up the numbers then divide the result by the number of numbers.
In physics, this is not the case for Average Velocity.
Definition of Average Velocity:
(Average Velocity) = (total distance)/(total time)
PERIOD
You can’t use (15+45)/2=30
Yes I did it with this formula only
Well can you please tell my why in average velocity we have to use this formula
It's a harmonic average, not a linear average.
@@DrDeuteron sorry I am not able to understand
@@tushar9655 harmonic average, c, of a and b satisfies
1/c = (1/a + 1/b)/2
linear average is c = (a+b)/2,
geometric average is c^2 = ab
quadrTIC mean is RMS: c^2 = (a^2+b^2)/2
Well, yes and no. The wording of the question is critical. Specifically, the "for the entire 2-mile trip" part.
While it's true that the speed for the two-mile trip _as a unit_ can never reach 30mph, the average speed _per one-mile leg_ of the trip is unbounded. This is because we calculate the former as (total) distance over (total) time, but we would calculate the latter as total speed over total legs. (To be pedantic, distance-over-time is just _speed,_ regardless of whether it's for an entire trip or just a segment. "Average speed" is either redundant or something different.)
I've never heard anyone describing average speed as the average of "speed per one-mile (or any distance) leg".
@@Stubbari It's what we do sometimes when referring to racers' performance. Admittedly, we say things like "trial", "heat", or "race" instead of "leg" in such cases, but the operation remains the same.
@@PedanticTwit Can you give an example?
Presh: Can you figured it out
Give this problem a try
My brain's inner core: NO! its not possible
*sees end result directly
This is the first and only Mind Your Decisions problem/puzzle I’ve ever got right! I did initially think 45mph, but then wrote it out and nailed it
Great job, bro!
I like this. My answer to thumbnail question (which i only thought twice about because it said something about fooling people) is that the car cant do it unless it completes the second mile in an instant, cause the time elapsed after 1 mile is the target time for the whole trip.
Definitely unintuitive though
The key to solving this problem is to know the definition of average velocity. Average velocity is the displacement of the object divided by the time it takes the object to travel that displacement.
yes thats how I solved it too. I nearly thought I made a mistake when calculating the time elepsed on the ascent (1/15 hour) and the full trip (1/15 hour), so speed had to be infinite. luckily it was a trick question.
WOW! ... ingenious
All you did here is define what mph is measuring. That's what it measures. Miles traveled over 1 hour, or mph = miles/hour ( velocity = distance/time). Your statement is still incorrect however. You said "average velocity", but then gave the definition for "velocity". Average velocity is essentially and average of averages. In angular acceleration the collection of times and distances are represented by delta, or av = Δd/Δt.
@@vvhitevvabbit6479 Wrong! I gave you the physics textbook definition of average velocity. Lookup every physics textbook out there. MPH is a unit. I didn't define any units! Miles and hours are units that need to be defined operationally. You could also solve this problem using the definition of a weighted average which when simplified ends up being the definition of average velocity that I just gave you.
I can relate to that. When I have to go to an appointment, but I leave the house late, it is very difficult to make up the lost time. Even when I run I often barely make it on time.
Evil joke the car just broke Einstein general relativity theory😂
I was dead on the answer being 45. I even replied to several asserting the answer was definitely 45mph.
Then it hit me. The time/velocity/time relationship is not linear. This question is a perfect example.
Of course it's linear. There are just different lines.
1 hour at 15 and 1 hour at 45 would be 30, but here its 1 mile at 15, and since its scaled on distances, the car doesn't spend half the time at 15 mph, but half the distance
Just looking at the thumbnail before clicking the video I instinctively thought 45 mph, of course, but knowing it would be a trick I decided to calculate it and came up with 'infinite', for the reasons laid out in this video.
But it got me thinking, why is there this difference? So I thought, if I were to travel at 15 mph for 1 hour and then 45 mph for 1 hour I would travel a total of 60 miles in 2 hours, which averages out to 30 mph. So this works over long times, why not over short times?
So I thought of breaking it up into 15mph + 45 mph segments to see why it didn't work and then it hit me. It's because we're not measuring in units of time. We're measuring in units of distance.
If you spend half the _time_ travelling at 15 mph and the other half the _time_ travelling at 45 mph you're going to average a speed of 30 mph. But if you spend half the _distance_ travelling at 15 mph then no matter how fast you go you'll never get up to even twice that speed on average across the whole distance. Because you travelled for that pace for that distance, not for a specific time.
Actually, I think you have missed out a trick here too. I've just posted this, but there is another way to consider what the average speed is over the DISTANCE.
