Can you answer this question about acceleration in physics?

Question

A cop is stopped on the side of a freeway when a lady in a car passes him while going 90km/hr. He waits for two seconds than starts to pursue her. What must his average acceleration be if he is to catch her exactly as she reaches the state line 2 km away? Please help with this and tell me how you did it if you can.

Answer 1

90 km/hr = 90,000 / 3600 = 25 m/s.

110 km/hr = 30.556 m/s.

The lady reaches the border

2 * 3600 / 90 = 80 sec.

after passing the cop.

The cop has to cover 2000 metres in 78 sec.

Let:

a m/s^2 be the cop's acceleration,

t sec. be the time he takes to reach top speed,

s metres be the distance he covers before reaching top speed.

The distance covered at top speed is 2000 - s metres, covered in time 78 - t sec.

30.556 = at ...(1)

s = 30.556 t / 2 ...(2)

30.556(78 - t) = 2000  - s ...(3)

Adding (2) and (3) eliminates s:

2000 - 30.556 t / 2 = 30.556(78 - t)

2000 - 15.278t = 2383.368 - 30.556t

383.368 = 15.278t

t = 25.093 sec.

From (1):

a = 30.556 / 25.093

= 1.22 m/s^2.