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March 30, 2017

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An unmarked police car traveling a constant 95km/h is passed by a speeder traveling 14km/h .Precisely 2.50s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.30m/s^2 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?

  • college physics 44 - ,

    The time of police car motion is t,
    the time of the speeder motion is (t+2.5).
    The distance covered by
    - the police car is v(p) •t+at^2/2,
    - by speeder is v(s)•(t+2.5).
    v(p) •t + at^2/2 = v(s)•(t+2.5).
    v(p) = 95 km/h = 26.4 m/s
    v(s) = 140 km/h = 38.9 m/s
    v(p) •t + at^2/2 = v(s)• t + v(s)• 2.5,
    at^2/2 + [v(p) – v(s)] •t -2.5•v(s) = 0,
    t^2 –(2/a)•[v(s) – v(p)] •t – 2.5•2/a = 0,
    t^2 – 10.9t - 84.6 = 0,
    t = 5.24 s.
    The total time is 5.24+2.5 = 7.74 s.

  • @elena - ,

    im still a bit confused, i still don't know the answer to the question

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