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28 June, 08:28

A 6110-kg bus traveling at 20.0 m/s can be stopped in 24.0 s by gently applying the brakes. If the driver slams on the brakes, the bus stops in 3.90 s. What is the average force exerted on the bus in both these stops?

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  1. 28 June, 08:39
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    For 24 seconds force exerted is 5092 N towards opposite direction of motion of bus. For 3.90 seconds force exerted is 31333 N towards opposite direction of motion of bus.

    Explanation:

    We have equation of motion v = u + at

    Initial velocity, u = 20 m/s

    Final velocity, v = 0 m/s

    Case 1:-

    Time, t = 24 s

    Substituting

    v = u + at

    0 = 20 + a x 24

    a = - 0.8333 m/s²

    Force = Mass x Acceleration = 6110 x - 0.8333 = - 5092 N

    Force exerted is 5092 N towards opposite direction of motion of bus.

    Case 2:-

    Time, t = 3.90 s

    Substituting

    v = u + at

    0 = 20 + a x 3.90

    a = - 5.13 m/s²

    Force = Mass x Acceleration = 6110 x - 5.13 = - 31333 N

    Force exerted is 31333 N towards opposite direction of motion of bus.
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