It's not actually a trick question.... You never specified whether the average was over time, or distance. If you consider the average being over time, then you have to calculate this as you did. But if the average is over distance, then the answer is 45mph.
If half of a journey was completed at 15 mph, and we know that the average speed over the distance of the whole journey was 30mph, then logically the remaining speed for the second half of the journeys distance was 45mph.
Think of it this way, if we split the distance into 1% chunks, and after each 1% of distance travelled, we drop 15 sweets for the first half, and "X" number of sweets for each remaining percent after 50%, for us to reach and average of 30 sweets per 1% travelled, we need to drop 45 sweets per percentage travelled. When time is not a factor, then the speed is irrelevant, only the value needed to be assigned to each percentage, no matter how much faster we may have travelled it.
@@Ste-The-Leo Yes, but speed is distance over time. They asked for the average speed. Which means they set the parameter as being over time.
Spot on!
Because averages are weighted - in this case by time. So when you have a set of speeds, in order to find the overall average, you have to multiply them by the times travelled at those speeds (to find the total distance travelled) before you can divide by the total time.
When the time intervals are all the same, the weighting is equal so they all cancel out. If the distance intervals are the same, but the speeds are not, the times cannot be the same, thus they must be weighted.
S = D / T = (d1 + d2 + ... + dn) / (t1 + t2 + ... + tn)
when t1 = t2 = ... = tn = t:
S = (d1 + d2 + ... + dn) / (n * t) = (1/n) * (d1 / t + d2 / t + ... + dn / t)
but s1 = d1 / t, s2 = d2 / t .... sn = dn / t =>
S = (1/n) * (s1 + s2 + ... + sn) = (s1 + s2 + ... + sn) / n
But, when d1 = d2 = ... = dn = d:
S = D / T = (n * d) / (t1 + t2 + ... + tn)
t1 = d / s1, t2 = d / s2, etc =>
S = D / T = (n * d) / (d / s1 + d / s2 + ... + d / sn)
It got messy after this, so I'll stop there. I hope it points you in the direction as to why it's different though.
Cheers.
but you didnt travel 15mph for an hour. u did it for a mile. same for 45. and the asked question is the average in two miles, so 45 is right.
@@diegomarxweiller1814 No, it says what speed would you need to go at to achieve a 30 mph _speed_ across the 2 miles. Speed is distance over time. So to average 30 mph across the two miles the second mile needs to be traversed instantaneously.
The issue, for those who still don't understand, is that the distance travelled is limited to 2 miles, total.
If there was no limit specified, then an answer of 45 mph for the same time as the 15 mph travel time would be acceptable.
In the 45 mph travel speed, the car would travel 3x further than the 15 mph travel speed time.
The problem here is that distance for the 15 mph section was defined as one mile AND the unknown speed section ALSO defined as one mile.
Now the answer is constrained.
The answer must account for a total distance of 2 miles, with one mile being 15 mph.
As you can intuit, since the car must travel the same distance as the 15 mph section, but faster, time is also a limiting factor.
Which is why figuring out the travel time for the first 15 mph section is important.
And so there is no value that will meet the multiple constraints, other than instantaneous teleportation over the second mile.
Actually there’s multiple answers to this question because the question states the car averaged 15/mph during that one mile time it didn’t say how long it took to travel that mile so it’s necessarily 4 min
@@hotshot2101 D = V x T
D is known, as is V (over the entire D as an aggregate).
Therefore you simply solve for T.
There is only one answer.
For the first section, D is one mile, while V is 15mph, therefore T is one fifteenth of an hour, or 4 minutes.
Now you figure for an average speed of 30 over two miles.
D = 2 miles, while V = 30
V divided by D is 15, therefore the car travelling 2 miles at an average of 30 mph takes one fifteenth of an hour, or 4 minutes...
If it took 4 minutes to travel one mile already, it cannot take ANY time to travel the second mile to meet the 4 minute constraint imposed by the constraints of the distance and the required average V over 2 miles when a V of 15mph over the first mile is defined by the question.
It requires instantaneous travel of the second mile, after taking 4 minutes to travel the first.
If it was possible to go down instantaneously wouldn't the total avg be 7.5 mph.?
15+0=15/2=7.5
@@Chili327 What even is this?
@@Chili327 V = D / T
D is 2 miles, T is 4 minutes, so V = 30 mph
1 mile every 2 minutes = 30 miles per hour
For those of you who are confused... averages and ratios (like mph) are essentially the same thing. There is no way to express velocity without using distance as a ratio relative to time. Think of it this way:
Imagine that there are TWO cars racing from point A to point B which is 2 miles away.
Car#1 has to drive at 30mph for the entire 2 miles.
