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11 December, 14:29

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.

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  1. 11 December, 16:27
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    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.
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