Question

A speeding car is traveling along a straight road with a constant speed of 100 km/hour. It passes a police car. The officer starts 5.00 s after the speeder passes. The police car accelerates uniformly (i.e., at a constant acceleration) for 10.0 s, at the end of which its speed is 120 km/hour. After that, the police car moves with the constant speed of 120 km/hour till it reaches the speeding car.

How far ahead of the police car is the speeding car when the police car stops accelerating? How long after it stops accelerating does the police car reach the speeding car?

Answer 1

When police car stops accelerating :

when the police car stops accelerating, the speeding car has traveled for 15 sec

v_s = speed of speeding car = 100 km/h

t_s = time of travel speeding car = 15 sec = 15/(3600) h

d_s = distance traveled by speeding car = v_s t_s = 100 (15/3600) = 0.4167 km = 416.7 m

when the police car stops accelerating, the police car has traveled 10 sec

v_o = initial speed of the police car = 0 km/hr

v = final speed of police car = 120 km/h

t = time = 10 sec = 10/3600 h

distance traveled by police car is given as

d_p = (v_o + v) t/2

d_p = (0 + 120) (10/3600) (1/2)

d_p = 0.1667 km

d_p = 166.7 m

d = distance of the speeding car ahead of police car = d_s - d_p = 416.7 - 166.7 = 250 m

v = final speed of police car = 120 km/h

speed of police car relative to speeding car

v_r = v - v_s = 120 - 100 = 20 km/h = 5.56 m/s

t' = time taken for the police car to reach speeding car after accelerating

time taken for the police car to reach speeding car after accelerating is given as

t = d/v_r = 250/5.56 = 45 sec