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Old Feb 10th 2009, 04:53 PM   #1
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A train travels...

Between two stations a train travels the first 1/nth of the distance with uniform acceleration, then with uniform speed, v, and for the last 1/nth of the distance with uniform retardation.
(1)What is the average speed?
(2)If the acceleration and the retardation each occupied 1/nth of the total time, what would be the average speed in this case?
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Old Feb 11th 2009, 02:13 AM   #2
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I think that you must introduce letters for the unknown quantities and hope that they cancel out, or that the problem expects an answer with letters.

if the acceleration is "a" then the train's position is given by x = (1/2) a t^2 while it is uniformly accelerating. Let L be the total distance. When the train reaches position x = L/n we have L/n = (1/2) a(t_1)^2. So you can solve that equation (in symbols) for t_1. The train's velocity at that time is v_1 = (a)(t_1) so you can write an expression for that.velocity. Then it proceeds at velocity v_1 for a distance of (L - 2/n) which takes time t_2 = (L - 2L/n)/ v1. Then it decelerates. By symmetry, I would assume that the time it takes to decelerate is also t_1. The average speed = total distance/total time = L / (t_1 + t_2 + t_1). See what cancels out in that expression.

The second part asks what happens if the acceleration takes 1/n of the trip time instead of 1/n of the trip distance. I'll think about that later, it's getting late.
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train, travels

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