Average speed

  • #1
thursdaytbs
53
0
You move from point 1 to point 2, and then from point 2 back to point1, using the identical path. if your average speed from point 1 to point 2 is S1 and the average speed from point 2 back to point 1 is S2. What's the average speed of the entire trip there and back, in terms of S1 and S2.

Wouldn't the average speed going to point 2, and coming back to point 1 sum divded 2, basically (S1+S2)/2 be the average speed for the whole trip?
 

Answers and Replies

  • #2
mattmns
1,118
6
Yes that is correct.
 
  • #3
thursdaytbs
53
0
But the problem should not be this easy. Our teacher had said that it's not what you think it is.

Therefore, I had come up with this, which doesn't come out to make sense for some reason. Maybe someone could point out why?

Average Speed = Distance Traveled / Time it took to travel

S1 = (1->2) / t1 and s2 = (2->1) / t2
*1->2 = 2->1, let's just call them X
S1 = X / t1 and s2 = X / t2

Total average speed = [(1->2) + (2->1)] / (t1 + t2)
= (X+X) / (t1+t2)
= 2X / (t1+t2)

now... t1 = X/S1 and t2 = X/S2
so...

2X / (X/S1 + X/S2) turns out to equal (2)(S1)(S2) / (S1+S2)

Would that be the correct answer? But... If I say S1 = 2, and S2 = 4, then the average speed should be (S1+S2)/2 = 3. But.. using (2)(S1)(S2) / (S1+S2) you get, 8/3... ?

Any help appreciated.
 
  • #4
Leong
382
2
i think your method is the correct one since you derived it from the definition of average speed.

average speed is not the same as the average in statistics ie arithmetic mean.
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
43,021
971
mattmns said:
Yes that is correct.


NO, that's completely wrong! "average speed" is not "arithmetic average of two numbers". It is, in fact, the "harmonic average".

The "average" speed is defined as the speed such that one would cover that distance in exactly the same time as with the varying speed.

Let D be the distance from point 1 to point 2. Then at speed S1, it would take D/S1 (time units). Going back at speed S2, it would take D/S2. The time to go both ways is D/S2+ D/S1, of course.

Let S be the average speed. To go the total distance 2D at speed S would require 2D/S (time units). Those two times must be the same:

2D/S= D/S1+ D/S2.

Fortunately, the "D" cancels out. 2/S= 1/S1+ 1/S2. Multiplying through by S1, S2, and S to get rid of the fractions, 2S1S2= S2S+ S1S= (S2+ S1)S so
S= 2S1S2/(S1+ S2).

Hey, that's what thursdaytbs got!

(Unfortunately, you spoil it by saying "But... If I say S1 = 2, and S2 = 4, then the average speed should be (S1+S2)/2 = 3." Didn't you start by saying you didn't think it was just the arithmetic average of two numbers!)
 

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