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the speed of the train is reduce uniformly from 15m/s to 7m/s while travelling a distance of 90m.
a) compute the acceleration
b) how much further will the train travel before coming to rest provided the acceleration remain constant.

  • Physics -

    a) The average speed during the interval is (1/2)(15+7) = 11 m/s

    The time required to travel 90 m while decelerating from 15 to 7 m/s is
    t = 90/11 = 8.18 s

    The acceleration rate during that interval is
    a = (7 - 15)/8.18 = -0.978 m/s^2

    b) To stop, the additional time required is (7 m/s)/(0.978 m/s^2) = 7.158 s
    Multiply that by the average speed during that interval, 3.5 m/s.
    25.06 m/s

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