Car #2 can only drive at 15mph for the first mile but can then go as fast as it wants to catch up to Car # 1.
How long does it take Car #1 to reach the finish line?
- going 30mph for 2 miles takes 4 minutes.
How long does it take Car #2 to reach the 1-mile point, and then the finish line?
- going half as fast as Car #1 for half the distance as Car #1 takes the same amount of time, which is 4 minutes.
- so, after 4 minutes, Car #2 has gone half the distance, and now has to cover another mile with NO time left on the clock.
So, if the restriction exists on the distance in which the average must be calculated, then there is also a restriction on the amount of time it can take.
Car 1 doesn't have to drive 30mph for the entire 2 miles. They just have to average 30mph over two miles, so that when they reach 2 miles their average speed was 30 mph. That is where the problem in the question lies. Saying the "entire" 2 mile trip may give it away to some but not to most imo. I get that it's a trick to fool people but i suppose it appears the writer of the question was trying to trick and therefore was purposefully not very clear in what was wanted. Just my opinion.
@@joeshawcroft7121 Well, there's not really a problem in the question because the unit for velocity is an equation that behaves like an averagee if any of the terma of the equation are given. So, regardless of how it is phrased, the point is this: if Car 1 takes 4 mins to travel 2 miles (30mph) and Car 2 takes 4 mins to travel 1 mile (15mph), it is impossible for Car 2 to travel the distance necessary in the amount of time required to also average 30 mph like Car 1. There's no time left in order for Car 2 to travel that 2nd mile, which means the equation/ratio for velocity has a denominator of zero.
The original question posed to Einstein just uses one car. Same result.
@@robjames4160 the question is about speed. And your answer is time ? That is like asking for the weigth of a given object and give the answer in feet. Nowhere in the question is there any limitation in time spent.
@@briankristiansen821 You have to calculate that time limitation - average speed of 30mph is 1 mile every 2 minutes, so 2 miles have to be done in 4 minutes to average that speed.
@@KasparsRumba8 you are not adressing my critizism. The question calls for an average speed V(velocity) and your answer is four minuts T(time). It is a fallecy of wrong units.
Now I can solve a problem that even Albert Einstein was confused to solve.
i can use a smartpone better than him... and i know that feel, i pretend to be a techie and sometimes ppl or owners of stuff teach me how to access some things
I got the answer right away in my head but since you said Einstein figured it out I assumed I was wrong somehow. Did the calculations 2 more times and said "it can't be done" Then watched the rest of the video. ya got me.
Did the same but also because I always doubt my head calculations 😂 Did it twice and watched the video to confirm I didn't miss anything conceptually.
I did actually get the answer on this one, though I have to admit when I first came up with infinite velocity for the trip down I dismissed this as impossible and had to keep working the problem for a while. Once I was sure that the total time for a one mile trip was actually equal to the time of the two miles trip did I come back to the conclusion and watch the rest of the video. Thanks Presh, it has been several decades since I'd encountered this particular puzzle. It was fun to work it out again.
Same here. Found the answer pretty quickly using mental calculus. I was almost certain I had fucked up and was curious how and why but... no i was correct lol!
But what is trick in it ....
I also found that its infinity
You’re all wrong. The driver only needs to travel the speed of light as time stops from the driver’s perspective at the speed of light. No frame of reference was stated the in riddle.
@@edwardwood3622 that is quite a steep hill to have accelerated the car to light speed. It would take infinite energy to do that so,...
@@edwardwood3622 Since you need time to stop ,true theoretically and false in reality.
... just spend 10 minutes on figuring out why my awnserd was 1800/v2 = 0. As in the car must have infinit speed. After admitting defeat i checked and i was correct 😅
Yeah same
2:07
" Einstein enjoyed easy problems"
"I enjoy easy problems"
Conclusion: I am Einstein!
Ohhh 2nd Einstein😂😂😂😂😂😂😂LOL
@@aryansundarsingh713 Exactly..
Okk Einstein ji ... U r my role model😉😉😉
@@bharath9019 lmao😂😂😂😂😂😂
@@bharath9019 hmmmmmm
The reason 45mph is wrong is because we cannot find the average of speed using the conventional average formula . It should be calculated using total distance / total time . For example , a car moves from 2 miles in 3 hrs and the next 17 miles in 6 hrs . Using the conventional formula we get the the average of speed as - ((2/3)+(17/6))/2 = 1.75mph . But if you use the right formula , you will get (2+17)/(3+6) = 2.1mph. Hence you cannot average the speed using the conventional method. If you follow the same logic in the given puzzle , you can see why there's no time left downway.
