Aspeeder passes a parked police car at a constant speed of 25.1 m/s. at that instant, the police car starts from rest with a uniform acceleration of 2.56 m/s 2 . how much time passes before the speeder is overtaken by the police call?

19.61 second

Explanation:

speed of speeder car, u = 25.1 m/s

initial speed of police car, u = 0

acceleration of police car, a = 2.56 m/s^2

let it take time t to overtake the speeder car.

For overtaking, the distance traveled by the speeder car in time t is same as the distance traveled by the police car.

Distance traveled by the speeder car, d = speed x time

d = 25.1 x t ..... (1)

Distance traveled by police car is given by second equation of motion

d = 1.28t^2     .... (2)

From equation (1) and (2) we get

25.1 t = 1.28 t^2

25.1 = 1.28 t

t = 19.61 second

you get 2.3=1.21*\frac{a}2 or 4.6=1.21a or a=3.802 (b)

explanation: