When a genius misses an easy question

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  • Опубліковано 4 гру 2024

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  • @gordonweir5474
    @gordonweir5474 6 годин тому +22

    The video and what I see in the comments make it seem as though Einstein was fooled by the question. And yet at 11:31 he correctly calculated that there was no time available for the trip down. I would say that he got it!

    • @juncheok8579
      @juncheok8579 5 годин тому +3

      "Not until calculating did i notice..." implies he got tricked, even if it was for a few moments

  • @solandri69
    @solandri69 5 годин тому +13

    Averaging speeds over the same distance requires taking a harmonic mean.
    2/v_avg = 1/v1 + 1/v2
    The reason is that you've defined the problem as saying the two legs of the trip cover the same _distance._ So to calculate it, the _distance_ needs to be in the denominator (since they're the same distance), just like when you add two fractions they need to have the same denominator. Since miles/hr has distance in the numerator, you need to flip it to put distance in the denominator before calculating the average.
    If the problem had been the car spends 1 hour driving 15 mph, and 1 hour driving v2 mph, then the time is equal for both legs, so time needs to go in the denominator. Which it already is for mph, so you can solve _that_ as v_avg = (v1 + v2)/2. And v2 = 45 mph would be correct.
    Same issue crops up with fuel efficiency in the U.S. (which uses miles per gallon). If you've got a car which gets 20 mpg and another car which gets 40 mpg and you drive both cars the same distance over the year, then the average mpg is the harmonic mean. Resulting in an average mileage of 26.7 mpg.
    OTOH, if you put 15 gallons in the 20 mpg car, and 15 gallons in the 40 mpg car (same gallons, instead of same distance), and drive both until their tank is empty, then the average mileage will be 30 mpg.
    Since the vast majority of driving you do is based on distance, not how much fuel you have, the rest of the world uses "liters per 100 km" to avoid this inversion issue (puts distance in the denominator).

  • @jcortese3300
    @jcortese3300 4 години тому +11

    Translation: "If I have to be at work by 8am, and I'm only halfway there when the clock strikes 8am, how fast do I have to drive to get to work on time?"
    YOU CAN'T. You will be late to work no matter how fast you go.

    • @Bigchickenburger
      @Bigchickenburger Годину тому

      If u go at infinite speed u can reach or just light speed

  • @nicholasharvey1232
    @nicholasharvey1232 8 годин тому +45

    6:57 I've seen this one before. In order to average 30 mph for a 2 mile trip, the car has to make that trip in 4 minutes. But the car already took 4 minutes to travel the first mile, therefore the car would have to teleport the rest of the way!
    If you average a speed of x for the first half of any trip, it is impossible to average a speed of 2x for the full trip.

    • @Sonny_McMacsson
      @Sonny_McMacsson 6 годин тому +2

      Einstein was figuring out HOW to do it, considering he's gonna need a wormhole.

    • @AlbertTheGamer-gk7sn
      @AlbertTheGamer-gk7sn 2 години тому

      @@Sonny_McMacsson And then, there has been rumors that Einstein's spirit got cloned by magical equation sheets, so when he was proclaimed dead, his clone spirit quantum materialized a clone body and teleported underground into a base, where he would then change his name to "Doc Brown", and then he would then suffer a blow to the head which will cause him to use his theory of relativity to figure out the way to create a flux capacitor to make wormholes, as well as a hyperinfinipotent liquid known as "Time Juice" to power it, and then created a DeLorean Time Machine to the future, where he then drank a potion of immortality.

