A train travels between stations 1 and 2 as shown in the figure. The engineer of

Question

A train travels between stations 1 and 2 as shown in the figure. The engineer of the train is instructed to start from rest at station 1 and accelerate uniformly between points A and B, then coast with a uniform velocity between points B and C, and finally decelerate uniformly between points C and C until the train stops at station 2. The distances AB, BC and CD are all equal and it takes 5:50 min. to travel between two stations. Assume that the uniform accelerations have the same magnitude, even when they are opposite in direction. (I can not draw figure but it is straight line Station 1 is at point A and Station 2 is at point D it is given AB=BC=CD)

How much of this 4:59 min. period does tht train spend 1.)between points A & B, 2.)between B and C, 3.) between C and D?

Explanation / Answer

Let the distance AB = BC = CD = d,
Acceleration = a.
Between A and B : ---------------------
d = (1/2)at^2
t = sqrt(2d/a)

v = at = sqrt(2ad) at B

Between B and C : ---------------------
a = 0
v = sqrt(2ad)
t = d/v = d/sqrt(2ad) = sqrt(d/2a)

Between C and D : ---------------------
Time is the same as between A and B
t = sqrt(2d/a)

Between A and D: --------------------
5:50 min = 2sqrt(2d/a) + sqrt(d/2a)
               = sqrt(8d/a) + sqrt(d/2a)
               = sqrt(16d/2a) + sqrt(d/2a)
               = {sqrt(16d) + sqrt(d)}/sqrt(2a)
               = {4sqrt(d) + sqrt(d)}/sqrt(2a)
               = 5sqrt(d)/sqrt(2a)
               = 5sqrt(d/2a)
5:50 min = 5sqrt(d/2a)
1:10 min = sqrt(d/2a)

                           Between B and C the time is 1:10 minute.
The remaining time, 4:40 minutes is divided equally between the other two sections;
                2:20 minutes each between A and B , and between C and D.

You gave two different values for time 5:50 min and 4:59 min. I take 5:50 min as the correct one.If not, please substitute 4:59 min inplace of 5:50 min in the solution. Thank You.


Thank You.

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