
A car is passing a parked police car at 160 km/h. after 2 seconds the police start chasing the passing car at an acceleration of 3 m/s^{2}, the maximum speed of the police car is 220 km/h. Find the distance that the police will catch the escaping car.

The distances traveled by both cars are the same and the time of escaping car is t + 2 sec.
After 2 seconds the police car start chasing at an acceleration of 3 m/s^{2}.
We have to verify if the police car catches the escaping car at the acceleration phase or at the top speed velocity of 220 km/h.
This will be done by assuming that the police car accelerates until it caches the escaping car, and compaing the distances traveled by both cars which should be the same.
For the police car: 
d = v_{0}t + a t^{2} / 2 
For the escaping car: 
d = v_{e} (t + 2) 

Comparing d of both cars: 
v_{e} (t + 2) = v_{0} t + a t^{2} / 2 
⟶ 
a t^{2} − 2 v_{e} t − 4 v_{e} = 0 
By solving the quadratic equation for the time we get t = 31.5 s and the police car reaches a velocity of:
v_{t} = v_{0} + a t = 0 + 3 * 31.5 * 3.6 = 340.2 km/h this velocity is more than the upper police car speed limit, so we can say that the police car accelerates untill it reach the top speed of 220 km/h and after that it continues chassing by a constant speed of 220 km/h.
The time of the police car to reach top speed of 220 km/h from eq. (2b) is:
t = v_{t} / a = 220 / (3 * 3.6) = 20.37 s
The distance traveled by the police car during 20.37 s of acceleration according to eq. (2a) is:
d = v_{0} t + a t^{2} / 2 = 0 + 3 * 20.37^{2} / 2 = 622.4 m
The distance traveled at the same time of 20.37 + 2 seconds by the escaping car is:
d = v t = 160 * (20.37 + 2) / 3.6 = 994.2 m

If we denote the additional time required to catch the escaping car as t_{a} then we can calculate the additional time by comparing the total traveled distances
Distance traveled by police car: 
d_{p} = 622.4 + (220 / 3.6) t_{a} 
Distance traveled by escaping car: 
d_{e} = 994.2 + (160 / 3.6) t_{a} 
Solving for t_{a} by setting d_{p} = d_{e} we get: t_{a} = 22.3 s
Escaping car total travel time is: 
T_{e} = 2 + 20.37 + 22.3 = 44.67 s 
Total chasing distance of the police car is:
D = v_{e} T_{e} = 160 * 44.67 / 3.6 = 1985.3 m


Schematic plot
Red curve is the acceleration of the police car until it reach top speed, after the max velocity reached the chasing is at the maximum police car speed the blue line.
Blue line police car chasing at top speed.
Green line Escaping car constant speed during the complete chasing.

Figure  1  

If the police car top speed is high enough it can catch the escaping car during the acceleration time, this case is simpler and is:
d_{e} = 160 * (2 + t_{a}) / 3.6 
d_{p} = a t_{a}^{2} / 2 
d_{e} = d_{p} 
1.5 t_{a}^{2} − 44.4 t_{a} − 88.9 = 0 
The result is: t_{a} = 31.5 s
And the catch time D is:
D = 140 (2 + 31.5) / 3.6 = 1302.8 m
Police top speed should be more than:
v_{max} = a t_{a} = 3 * 31.5 
= 94.5 m/s 

= 340.2 km/h 

Figure  2  
