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March 26, 2017

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An unmarked police car traveling a constant 95 km/h is passed by a speeder traveling 120 km/h.

Precisely 1.00 sec. after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 1.90 m/s^2 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?

  • Physics. - ,

    Write two equations for distance travelled vs time after the speeder passes. One equation for each car.

    Set the distances equal and solve for the time t.

    V1 = 120 km/h = 33.33 m/s is the speeder's speed

    V2 = 26.39 m/s + 1.90 (t-1)
    is the policeman's speed,for t>1 s. For 0<t<1, it is 26.39 m/s

    X1 (the speeder) = 33.33*t
    X2 (the cop) = 26.39*t + 0.95(t-1)^2

    Use X1 = X2 to solve for t

  • Physics. - ,

    6.1 seconds

  • Physics. - ,

    6.1

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