At first glance you think "oh, (15 + 45) / 2 = 30, easy!" And then you remember(maybe, some people aren't familiar with this concept) that speed, distance, and time are factors! haha I do like even the "simple" problems. I've seen similar questions posed in the form of "If I drive for 1 mile at 15 miles per hour, how long (or far) do I have to drive at 45 miles per hour in order to average 30 miles per hour?" This question helps people understand the relationship of speed, distance, and time. This question is really only a riddle if you aren't familiar with this relationship or don't use it often so it doesn't pop into your head immediately. I guess that applies to a lot of riddles though.
The assumption here is sampling w.r.t. time which is how average speed is usually calculated (it is literally distance over time) but you can technically sample w.r.t. distance, something you might do if you wanted speed to vary as a function of distance
This is an evil problem: before even watching the video I noticed it wasn’t 45 because the car would spend less time on the 2nd mile, pulling the average closer to 15 than 45.
My thought process was: Oh Einstein almost messed this up gotta pay attention. 1/15 hours to travel the first mile because 1 mile 15 mph. Avg trip is 30mph so 2/30hours spent for the whole trip, which is already all spent on the first mile
If you got 45 mph for the 2nd mile, you are correct. The solution presented here, comparing v=d/t resulting in 4 minutes and rendering the riddle unsolvable is not the correct formula given the information we have about the entire trip. We already know the formula doesn't address the riddle because we've been given a speed for the first mile and the average speed for the trip. We're not solving for the average speed because we already have it. If you think I'm lying, or must be incorrect because Einstein or whatever, take it up with the PhDs and experts at Omnicalculator. They too will say if you have an initial speed and an average speed, you find the final speed by doubling the average and subtracting the initial, directly answering the riddles question. Obviously, to verify the average speed, you add your initial and final speeds together, and divide by 2, just as you've likely assumed. Your total trip takes 5.33 minutes, 4 minutes uphill at 15 mph, 1 1/3 minutes downhill at 45 mph. You only get it done in 4 minutes if your speed is 30mph both miles. Either this amusing Einstein anecdote is misleading, or the experts are. Check it out at their average speed page and feel free to prove them wrong.
@@underscoredfrisk but the question doesn't say that the trip has to be defined by an amount of time, it could just be an amount of distance, and if the trip is just a 2 mile stretch, then 45 is the correct answer
@@mikelarrivee5115 How long does it take to drive 2 miles at an average speed of 30mph?
@@Stubbari what does that matter? That has nothing to do with the problem. First tell me why that matters and then I'll tell you the answer
If anyone's looking for an idiots explanation to this (I could have used one while struggling through comments from people who know a lot more maths/physics than me!) :
If you travel for 1 minute at 15mph then 1 minute at 45mph you'll average 30mph, but you'll have traveled a lot further during your minute at 45mph.
But here we're limited by distance as well as time, so we can't travel further to compensate. To do 2 miles at 30mph will always take 4 mins. However, you've already traveled for 4 mins doing 1 mile at 15mph. So it doesn't matter how fast you go, you've already used all your time!
To reach an average speed of 30mph after doing 1 mile at 15mph (which takes 4mins) you'd need to travel for another 4 mins at 45mph, which would mean you traveling 3 more miles instead of the 1 allowed in the question.
but... why you can apply 15 + 45 = 60/2 = 30 when the unit is a minute, but you can´t when is applied to miles?
i think i almost got it, but that is the thing that keep me from solving it, also, someone tell me what the average is 22.5, but i don´t understand the formula and why is planted the way it is.
could you tell why is wrong?
I love this, despite the fact that I was clearly wrong in my initial assumption! It's not until I locked into the parts of the puzzle which were constrained that I realized that the answer was an impossibility. It's so simple, but my brain still wanted to ignore the obvious. (Solve for x... but x must equal zero) As a musician, I translated the terms of reference to musical tempos, which made no more sense, but proved the futility of the endeavour. Thanks so much.
The 'trick' here is that the information given to you is using an average...which messes things up. So you have to get back to the numbers that allow you to calculate. This is similar to the question: a car gets 10 mpg's going up the one mile hill...then 100 mpg's going back down again....what's the average for the round trip? You can't use either mpg number to find the answer other than getting back to actual amount of fuel consumed...which then applied over the distance traveled ends up being in the high 18's. Good stuff!
it isn't impossible you just have to achieve light speed so that time dilation kicks in.
I've never seen a more pointless "As a ____" than this As a musician.
How fascinating that you converted the concept into the world of music. Isn't the human mind wonderful??