    • @logx-ow1us
      @logx-ow1us 2 години тому

      Oh that’s why I was so confused. I kept using d=rt while looking at he thumbnail

    • @stevehorne5536
      @stevehorne5536 2 години тому

      I've seen this before too, and my answer is that "average" is ambiguous for this - I'll stick with the arithmetic mean anyway, but if you average WRT distance instead of time you get that for the second mile the speed must have been 45mph. Remember - you can average anything WRT any variable you want. For what averaging speed WRT distance is (in the general case) that you integrate the (varying) speed WRT distance, then divide by total distance. When you're given averages for intervals that sum to the overall interval, that simplifies to a weighted average of those discrete intervals. When those smaller intervals are all equal, that simplifies to the sum-all-the-values-and-divide-by-the-number-of-values average - in this case (15+k)/2 = 30 which solves to k=45.
      Just because averaging WRT time is what you'd normally expect for problems involving speeds, doesn't mean it's the only possible option. People will say that speed is by definition an average given by distance per unit time, but they're wrong - speed is by definition the (calculus) derivative of distance WRT time, which can vary continuously, and while it's natural to average WRT time when using a derivative WRT time (giving you a convenient integral-is-the-inverse-of-derivative simplification), you can average WRT any variable you like.
      It's not a popular answer, based on the last time I gave it, but it's correct - averages aren't even always arithmetic means (median, mode, geometric mean, harminic mean, many other kinds of mean) but, assuming the arithmetic mean, you can average over intervals of any kind you want, or you can even have averages of a list of numbers with no intervals specified (so assuming equal weight to each number) - again (15+45)/2 = 30. Personally, I'd consider my answer wrong normally (we generally know what is meant by "average speed" - ie. WRT time), but if someone asks an absurd question, I say exploit any ambiguity because ignoring conventional interpretation is the lesser absurdity.
      Averaging per distance travelled makes a lot more sense in reality when looking at fuel use - many road conditions vary with the interval of distance you're travelling on (others varying with time due to weather and traffic, others varying depending on the mood of the driver). Gallons per mile times miles = total gallons.

    • @nicholasharvey1232
      @nicholasharvey1232 Годину тому

      @@stevehorne5536 He explained why 45 is incorrect, you must go by total distance divided by total time to get average speed, even if the car drove 200 mph down the hill, the 4 minutes required for a 30 mph average speed are already "used up" so as additional seconds tick by, so the maximum possible average speed of the trip ticks down from that 30 mph target. After 8 minutes, it similarly becomes impossible to average 15 mph for the trip, because the required time threshold has already been passed. And as additional minutes pass from there, the maximum possible average speed gets lower and lower, no matter how fast the trip is completed.

  • @plentyofpaper
    @plentyofpaper 7 годин тому +12

    The hill problem is something that I've often thought about as both a bicyclist and a runner.
    The problem is intuitive if you're talking about differences in speed over fixed amounts of time, but not distance.
    This is basically why dealing with hills, even on a round trip is more of a pain than just a level path.

  • @reinisliepins6177
    @reinisliepins6177 6 годин тому +12

    I feel smart when i instantly got the dice problem.

  • @TheRealMirCat
    @TheRealMirCat 4 години тому +4

    Therefore, Officer, it was impossible for me to be speeding.

  • @dache85
    @dache85 7 годин тому +15

    7:59 i remember one video where you argued "half of two plus two" is 1/2 * 2 + 2 and not 1/2 * (2 + 2), because 1/2 (2 + 2) would be said as "half of *the sum* of two and two" or "half of *the quantity* two plus two" but look where we are now

  • @andrewewart7166
    @andrewewart7166 6 годин тому +2

    For the first puzzle, given the 2 results you can get a third result for r in terms of a and c only which comes to r = (a+c) + sqrt(2ac) or r = (a+c) - sqrt(2ac) (both values work as a,c > 0, so a+c > sqrt(2ac))

  • @timsmith8489
    @timsmith8489 6 годин тому +4

    There's a story about someone giving John von Neumann this puzzle: Two trains 180 miles apart are on the same track going opposite directions toward each other, one traveling 10 miles/hour and the other 20 miles/hour. A fly starts on the front of one train and flies toward the other at 60 miles/hour. As soon as it reaches the other it turns around and flies back at that same speed to the first train. When it reaches that it reverses, and so on, until the fly finally is crushed when the two train collide. Question: how far does the fly fly?
    von Neumann instantly gives the correct answer. The person then said "Oh, you saw the trick! Most people just try to sum an infinite series!".
    von Neumann than said "I did sum the infinite series! There's a trick?"

    • @michaelz6555
      @michaelz6555 5 годин тому

      Just so we're all on the same page, it's 360 miles, right?
      Of course I solved the answer in my head and I also summed the series. But the series was the sum of the two speeds giving me the relative speed between the trains (30mph), giving me a time to collision (6hr), and the total distance traveled by the fly (60mph x 6hr = 360miles).

    • @akuunreach
      @akuunreach 5 годин тому

      I did it by combining the speeds, 30mph, 180miles, 6hrs, then how far the fly can get at 60mph.
      After solving, I realized, I could have solve as follows.
      Combined the speed, observe that the fly is traveling at double the speed, and note that it would cover double the distance, and arrived at 360mi

    • @markbothum4338
      @markbothum4338 3 години тому

      Well, 180 miles covered at a combined speed of 30 mph takes 6 hours. Fly always travels at 60 mph during that time = 360 miles. Easy peasy.

