A funny incident happened over 30 when a statistics teacher tried to trick me in to fing average speed of 40 and 50 and distance of 150. My friend sitting next to me whispered "say 40" but I knew that that teacher wouldn't have have asked me if it was that simple. My initial guess was that the answer should be between 30 and 40 and I calculated 37.5. I still remember the look on the teacher's face and he said that in his 25 years of teaching the course no one else came up with the correct answer.
I just said “rates average harmonically”, so 2x80x40/(40+80). Works for speed, resistors in parallel, capacitors in series, springs in series, nuclear decay rates through different channels, other relaxation processes.
Nice problem. The 10 miles is irrelevant. The same answer will occur at any distance. This can be solved using alligation methods. Rate and time are inversely proportional. He spends twice the time going 40mph as he goes 80mph. The final answer is between 40 and 80 mph and the 40 receives a weight of 2 and 80 a weight of 1. It’s like a lever problem. There is a 2 pound weight on the 40 marker and a 1 pound weight on the 80 marker. The fulcrum should be placed at 40+ 1/3(80-40) .
s=d/t => 1/s = t/d, so 1/s1 = t1/d and 1/s2 = t2/d (t1+t2)/d = 1/s1 + 1/s2 => (t1+t2)/(1/s1 + 1/s2) = d here total time T = t1+t2 and total distance D = 2d, so we multiply by 2 and get 2T/(1/s1 + 1/s2) = D average speed : S = D/T = 2/(1/s1 + 1/s2) S = 2/(1/40 +1/80) = 2/(3/80) = 160/3 mph
I knew this already I just like watching his videos cause he's smart.
Very important to “simplify” the problem and do one step at the time. You are an excellent teacher, making it easy to understand! Thank you.
Thanks,
Today I solved a similar question and now I can confirm I have done it correctly
Well done.
👍👍 happy i solve the answer in my head correctly in less than minute before end of lesson. Great mind exercise Mr H!
Excellent!
Thanks for the tutorial..i made a mistake in such a question in my test last time .Now I know how to solve it
Really love his tutorial video, using simple explanation to teach.
That was a good problem!!! I intuitively went to 60 mph, and 20 seconds later realized I was wrong!
A funny incident happened over 30 when a statistics teacher tried to trick me in to fing average speed of 40 and 50 and distance of 150. My friend sitting next to me whispered "say 40" but I knew that that teacher wouldn't have have asked me if it was that simple. My initial guess was that the answer should be between 30 and 40 and I calculated 37.5. I still remember the look on the teacher's face and he said that in his 25 years of teaching the course no one else came up with the correct answer.
It's difficult. Simple use 2xy/X+y.
X= first speed
Y= second speed.
This trick will use when distance will be same but distance will be different
For equal time intervals ,we have, v avg.=(V1+V2+...v n)/n
I just said “rates average harmonically”, so 2x80x40/(40+80).
Works for speed, resistors in parallel, capacitors in series, springs in series, nuclear decay rates through different channels, other relaxation processes.
If you are yet to, you should consider tutoring for standardized tests like GMAT and GRE. I think you would make a great tutor, sir.
Nice problem. The 10 miles is irrelevant. The same answer will occur at any distance. This can be solved using alligation methods. Rate and time are inversely proportional. He spends twice the time going 40mph as he goes 80mph. The final answer is between 40 and 80 mph and the 40 receives a weight of 2 and 80 a weight of 1. It’s like a lever problem. There is a 2 pound weight on the 40 marker and a 1 pound weight on the 80 marker. The fulcrum should be placed at 40+ 1/3(80-40) .
You're right.
The method used in the video is more useful if two distances are different.
Math is not that complicated. And I like the attitude of the author.
I was scared for a while that he would leave 160/3 as the final answer. :)))
thanks for the good content on youtube!
Driving back, Chris is stopped by the police for speeding. The police take one hour to issue the citation. What is Chris' new average speed?
From India!
Videos for functions please, i need them
ขอบคุณครับ
Harmonic mean.
GRATE 👍
s=d/t => 1/s = t/d, so 1/s1 = t1/d and 1/s2 = t2/d
(t1+t2)/d = 1/s1 + 1/s2 => (t1+t2)/(1/s1 + 1/s2) = d
here total time T = t1+t2 and total distance D = 2d, so we multiply by 2 and get
2T/(1/s1 + 1/s2) = D
average speed : S = D/T = 2/(1/s1 + 1/s2)
S = 2/(1/40 +1/80) = 2/(3/80) = 160/3 mph
That's right
Actually, you do not need to know the exact distance. If S1 = 40 and S2 = 80, we have
Sav = 2*d/(t1+t2) = 2*d/(d/S1+d/S2) = 2*S1*S2/(S1+S2) = 53.3
Great observation!
A question for you.
What if the 2 distances are not equal to each other?
In that case, you need to know the exact distances to calculate the average speed.
👍👍
Std 😂
2(40x80) / (40+80)=
2(3200) / 120 =
6400/120
=53.333333333333333333
Sneaky easy problem