Aspeeder passes a parked police car at a constant speed of 28.1 m/s. at that instant, the police car starts from rest with a uniform acceleration of 2.17 m/s 2 . how much time passes before the speeder is overtaken by the police car? how far does the speeder travel before being overtaken by the police car?

(A) after 12.94 sec speeder will be overtaken by police car

(B) speeder will travel 181.9377 m before coming to rest

Explanation:

We have given that speeder passes a parked police car with a constant speed of 28.1 m/sec

As the police car starts from rest so its initial velocity u = 0 m/sec

It is given that police car is travelling with a constant acceleration of

So acceleration a =

From first equation of motion v = u+at

So

t = 12.94 sec

So after 12.94 sec speeder will be overtaken by police car

(b) From second equation of motion

So speeder will travel 181.9377 m before coming to rest

cant without the picture! ?

what is the actual question though?