@@needdamemes2757 yeah
Once you get up, the hill collapses and it takes zero min
And what about the time which will be spent while the hill is collapsing
Only if the hill collapses with a speed of infinity then only it would work
@@Flippyy cmon I got it that it was a joke duh🙄
Can't someone just explain things
@@prakashsingh8419 even if hill collapses in no time , the time to fall down will be more than 0. Ik ur jk but whatever
air resistance
iFireliar No, because the initial vertical velocity would be zero. Air resistance can be neglected.
Veritasium explained a different version of this puzzle a few years ago as one of the “4 Revolutionary Riddles”
I remembered it from there to!
That explains why I was able to do it first try 😂 my brain went I feel like I need to calculate the total time and time already taken, but idk why.... thanks Derek!
Got it - I assumed averaging velocities would not be reasonable, so I look at time taken and discovered we'd need to travel 1 mile in 0 seconds. Cool puzzle.
The way I solved it was that the total time is 1/15 + 1/v, where v is the downhill velocity.
Average speed is therefore 2/(1/15+1/v) = 30 ==> (1/15+1/v) = 1/15 ==> 1/v = 0 which is impossible!
I knew the answer immediately, but only because I'd seen a variant before. It's a really good riddle.
same for me
Veritasium maybe by any chance? That's where I know this one from.
@@SmileyMPV I don't remember where I first heard it, but it was long before UA-cam.
I've seen a total of 3 videos from this channel. 1 had an obvious answer that wasn't even a challenge, two had no answer and were time wasters. Sometimes there is such a thing as a bad question.
Well, you could have just not commented! Commenting on this video fuels the algorithm, and disliking the video (which you probably did), ALSO fuels the algorithm.
@@solicoli It must be hard for people to like you. You add zero value to the world and your presence makes people uncomfortable and want to leave.
60 mph = 1 minute for each mile
30 mph = 2 minutes for each mile
15 mph = 4 minutes for each mile
At 30 mph the entire 2 mile trip = 4 minutes
At 15 miles per hour during the first mile,
the entire 4 minutes is used up,
making it impossible to achieve
a 30 mph average over the 2 mile trip.
If you went down the hill faster than
the speed of light, it would still take you
longer than 4 minutes to make the trip.
I did this in my head. Well first I came up with 45 mph after seeing the thumb but after he laid the problem out I used exactly this process to realize the answer is infinity. Pretty cool. Because of the nice round numbers very easy to work out in your head. You could force someone to calculate using non-round numbers but I like the simplicity of it.
I did the same thing, at first glance you think "oh, (15 + 45) / 2 = 30, easy!" And then you remember that speed, distance, and time are factors! haha I do like even the "simple" problems. I've seen similar questions posed in the form of "If I drive for 1 mile at 15 miles per hour, how long (or far) do I have to drive at 45 miles per hour in order to average 30 miles per hour?" This question helps people understand the relationship of speed, distance, and time. This question is really only a riddle if you aren't familiar with this relationship or don't use it often so it doesn't pop into your head immediately.
When you have a question and the answer doesn't answer the same question.
I didn't pick the answer being "infinite" but I figured out very quickly in my head that the trip down would need to be excessively fast... after reaching what I thought was an answer, and immediately saying to myself "no, that's still not fast enough" I clicked play and discovered that yes, infinite speed is required, which is indeed, very fast :)
Infinity doesn't exist.
@@carlosoliveira-rc2xt Just because the human mind can't comprehend it, doesn't mean it doesn't exist.
Technically the speed is the speed of light so relativistic time dilation takes effect.
@@soldjahboy Tell me where it exists ? There is always a number to everything. For example you can say there is infinite water on earth or there are infinite number of atoms in the universe but it never is. It’s just too large of a number but still it’s not infinite.
@@zarifshoeb How do you expect me to explain something to you, that you claim doesn't exist? What's the biggest number you can think of? What comes after that? What about after that, and after that.... and, after that? Where do they end? They do not. Ergo, infinite.
This feels fundamentally flawed. I am pretty sure its misdirecting the flow into separate pools that have nothing to do with the vehicles overall average over a given range. Regardless of direction the car traveled (up or down), the average speed over two miles, even if it took more time than the 4 minutes, would still be able to be reached at 30mph per traveling at 45mph. I get how its a psuedo riddle, but I think it was answered in a flawed way. If a mile at 15mph, then travel a mile at 45mph, then my average speed over those two miles is 30mph. It has nothing to do with time here, but everything to do with distance and speed.
Speed is measured in units of distance divided by units of time (miles per hour) and if you need 4 minutes to travel 2 miles at an average speed of 30mph, but take 4 minutes for the first mile, even if you could travel down that hill at light speed, you would get only get very close to the 30mph average, but not quite, even if the difference would be miniscule
No. If you want to add two speeds together and divide by 2, it is only valid as an average speed if you traveled for the same interval of time at each speed. It is not a pseudo riddle in anyway but, respectfully, your 'average' is a pseudo average.
