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16 November, 10:36

You are at a train station, standing next to the train at the front of the first car. The train starts moving with constant acceleration, and 9.0 s later the back of the first car passes you. How long does it take after the train starts moving until the back of the seventh car passes you? All cars are the same length. t=?

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  1. 16 November, 11:02
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    The answer is t = 23.81 s

    Explanation:

    A car in this question means those partition that make up the train

    Let denote the the of each car as Z

    The equation of motion that has to do with distance is given as

    S = ut + (1/2) at²

    Where u = initial velocity = 0

    t = time = 9 s

    Substituting these values we have

    Z = (1/2) a (9) ²

    = 40.5 a

    making acceleration the subject we have that a = Z / 40.5

    Let us denote the length of the 7th car as = 7Z

    Considering the distance at the 7th car we have

    S₇ = ut₇ + (1/2) at²₇

    Where u = initial velocity = 0

    Substituting values we have

    7Z = (1/2) (Z/40.5) (t₇) ²

    7 = {1/2 * 40.5} (t₇) ²

    t₇² = 7 * 2 * 40.5

    t₇² = 567

    t₇ = 23.81 s
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