A sports car is passing a police check post at ` 60 km h^(-1)` , immediately started slowing down uniformly until its speed was ` 40 km h^(-1)`. It continued to move at the same speed until it was passed by a police car ` 1 km` from the check post. The police car had started from rest at the check post at the same instant as the sports car had passed the check post. The police car had moved with a constant acceleraation until it had passed sports car. Assuming that the time taken by the sports car in slowing down from `60 km h^(-1)` to ` 40 km h^(-1)` was equal to the time that it travelled at constant speed before passed by the police car, find
(a) the time taken by the police car to reach the sports car
(b) the speed of the police car at the instant when it passed the sports car
(c ) the time measured from the check post when the speeds of the two cars were equal.

Explanation

To solve the problem step by step, we will break it down into parts as per the questions asked.

### Given Data:
- Initial speed of the sports car, \( U_S = 60 \, \text{km/h} = \frac{60 \times 1000}{3600} = 16.67 \, \text{m/s} \)
- Final speed of the sports car, \( V_S = 40 \, \text{km/h} = \frac{40 \times 1000}{3600} = 11.11 \, \text{m/s} \)
- Distance from the check post to where the police car passes the sports car, \( S = 1 \, \text{km} = 1000 \, \text{m} \)

### Part (a): Time taken by the police car to reach the sports car
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