A police car is traveling at a velocity of 17.0 m/s due north, when a car zooms by at a constant velocity of 40.0 m/s due north. After a reaction time 0.300 s the policeman begins to pursue the speeder with an acceleration of 4.00 m/s^2. Including the reaction time, how long does it take for the police car to catch up with the speeder

8.13 seconds

Explanation:

Given that a police car is traveling at a velocity of 17.0 m/s due north, when a car zooms by at a constant velocity of 40.0 m/s due north. After a reaction time 0.300 s the policeman begins to pursue the speeder with an acceleration of 4.00 m/s^2. Including the reaction time, how long does it take for the police car to catch up with the speeder

First calculate the acceleration of the speeder

V = U - at

0 = 40 - a × 0.3

0.3a = 40

a = 40/0.3

a = 133.3 m/s^2

The speeder distance covered will be

V^2 = U^2 - 2as

0 = 40^2 - 2 × 133.3 × S

266.67S = 1600

S = 1600/266.7

S = 6 m

Let's calculate the time to reach the distance by using second equation of motion

S = Ut - 1/2at^2

6 = 17t - 0.5 × 4 t^2

6 = 17t - 2t^2

2t^2 - 17t + 6 = 0

t^2 - 17/2t = - 3

(t - 17/4)^2 = 18.06 - 3

(t - 17/4)^2 = 15.06

t - 17/4 = 3.88

t = 17/4 + 3.88

t = 8.13 seconds

Therefore, it take 8.13 seconds for the police car to catch up with the speeder

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