Speed is never "nothing to do with time" - speed a measurement of distance travelled and time taken. Without time, speed as a concept doesn't really exist.
If you drive a mile at 15mph, it's impossible to do it in anything other than 4 minutes.
First i assume it's 45 mph, then i solved it to find average speed of 22.5mph
Then i took the value of 60mpg, average speed became 24mph
I assume 120mph, average speed became 26.66mph
I assume 240mph, average speed became 28.25 mph
I assumed 2400mph, average speed became 29.998mph
I understood.
The car came down at warp speed.
i cant figure how to reset my clock in my car when daylight savings happens - usually i just wait
Oh, you solved the problem! The trick is to plan this drive for when Daylight Savings takes effect, then drive 1mph down the backside of the hill, taking one hour, then set your clock back when you reach the bottom, thus arriving at the same time you left!
Great riddle. Solved it before going through the solution. People tend to calculate average speed as the arithmetic mean of all the speeds, however average speed is actually calculated as the total distance covered in the trip divided by the total time taken.
Ah, I see you are a man of culture as well.
Me thinking it's calculated by the gauge in the dash. 🤦♂️
@@titaniumwormYT *woman, actually =)
If you got 45 mph for the 2nd mile, you are correct. The solution presented here, comparing v=d/t resulting in 4 minutes and rendering the riddle unsolvable is not the correct formula given the information we have about the entire trip. We already know the formula doesn't address the riddle because we've been given a speed for the first mile and the average speed for the trip. We're not solving for the average speed because we already have it. If you think I'm lying, or must be incorrect because Einstein or whatever, take it up with the PhDs and experts at Omnicalculator. They too will say if you have an initial speed and an average speed, you find the final speed by doubling the average and subtracting the initial, directly answering the riddles question. Obviously, to verify the average speed, you add your initial and final speeds together, and divide by 2, just as you've likely assumed. Your total trip takes 5.33 minutes, 4 minutes uphill at 15 mph, 1 1/3 minutes downhill at 45 mph. You only get it done in 4 minutes if your speed is 30mph both miles. Either this amusing Einstein anecdote is misleading, or the experts are. Check it out at their average speed page and feel free to prove them wrong.
@@JL-fg2vk I see. However, I think you might be confusing velocity with speed. While calculating velocity, since the direction of displacement (shortest distance) is also considered, we may calculate the average velocity as the arithmetic mean of initial and final velocities, just like you have explained. This can be proved using v=u+at and s=ut+1/2at^2. As far as speed is concerned, doing that won't be right as speed is a scalar quantity as distance does not have direction. So, when we calculate the avg speed as the arithmetic mean, we completely rule out the possibility of speed in opposite direction.
Me, who calculated 1 minutes and 20 seconds: yeah, i totally knew that it was impossible.
I feel like some people still don’t understand because of the iffy explanation in the video. Here’s mine: The average of 15 and 45 is 30, so from that math alone, 45mph. However, if the car traveled 2 miles at 30mph, it would take 4 minutes. But because the car has already gone for 4 minutes, the 45mph solution would theoretically work, but in our universe you have to factor in the laws of time, meaning you could not go 1 mile instantly. If the car were to descend the hill at 45mph, it would actually work out so that the average speed of the car is 22.5mph because speed equals distance over time, and distance is 2 miles, and it would take 4 minutes to go up the hill and 4/3 minutes to descend, which means that speed equals 2/(16/3) which is 0.375 miles per minute, or 22.5 miles per hour
Hey, I was among those telling 45;) It seems simple: 1st minute 15mph, 2nd minute 45 mph, gives on average 30mph UNTIL you notice going uphill takes 4 minutes, the entire time you have to go up&down. Nice puzzle:)
Except the 4 minute time constraint wasn’t included in puzzle as presented at the beginning. It was only added after, during the presentation showing everyone how foolish they were thinking this was just a math problem of average speed, where the correct answer for the puzzle, as presented, was 45, as most people assumed. This would have been a better presentation if the full puzzle constraints were made available at the beginning. That way people wouldn’t be puzzled at the confusion of a renowned physicist over an elementary arithmetic problem.
@@tcorris The 4 minutes is not an added constraint. You calculate it from facts given from the start.
@@tcorris
If you travel one mile at a speed of 15 mph, it will take 4 minutes. That is a calculated fact, not an added constraint.
If it took you any time OTHER than 4 minutes to get to the top of the hill, you couldn't have driven exactly one mile, or you must have been going a different speed.