  • @CatholicSatan
    @CatholicSatan 7 годин тому +9

    Hah! That Einstein problem... I was getting annoyed with myself because I was getting an infinite speed/zero time for the descent and thought I was calculating it wrong. And then I thought, nah, it's a trick question 🙂

    • @hrayz
      @hrayz 13 хвилин тому

      Like reverse psychology, it's a reverse trick question. Ie: Not the trick you're thinking of!! 🤣

  • @johng.1703
    @johng.1703 7 годин тому +3

    looking at the car problem, to get a 30mph average, and you can only do 15mph for the 1st half of the trip you would have 0 time to do the 2nd mile, it takes the same amount of time to travel 2*D at 2*T as it does to travel 1*D at 1*T.
    this problem is deceptive in that it is really a time problem, not a speed problem.

  • @Kenan-79
    @Kenan-79 7 годин тому +4

    6:37 it's interesting how the highest possible number to get is 7, while 2 or 12 are the lowest

    • @TheRealScooterGuy
      @TheRealScooterGuy 6 годин тому

      Casinos rely on this not being intuitive.

    • @angrytedtalks
      @angrytedtalks 6 годин тому

      Yes, every child learns this in 2nd or 3rd grade, but probably already knew it from playing board games.

  • @acem82
    @acem82 39 хвилин тому

    6:46 ...I "invented" this problem a few years ago, trying to figure out why if I could ride my bike at 20 mph on the flat, why it was so hard to do so on a hilly course, if you get back all the energy you gained by going up the hill. I realized, that if the first half of the trip was up a hill you could only go up at 10 mph, you'd need to teleport down the hill to average 20 mph, and that allowed me to realize why the 20 mph assumption was wrong!

  • @kenmore01
    @kenmore01 7 годин тому +3

    Einstein got it right. Thats what the video said. Also, he had time left to make bagels!

  • @WillRennar
    @WillRennar 2 години тому

    I refer to it as "a 5th-gear brain working on a 2nd-gear problem." If you think of the brain as a car engine, you can think of working on higher-end problems as being in a higher gear. Solving a simple problem with your brain in low gear is like trying to drive at a slow speed in a low gear; the engine handles it just fine. Trying to drive slow in a high gear, however, causes so few RPMs that the engine stalls and dies out, so it can't do it. ...A bit of a weird analogy, I admit, but it seems to get the point across.

  • @angrytedtalks
    @angrytedtalks 6 годин тому

    I didn't overthink the first one, it seemed obvious using trig that r=b.
    The second one is deeply obvious unless you think 6+5 and 5+6 are different.
    The last one takes a moment to realise the average speed is already sucked up by the ascent time.
    Sometimes overthinking or expectation stops the otherwise obvious from being apparent.

  • @samuelgionet529
    @samuelgionet529 2 години тому

    My intuition said 60 mph, because if we go half the speed in the ascent, then we must double the speed in the descent. Alas, calculating it gave an overall speed of 24 mph. Very interesting problem!

  • @RabbitInAHumanWoild
    @RabbitInAHumanWoild 7 годин тому +6

    So Feynman overthinks and Leibnitz & Einstein underthink. Neither do I wear socks. I'm glad to be in good company.
    The car might not have to go infinitely fast; the speed of light would do for those in the car as they would experience no time for the downward part. Those watching would 'see' the car going the speed of light however.
    BTW, die is the singular of dice. I still remember my Grade 3 teacher correcting me.

    • @hyperboloidofonesheet1036
      @hyperboloidofonesheet1036 7 годин тому

      One issue here is from the point of view of the light speed car, it hasn't actually traveled a mile downhill. No time has passed, and due to length contraction, no distance was traveled. The other issue is that no object with mass can reach the speed of light.

    • @kenmore01
      @kenmore01 7 годин тому

      ​@@hyperboloidofonesheet1036 Length contraction is why it would take no time. To those observing, you wouldn't "see the car going the speed of light" because you wouldn't be able to see it since it's going the speed of light. It would appear at the start and the end at the same instant.
      All this is silly though, of course, because a car that can't go faster than 15 mph average uphill wouldn't be able to go the speed of light downhill, because it's in need of repairs. 😅

    • @OrbitTheSun
      @OrbitTheSun 6 годин тому +1

      @@hyperboloidofonesheet1036 You are correct that the traveler's perceived distance is shorter, but he actually covered the two miles in four minutes (own time). This works because a very small time dilation occurs in the first half of the route. And this makes it possible to stay below the speed of light for the second half of the route in order to reach 30 mph. So physically feasible.

