A subway train accelerates from rest at one station at a rate of 1.50 m/s 2 for half of the distance to the next station, then decelerates at this same rate for the second half of the distance. If the stations are 1200 m apart, find the time of travel (in seconds) between the two stations.

The time of travel between the two stations is 56.6 s.

Explanation:

The equation for the position and velocity of the train will be as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where

x = position of the train at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

First let's calculate the time of travel for the first half of the distance:

In this case x0 and v0 = 0. Then:

x = 1/2 · a · t²

600 m = 1/2 · 1.50 m/s² · t²

2 · 600 m / 1.50 m/s² = t²

t = 28.3 s

The velocity at the end of this part of the travel will be:

v = v0 + a · t (v0 = 0)

v = a · t

v = 1.50 m/s² · 28.3 s = 42.5 m/s

Now, let's calculate the time for the second half of the travel. The initial velocity will be the final velocity of the first part of the travel (42.5 m/s).

Using the equation for velocity:

v = v0 + a · t

0 m/s = 42.5 m/s - 1.50 m/s² · t

- 42.5 m/s / - 1.50 m/s² = t

t = 28.3 s

This makes sense because the acceleration is of the same magnitude.

The time of travel between the two stations is (28.3 s + 28.3 s) 56.6 s.