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15 September, 22:46

A speeder passes a parked police car at a constant speed of 23.3 m/s. At that instant, the police car starts from rest with a uniform acceleration of 2.75 m/s 2. How much time passes before the speeder is overtaken by the police car? Answer in units of s.

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  1. 15 September, 22:47
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    t = 16.94 s

    Explanation:

    t is the time passes before police catch the speeder

    speed of speeder Vo = V = 23.3 m/s

    T = t

    Police Info

    Vo = 0 m/s

    a = 2.75 m/s^2

    t = t

    Now,

    displacement of the police car = displacement of the speeder.

    x_{police} = Vo * t + 1/2 at^2

    since Vo = 0

    x police = 1/2 at^2

    x police = 1/2 (2.75) (t) ^2

    Now the displacement of speeder is

    x_{speeder} = Vt

    x_{speeder} = 23.3 t

    x_{speeder} = x_{police}

    23.3 t = 1/2 * 2.75 t^2

    23.3 t = 1.375 t^2

    t = 23.3/1.375

    t = 16.94

    t = 16.94 s
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