    • @asdfqwerty14587
      @asdfqwerty14587 6 годин тому

      @@OrbitTheSun It makes no sense to be using the observers point of view to judge what the distance is while using the car's point of view for determining the time. That will just give a nonsense calculation that doesn't actually mean anything - the speed could be literally anything in that case without any extra information (why stop at just looking at the car's frame of reference and the frame of reference of the mountain? Maybe one of the frames of reference is right next to a black hole millions of light years away from the problem for all we know).

    • @RabbitInAHumanWoild
      @RabbitInAHumanWoild 6 годин тому

      @@hyperboloidofonesheet1036 Of course you're right but, well, I wasn't writing a paper. Thanks for the reply.😀

  • @d007ization
    @d007ization Годину тому

    I vaguely knew about the Einstein riddle but I never figured out why it truly was the case.
    Of course, students might have an easier time understanding it by considering that a single B on their record means they'll never have true straight As.

  • @danielbranscombe6662
    @danielbranscombe6662 7 годин тому +4

    So Feynman's solution actually presents a restriction on viable values for a,b,c. If you instead use b for r in his equation and then solve for b you get that
    b=a+c-sqrt(2ac)
    so for any other value of b given a,c then the problem is unsolvable because the combination is invalid.
    For example if a=2,c=9 then we get b=17. If instead we are given instead b=19 then there is no possible rectangle with diagonal 19 such that to given quarter circle has the extra lengths a,c as given.

  • @markofdistinction6094
    @markofdistinction6094 20 хвилин тому +1

    I once thought that I made a mistake ... but I was wrong.

  • @robertdubard7959
    @robertdubard7959 3 години тому +1

    Every one of these is trivially easy, IF you look at them a certain way.

  • @ashtonaughtband
    @ashtonaughtband Годину тому

    Technically with time dilation it is possible and you wouldn't need to hit the speed of light because there is also very very very slight time dilation on the ascent as well. So it would be like (4 -(10^-32)) mins for the accent and 1O^-32 mins for the descent.

  • @robertrisk93
    @robertrisk93 Годину тому

    The answer is simple.
    Just reach the speed that allows you to run the next mile in no time. And there you have it, you have just averaged 30 miles per hour.

  • @Omnifarious0
    @Omnifarious0 2 години тому

    Thumbnail pepblem: Infinity. The car would have to instantly travel the remaining mile in order for it to average 30mph for the whole trip.
    Circle radius problem. The answer is just r = b. The two diagonals of a rectangle are of equal length. One diagonal is the radius. The other you've labeled b.

  • @malcolmtaylor1224
    @malcolmtaylor1224 25 хвилин тому

    It isn't necessary to travel at an infinite speed for the trip down. Just travel at the speed of light and time stands still.

  • @kevinmcknight5000
    @kevinmcknight5000 2 години тому

    Due to time dialiation, I think the car only needs to go the speed of light (with reference to whoever observes it go an average of 15 mph for the first half) to make it down the hill with an average of 30 mph.

  • @rgmolpus
    @rgmolpus 3 години тому

    If the back side of the hill is a cliff, then the car can make the trip in the required time - 1 mile down won't take that long.

  • @UncleKennysPlace
    @UncleKennysPlace 8 годин тому +2

    My father showed us the first one, a half-century back.

  • @chinareds54
    @chinareds54 5 годин тому

    Clearly Leibniz never played Settlers of Catan. The probabilities (out of 36) are printed as dots on the hex numbers. :p

  • @ErikBongers
    @ErikBongers 6 годин тому +1

    I doubt that Einstein discussed this in miles instead of kilometres.

    • @TheRealScooterGuy
      @TheRealScooterGuy 6 годин тому

      But if the problem was sent to him in miles per hour, mightn't he respond using the same units?

    • @necrorebel5718
      @necrorebel5718 2 години тому +2

      It turns out that intelligent people are entirely capable of working in different units. An intelligent person, such as Einstein, could have been given the problem in brags per spuds, and he could've solved it - the units don't actually affect the math at all, so even totally meaningless units can be worked with.
      Furthermore, Einstein spent much of his life living in the USA. He definitely had decent familiarity with the mile as a unit of distance and the mile per hour as a unit of speed.

  • @SiqueScarface
    @SiqueScarface 5 годин тому

    When looking at the radius problem, I didn't notice the r=b shortcut at first. But when I looked at the quadratic solution, I immediately saw the 2ac - a² - c², and in my head converted it to -(a-c)², and now I had 2b²-(a-c)² and thus b² + b²-(a-c)² = b² + (a+b-c)(a+b+c), and slowly, it dawned me, that I missed something.