I remember the different wording "you are driving to your friends house 30km away, you drive the first half of the trip at 15km per hour, how fast would you have to drive for the second half to average 30km per hour." Same concept but slightly more obvious.
It's actually the exact same concept and wording; only tiny distinctions are the removal of the hill (the framing device that explains why the car might be able to go faster in specifically the second half) and the change of the units.
-"For -_-the second half-_- to average 30 kph." That's a big difference. This problem asks for the average of -_-the entire trip.-_ I see my mistake now
@@mordekaihorowitz It's a 30km trip, the first half, naturally would be 15km. If you travel for the first half at 15km per hour, how long have you been driving? The answer to that question demonstrates why it is impossible to average 30km/h for the entire trip, regardless of how fast the second half is.
@@trondordoesstuff yeah... that's basically what I said. Same concept, similar wording, slightly more obvious by reducing superfluous factors.
@@mordekaihorowitz No it still asks for the average of the entire trip it’s just poorly worded. It’s not “How fast would you have to drive for: The second half to average 30km/h”, it’s “How fast would you have to drive for the second half: To average 30km/h (For the entire trip)”.
I spend several minutes trying to understand why my answer was 0 minute. There is another way to see that it's impossible. If you traveled x mile with a velocity v in average in t hours, and what to double the average velocity v'=2v of the whole travel in the next x miles, then
v'=2x/t'
2v=2x/t'
v=x/t'
therefore t=t', so it's impossible to that because there is no time left to make the second part.
Fu**ing nerd
Hi Bi U mean a smart person
@@hibi1440 Wym? He simply just doesn't have the IQ of a table unlike some persons.
Pink is a pretty good tasting color on one of those expensive suckers that they used to sell at school during Valentine’s Day.
I remember having to do a problem very similar to this in my physics class. The professor gave this to us as a group assignment. We were all stumped. If I remember correctly one of us did get zero but it was rejected thinking it must be wrong.
You didn't got zero, you might have got one by zero
@@VIShesh_Agrawal_happy how do you know he didn't get zero though? i think the original comment meant one of them got zero, but they thought it was wrong so they didn't submit it, it didn't mean the *professor* thought it was wrong
You've reached a hilltop , stay there for a while and enjoy the view.
On your way down just put your car in neutral and decelerate slowly.
I was on the right track thinking that the second half will take a shorter amount of time at a higher speed and to find the average speed one has to take that into account. I wrote a whole equation where I tried to solve for x and got a nonsensical answer. Only then I checked how long a trip of 2 miles at 30 mph would actually take. When I found it was also 4 minutes, the same as the first half, I realized what was going on.
Doable in your head if you know the key to these problems is often total time driven, and at 60 mph you cover 1m/min.
Total distance is 2 miles - at 60 mph average that's 2 min, at 30 mph its 4 min.
1 mile at 15 mph = 4 min - so you've no time left
Though travelling the second mile at an appreciable portion of the speed of light would probably "allow" You to do it via rounding. There isn't much difference between 240 seconds and 240.0000053 seconds
Intuitively, I guessed 60 for the second leg. Then did the math and chuckled at myself.
Does anyone have a good formulation to generalize this puzzle? "You cannot double your average trip speed if you are past the half-way point" or something like that? Not sure it's very practical, but an interesting thing to consider.
Your max average speed is your current average speed times one over the part of the journey you've already completed.
I.e. 10mph times 1-over-1/3(=3) makes a max average speed of 30moh
Average speed works "normally" over equal time, not equal distance. Therefore, distance needs to be proportional to speed. If you go 10 mph for 10 miles and then 40 mph for 40 miles, average will be 25 mph over 50 miles. But if you reduce the second part of the distance and make it equal to the first one, you will reduce the time spent on it which will reduce the impact that second part has on the average speed: (10+10)/(10/10+10/40) = 16 mph over 20 miles.
The formula is (X1+X2) / (X1/Y1+X2/Y2), where X1, X2 - first and second part of the way (might not necessarily be equal halfs);
Y1, Y2 - first and second speed.
Which creates the "limit". Even if you assume that second part of the way is just teleported over, it just means that X2/Y2=0. You are still left with (X1+X2)/(X1/Y1). This is the limit.
In simpler words, limit is equal to the entire way divided by the time that was already wasted. If you go 10 mph for 10 miles and teleport another 10 miles, the average speed is 20. If you stand still for an hour, then teleport through all the 20 miles instead, the average is still 20 mph.
So what is the mph if the second leg was traveled at 60mph? Since at this point it's obviously less then 30mph, but haven't worked out by how much.
