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

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 D until the train stops at station 2. The distances AB, BC, and CD are all equal, and it takes 5.63 min to travel between the two stations. Assume that the uniform accelerations have the same magnitude, even when they are opposite in direction. How much of this 5.63 min period does the train spend between points A and B? How much of this 5.63 min period does the train spend between points B and C? And how much much of this 5.63 min period does the train spend between points C and D? Answer in units of minutes.

Expert Solution

Let the distance AB = BC = CD = d,
the 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
5.63 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)
5.63 min = 5sqrt(d/2a)
1 min = sqrt(d/2a)

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

There must be an easier way. Let t1 be the time between A and B , and between C and D.
Let t2 be the time between B and C. Let v be the cruising velocity between B and C.
Between A and B , and between C and D d = average velocity x t1 = (v/2)t1; t1 = 2d/v.
Between B and C d = vt2; t2 = d/v
So t1 = 2t2.
The total time = t1 + t2 + t1 = 2t2 + t2 + 2t2 = 5t2 = 5 min.
t2 = 1 min. t1 = 2 min.

The cruising velocity between B and C is v = d/(1 min).

genius_generous answered 2 years ago

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