  • @Vex-MTG
    @Vex-MTG 2 години тому

    The first question felt so obvious that I was SURE I was doing it wrong, especially when you went into your explanation. Nope, turns out it was just obvious. ;)

  • @daveincognito
    @daveincognito 2 години тому

    I remember the car riddle from an old Encyclopedia Brown book.

  • @robertjarman3703
    @robertjarman3703 Годину тому

    You know, I actually thought about it using the second method you were talking about, with the idea of 4 minutes. I am not the best at math in my head for equations like this, but I did notice that it seemed like there wasn't enough time remaining to go that fast although I wasn't sure exactly why.

  • @banibratamanna5446
    @banibratamanna5446 Годину тому

    for the last riddle actually we have to solve for x in 30=2/(1/15+1/x) ......it has the solution x=∞ which is not practical....

  • @ErikBongers
    @ErikBongers 6 годин тому

    I also came up with the same answer as Feinman, so yeah, we geniuses make mistakes.

  • @marioisawesome8218
    @marioisawesome8218 5 годин тому

    if you want an average speed of 30 mph then you can get the average of average speed per one mile segments. i was tired but saw this and kept on ending up with that and had to watch the vid for the solution

  • @andyiswonderful
    @andyiswonderful 5 годин тому +1

    You didn't define what you meant by average speed. It is perfectly valid to define it as average of the speed in the first section and the speed in the second section, regardless of how much time was spent in each section.

    • @foogod4237
      @foogod4237 5 годин тому

      No, that is the "average of the two speeds". That is not the same thing as the "average speed". "Average speed" is a fairly well-defined and standard term which is always interpreted to mean the total distance divided by the total time.

    • @Kevlar187
      @Kevlar187 56 хвилин тому

      The formula for calculating miles per hour is miles / hours. If you're driving 2 miles and you want to average 30mph, you'd need to do so in 4 minutes. You can't just disregard that the first half of the trip took too long when the question is asking about the whole trip.

    • @michaelh1832
      @michaelh1832 2 хвилини тому

      @@Kevlar187To do the trip in four minutes at 30mph, that would be your actual speed, but duration aside, it is not necessarily your average speed; they are smuggling in an undefined criteria of a 4 minute duration. The problem does point to the limitations of averages providing useful or meaningful explanatory power.

  • @vcvartak7111
    @vcvartak7111 2 години тому

    First two problems are trivial, only in the third problem more person make mistakes

  • @justinepaula-robilliard
    @justinepaula-robilliard Годину тому +1

    But
    Dice 1: 5. Dice 2: 6,
    Dice 1: 6, Dice2: 5...
    Dice 1: 6: Dice 2 :6
    Dice 1: 6: Dice 2: 6...
    2 dice, each has a 1 in 6 chance of being 5, a 1 in 6 chance of being 6... they are independent of each other...
    You goofed badly, you assumed a false fact.. The sum of 2 dice to 12 is 2 ways, 6+6 and 6+6.. Your table should for the 12 show both blue and green.. You cannot say 2 dice for 11, 2 ways, 5 6 then 6 5 , and not say the same for 12, 6 6 and 6 6.. 2 dice the same rules apply.. You goofed.. Badly..

  • @quigonkenny
    @quigonkenny 5 годин тому

    But from whose point of reference are we observing the car going up and down the hill? If that observer is moving fast enough, its quite easy to only take 4 min for the entire trip...

  • @jeffeloso
    @jeffeloso 7 годин тому

    So you can replace r in the first answer, by b, and get a relationship between a, b and c. It looks very messy.

  • @dhpbear2
    @dhpbear2 4 години тому

    "I was going 70 miles an hour and got stopped by a cop who said, "Do you know the speed limit is 55 miles per hour?" "Yes, officer, but I wasn't going to be out that long..."
    -- Stephen Wright

  • @denelson83
    @denelson83 5 годин тому

    Speed of light.

  • @marvhollingworth663
    @marvhollingworth663 4 години тому

    I got the dice thing & I said the car would have to teleport instantaneously to the bottom. (I've heard a version of the last puzzle before, but I got the answer then too.) My answer to the 1st puzzle was a bit more complicated, I should have noticed r = b. I said r = sqrt (b^2- (r-a) ^2) +c & r = sqrt (b^2- (r-c) ^2) +a. This seems a valid answer to me, but what do you think?

  • @drakkondarkspell
    @drakkondarkspell 2 години тому

    Actually, time dilation indicates that the proper answer to the Einstein riddle is C. When an object moves at the speed of light, time stops, and the journey down the hill will take exactly 0s. Mathematically, this is the correct answer, and now we simply need to wait for engineering to catch up and figure out how to make the car travel at the speed of light.