@@LazyOtaku So, we know the first mile takes 4 minutes. at 60mph the second mile takes 1 minute. so the mph for the whole trip is 2 x 60/(4+1) = 24 mph
45 mph for second leg is about 19 mph average using that formula
I got it right! At first I thought it would be 45mph, until I realized that the real question was how long it would take to do that first mile. At half the speed (15mph) it takes the same amount of time to do the first half of the trip as you are allowed to use for the full trip at twice the speed (30mph). 45mph would be the right answer if the question were a two minute full trip, rather than a two mile trip.
wat? nowere is talks about time other than in the distance/time. of course it takes less time, but you want the average speed in set distance. it is 45
In my head I said "oh, easy. 15 + 45 = 60, 60/2=30" But somehow that extra mile changed things about how I thought about it because there was a D involved. Then time became important. I really didn't see it at first and was convinced 45 was the right answer.
@Diego Marx Weller If you go 45mph for the second mile then your average speed will only be 22.5mph
@@diegomarxweiller1814 Time is sorta built in to the problem. MILES PER HOUR. If it takes too long to go the first mile, there is no way to make it up in the second mile. @30 MPH, you have to be doing 2 minute miles. It took 4 minutes to do the first mile...that is the total time you would have had to complete the whole trip. That you have another mile doesn't matter.
It is definitely counter-intuitive. But my guess is you shouldn't be arguing with Einstein.
@@TR1CK I don't think that's right either. The point of the problem is that you are already out of time. it took 4 minutes to do the first mile. you have to average 2 minute miles to do 30 MPH. You already used all the time you have. Speed has to be infinite.
“Nothing can travel faster than the speed of light.” - MindYourDecisions
“HolD mY aPple jUicE.” - Vsauce
Can his apple juice travel faster than light?
Also highlights the importance of double-checking your solutions!
Inputting 45 mph as the second speed yields an average speed of 22.5 mph.
AND, this seems like a good way to exercise determination of limits---in this case, there IS a limit to the possible average velocity.
Average speed, not the average speed over the distance. You guys are all over thinking your answer. I've just posted this, but you're welcome to check it and give me your thoughts.
It's not actually a trick question.... You never specified whether the average was over time, or distance. If you consider the average being over time, then you have to calculate this as you did. But if the average is over distance, then the answer is 45mph.
If half of a journey was completed at 15 mph, and we know that the average speed over the distance of the whole journey was 30mph, then logically the remaining speed for the second half of the journeys distance was 45mph.
Think of it this way, if we split the distance into 1% chunks, and after each 1% of distance travelled, we drop 15 sweets for the first half, and "X" number of sweets for each remaining percent after 50%, for us to reach and average of 30 sweets per 1% travelled, we need to drop 45 sweets per percentage travelled. When time is not a factor, then the speed is irrelevant, only the value needed to be assigned to each percentage, no matter how much faster we may have travelled it.
@@Ste-The-Leo speed is always distance / time, so your second answer doesn't make any sense. And speed cannot be "irrelevant" because speed is in the question - what (average) *speed" does the car need to travel.... to make the overall average *speed*...
@@iandavidson471 Think of it another way then. A simpler way.
I travelled 2 miles. For the first mile I travelled at 15mph, and for the second half of the distance I travelled at 45mph.
You created fixed distance, or even infinite, data collection points across this distance to measure the speed I was going. What would the average speed I travelled across all of these data points come to?
@@Ste-The-Leo That would work... if you were measuring based on time. Sure, spend half the time going 15mph and the other half going 45mph, and you get a 30mph average.
The problem is that in physics, time is always related to speed and distance. When you take distance into consideration, even if you take half the distance (a mile) at 15mph, there is no way you can travel 45mph for the other mile of the distance unless your car is traveling at light speed, since you've already spent all your time going uphill and you've got 0 seconds to complete that other half...
@@seisosimp Not quite, time is not something being measured in the example I provided. Actually, time is completely irrelevant.
The only unit being measured is speed, and the points it is being measured is across a distance. How the speed has been calculated is arbitrary and for the purpose of providing a figure that can be considered "an average speed" (not necessarily the true average speed the vehicle would travel), then averaging the speed over the distance is the factor.
Your calculation of the true average speed is, of course correct, but there is another way to interpret what an average can be considered to be, one of which is taking the known results, adding them together and dividing by the number of results.
Think of it this way, a row of speed cameras across the distance measure the speed. 10 cameras across the first half measure 15, and ten cameras across the second half measure 45.
Of all the results taken that day, the average speed measured by those cameras would be 30.
Heck, if the first mile was fraction of a centimetre longer than the second mile, I could just as logically say the average speed was 15, if considering a median average across the distance.