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

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