    • @drakkondarkspell
      @drakkondarkspell 2 години тому

      I'll be accepting my Nobel Prize in Physics for this, thank you.

  • @dumbsamalam
    @dumbsamalam 6 годин тому

    If I record the vehicle's speed every 10 ft traveled, then the average of the data set could conceivably reach 30 mph

  • @SarahAbramova
    @SarahAbramova 5 годин тому

    Thank you for this video!

  • @RabinSaidÖsteränggymnasietNA1C
    @RabinSaidÖsteränggymnasietNA1C 5 годин тому

    Intiutively I thought it must be 45 mph but then I started to think about it. I thought of it like this. If it takes 30 mph to through this in 2 unite time. Then 15 mph will take also 2 unite time for 1 mile so it must go infinite speed.

  • @richhughes5784
    @richhughes5784 Годину тому

    Dice is the plural of die. You don't say "one dice", you say "one die".

  • @bikerrick
    @bikerrick 3 години тому

    You add a time to the problem that was not part of the question.

  • @joelleet554
    @joelleet554 4 години тому

    I wonder if Pythagoras could have got these.

  • @n085fs
    @n085fs 6 годин тому

    In the Order of Operations, PEMDAS, where do square roots go? Why aren't we ever taught that?

    • @dumbsamalam
      @dumbsamalam 6 годин тому +1

      Pretty sure square root goes with exponents... but also putting an expression under the radical is a shortcut for parentheses, I guess

  • @suhaasvemuri7980
    @suhaasvemuri7980 7 годин тому +2

    This has got to be the easiest problem I've seen lol. It's impossible because it takes 4 minutes for the first mile, and 30mph for 2 miles means 4 minutes of travel time for 2 miles. Since you already traveled for 4 mintues for the first mile, you gotta teleport.

  • @OrbitTheSun
    @OrbitTheSun 7 годин тому +5

    Einstein would have answered the question correctly if he had named the speed of descent as the speed of light. Because of time dilation, a photon does not need any (internal) time to travel downhill and this would mean an average speed of 30 mph for the entire route, from the traveler's perspective.

    • @akuunreach
      @akuunreach 5 годин тому +1

      This is a mathematical problem, even light would take 5.368 microseconds, and thus skew the total time taken, even if it was fractional.
      You would need to descend the hill instantaneously, which is simply not possible.

    • @artjackson8360
      @artjackson8360 13 хвилин тому

      @@akuunreachIt depends on your reference frame. From the perspective of the driver of the car, the trip would take exactly 4 minutes even though the first half of the trip took all of that time. The second half, from the reference frame of the driver, takes zero time because traveling at the speed of light is instantaneous in that frame. From a stationary reference frame, it would take e a finite amount of time.

  • @lidarman2
    @lidarman2 3 години тому

    No wonder Einstein got it wrong. :P But if you were in the car, I bet it would feel like only 4 minutes.

  • @andyespinozam
    @andyespinozam 4 години тому

    Damn; you had me with the last one.

  • @ManjeetRani-v5n
    @ManjeetRani-v5n 7 годин тому

    hey presh! um.. sorry to ask this way but can you please guide me about how you got into stanford and how should i apply as an international student.

  • @RazvanMihaeanu
    @RazvanMihaeanu 8 годин тому +1

    Infinity?

  • @tomq6491
    @tomq6491 5 годин тому

    Am I the only one that kept trying the second one and thinking I was doing something wrong as it kept giving an infinite speed?

  • @johnyoung9649
    @johnyoung9649 3 години тому

    If I do the second half of the trip at the speed of light, then time would stop for me.

  • @Songfugel
    @Songfugel 6 годин тому

    Lightspeed is the correct answer due to time dilation

    • @gasparsigma
      @gasparsigma 6 годин тому

      But for the observer the time will still pass between point A and B, it'll just take 1/c time

  • @otakurocklee
    @otakurocklee 4 години тому

    So he got it right.

  • @pramodsingh7569
    @pramodsingh7569 4 години тому

    Thanks

  • @classydays43
    @classydays43 3 години тому

    Oh hell yeah, I got the last one right.

  • @emilie375
    @emilie375 6 годин тому

    2:50 🤤
    So now my question is : how to simplify this awful expression to get b ?

  • @Violaetor
    @Violaetor 3 години тому

    It's easier as a logic puzzle rather than a math problem. If you propose the required solution it solves itself. You want a 2 mile trip to take 4 minutes (2 miles at 30mph) AND the first mile already takes 4 minutes (1 mile at 15mph), it's impossible. Skip all the non-sense in the middle.

