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26 May, 17:16

2.85 A police car is traveling at a velocity of 18.0 m/s due north, when a car zooms by at a constant velocity of 42.0 m/s due north. The police officer begins to pursue the speeder - first there is a 0.800 s reaction time when the officer has no change in speed, then the officer accelerates at 5.00 m/s2. Including the reaction time, how long does it take for the police car to reach the same position as the speeding car

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  1. 26 May, 21:10
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    11.1 s

    Explanation:

    Speed of the police car as given = v = 18 m/s

    Speed of the car = V = 42 m/s

    Reaction time = t = 0.8 s

    Distance traveled by the police car during the reaction time = d₁ = 0.8 x 18 = 14.4 m

    Distance traveled by speeding car = d₂ = 0.8 x 42 = 33.6 m

    Acceleration of the police car = a = 5 m/s/s

    The police car can catch the speeding car only if it travels a distance equal to the speeding car in a time t.

    Distance traveled by the police car = D = d₁ + v t + 0.5 at², according to the kinematic equation.

    ⇒ D = 14.4 + 18 t + 0.5 (5) t²

    ⇒ D = 14.4 + 18 t+2.5 t² → (1)

    For the speeding car, distance traveled is D = 33.6 + 42 t, since it is constant velocity. Substitute for D from the above equation (1).

    ⇒ 14.4 + 18 t+2.5 t² = 33.6 + 42 t

    ⇒ 2.5 t² - 24 t - 19.2 = 0

    ⇒ t = 10.3 s

    Total time = t + 0.8 s

    ⇒ Time taken for the police car to reach the speeding car = 10.3+0.8 = 11.1 s
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