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4 September, 04:42

A fugitive tries to hop on a freight train traveling at a constant speed of 5.0 m/s. Just as a box car passes him, the fugitive starts from rest and accelerates at a=1.4 m/s^2 to his maximum speed of 6.0 m/s, which he then maintains. (a) How long does he take to catch the box car? (b) how far does he take to reach the box car?

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  1. 4 September, 08:18
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    Answer:Given:

    Initial speed of fugitive, v0 = 0 m/s

    Final speed, vf = 6.1 m/s

    acceleration, a = 1.4 m/s^2

    Speed of train, v = 5.0 m/s

    Solution:

    t = (vf-v0) / a

    t = (6.1-0) / 1.4

    t = 4.36 s

    Distance traveled by train, x_T = v*t

    x_T = 5*4.36 = 21.8 m

    Distance travelled by fugitive, x_f = v0*t+1/2at^2

    x_f = 0*4.36+1/2*1.4*4.36^2

    x_f = 13.31 m

    5*t = v (t-4.36) + x_f

    5*t=6.1 * (t-4.36) + 13.31

    solve for t, we get

    t = 12.08 s

    The fugitive takes 12.08 s to catch up to the empty box car.

    Distance traveled to reach the box car is

    X_T = v*t

    X_T = 5*12.08 s

    X_T = 60.4 m
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