  • @willzhao5889
    @willzhao5889 7 годин тому

    Cant subtracting the radical still result in a positive answer?

  • @edmundgerald5764
    @edmundgerald5764 3 години тому

    I propose that the Right Angle Theorem be properly and officially designated as the RAT Theorem ie Right-Angle Triangle Theorem. Enough with the Pythagoras nonsense.

  • @empmachine
    @empmachine 3 години тому

    so: (a+c+b^2)/2 == b ??? I solved the 1st by saying a+(b^2)+c covers 2r, and then just divided.. am I wrong? (besides being overly complicated?)

  • @Santrix125
    @Santrix125 8 годин тому +88

    Stop reuploading your old videos, this is a reupload of a 5 year old video, smh.

  • @benlap1977
    @benlap1977 7 годин тому

    I'm smarter than Feynman! Yay!
    Edit: And Leibniz too!
    Edit 2: And Einstein too!

  • @picardcook7569
    @picardcook7569 15 хвилин тому

    it's 45

  • @carly09et
    @carly09et 6 годин тому

    You are a burke, this is about the oneway speed of light and is a deep philosophical problem. Your missing the 'trees for the wood' here. The implication of this question is SUREAL time travel - which is allowed but not liked.

  • @loiscampos4553
    @loiscampos4553 5 годин тому

    When i got the first two
    Oh-Oh!
    But then fail the last
    Hiuffff!!!

  • @Rickety3263
    @Rickety3263 6 годин тому

    Lolshuddup we all know you can’t travel faster than light 😂

  • @radhaprabhu7371
    @radhaprabhu7371 Годину тому

    Bro solve my question in previous video's previous video

  • @cheriem432
    @cheriem432 6 годин тому

    (pssst - it's one *die*!)

  • @natviolen4021
    @natviolen4021 6 годин тому

    guess I'm a genius, then 🤣

  • @Morbius_Official
    @Morbius_Official 7 годин тому

    6:33 Adolf: ಠ_ಠ

  • @Geslot
    @Geslot 6 годин тому

    Today I learned I am not a genius

  • @blackmancer
    @blackmancer 6 годин тому

    For that square to be a real square it would need a 90 degree symbol in the opposite corner of the already placed 90 degree symbol. There are no angle indicators on any other corner of this 4 sided object. You've made an assumption about the shape.

    • @noahblack914
      @noahblack914 6 годин тому +5

      The problem literally defines the shape as a rectangle. So not only is it not a square, but it requires _no_ angle indicators to be identified as a rectangle, because it _is defined as a rectangle by the problem._

  • @AdelBazzineFredrikaBremergymna
    @AdelBazzineFredrikaBremergymna 7 годин тому

    So i guess then it turns out that I'm smarter than both Leibniz and Einstein combined 😎

  • @ironshoes1720
    @ironshoes1720 10 хвилин тому

    Next time, it would be nice if you could spend more time showing all the steps you took to come up with the equation r = ...
    It's harder for us older folks who haven't touch those equations in forever to figure out all that's going on during the simplification phase.
    Yeah, I know, I'm slowing the group, sorry. 🤷🏼‍♂️😞

  • @tomekiriazi8736
    @tomekiriazi8736 5 годин тому

    The "famous right triangle theorem" made me LOL 😂😂

  • @stevemonkey6666
    @stevemonkey6666 7 годин тому +1

    It's always a good day when I end up being smarter than Einstein😂

    • @kenmore01
      @kenmore01 7 годин тому

      Ah, but Einstein figured it out, according to the video I just saw.

  • @ClausB252
    @ClausB252 7 годин тому +1

    Richard over-thought the problem. Presh over-explained the solution.

    • @noahblack914
      @noahblack914 6 годин тому +1

      There seem to be a whole slew of people who watch these videos just to be mad. Don't you have better ways of spending your time?

  • @aakashgupta4230
    @aakashgupta4230 8 годин тому +1

    🙂🙃

  • @affilinet
    @affilinet 7 годин тому +2

    You can only get very close to 30 mph. If you drive the second mile at the speed of light, your average speed will be 29.999999999... mph.

    • @hhhhhh0175
      @hhhhhh0175 7 годин тому +2

      29.999... equals 30
      (me personally, i have mass, so i can't drive the speed of light, but who's to say a photon can't go from a really slow medium into a vacuum)

    • @OrbitTheSun
      @OrbitTheSun 7 годин тому

      Through Lorentz contraction, the descent distance is shortened to zero, and thus a speed higher than 30 mph can be achieved, from the traveler's perspective.

    • @jungle_run
      @jungle_run 6 годин тому

      From the perspective of the traveler at the speed of light, the trip takes no time, it is instantaneous.

    • @normalchannel2185
      @normalchannel2185 6 годин тому

      ​@@hhhhhh0175 Except the problem is that c is finite, and thus the 29.99999 number will eventually end.
      To be exact, since speed of light is 11,160,000 miles per minute, and the first speed is 0.25 miles per minute, it will take 4 minutes for the first part, and it would take 0.000000089605734767025 repeating (089605734767025 is repeating part). Or if you round down 0.00000008 minutes.
      And we need the avg time to be 4 minutes, which is gotten from 30 mph = 0.5 mpm, and we need to go 2 m.
      So the avg speed at speed of light would be....
      2 (miles)
      ---------------------
      4.00,000,008 (minutes)
      or 0.4999999900000 miles per minute, or 29.999999400000 mph. So yea, it could ROUND upto 30, but its not the same as 0.9999.... = 1

    • @OrbitTheSun
      @OrbitTheSun 6 годин тому

      @@normalchannel2185 But you can cover the two miles in exactly four minutes, using the traveler's time as a measure. This is because a small time dilation occurs on the first mile, which shortens your travel time so much that you would measure under 4 minutes for this first mile. And then you don't even need to reach the speed of light for the second mile to complete the entire route in exactly 4 minutes (own time), which puts you at exactly 30 mph as your experienced average speed.

  • @ccash3290
    @ccash3290 6 годин тому

    Label your reposts so nobody is confused

  • @peterpumpkineater6928
    @peterpumpkineater6928 6 годин тому

    Theres no way Einstein „got this wrong“ he was probably just briefly asked in an interview and didn’t care.

    • @TheRealScooterGuy
      @TheRealScooterGuy 6 годин тому

      He didn't get it wrong. He said he didn't realize it was a trick question until he was doing the calculations. Obviously, once he realized it was a trick, he had no reason to finish the math.

  • @vaughanwj
    @vaughanwj 5 годин тому

    Leibniz wasn't wrong. The "proof" is wrong. You can't use that grid. There are two dice..Dice 1 has a 1/6 chance of being five or six. Dice 2 likewise has the same chance. Outcomes for 11 are as follows:
    1) 5 & 6
    2) 6 & 5
    For 12:
    1) 6 & 6
    2) 6 & 6
    If you don't believe me, make the dice different colors. You cannot use the grid and magically pretend there's only one dice multiplied by 2 when looking for 12.

    • @annt229
      @annt229 Годину тому

      Dude, colors or not, doesn't matter at all.
      Let's say you have a red and a blue dice.
      For 12: only combination is [R6, B6]
      For 11: there're [R5, B6] and [R6, B5]

  • @abhigyanghosh9330
    @abhigyanghosh9330 8 годин тому +1

    The second one is correct only if the dice are distinguishable. If they are not, Leibniz was correct

    • @verkuilb
      @verkuilb 7 годин тому +5

      Incorrect. If the dice are fair, whether they are distinguishable is irrelevant. The best way to prove this to you is to tell you to get a pair of indistinguishable fair dice and try it yourself.

    • @MadaraUchihaSecondRikudo
      @MadaraUchihaSecondRikudo 7 годин тому +1

      No, that is incorrect. Even if the dice are indistinguishable, you still have two ways of reaching 11, and only one way of reaching 12. A more stark example would be to compare the probability of getting a sum of 7, with getting a sum of 12. Just roll some (identical) dice and you'll observe that 7 comes up a lot more often.

    • @UglyRooks
      @UglyRooks 7 годин тому +1

      when you say distinguishable what do you mean/imagine? when you roll two dice in a game of backgammon or craps though you cannot tell the dice apart (if for example you had named - before the roll - the one a and the other b) they are still two different dice. Try some hundreds (at least) of rolls and you will get twice as many elevens than twelves!

    • @UglyRooks
      @UglyRooks 7 годин тому

      @@MadaraUchihaSecondRikudo correct, but according to Leibniz's (erroneous) logic there are more ways to get a sum of seven (1+6, 2+5, 3+4) than 11 so he wouldn't say it's equally probable to get 7... still, I repeat, you are correct, it's just not a suitable example.

    • @Pyrogecko08
      @Pyrogecko08 7 годин тому +4

      Suppose you have two dice that are different colors, and are therefore distinguishable, but the colors happen to be indistinguishable to a friend of yours who is colorblind, do the sums of the dice appear in different distributions depending on which of you uses them?

  • @GourangaPL
    @GourangaPL 6 годин тому

    but r=b, so what's the point of the problem?