Question

A speeder 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?

Answer 1

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

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

Step-by-step 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
`2.17m/sec^2`

So acceleration a =
`2.17m/sc^2`

From first equation of motion v = u+at

So
`28.1=0+2.17* t`

t = 12.94 sec

So after 12.94 sec speeder will be overtaken by police car

(b) From second equation of motion
`s=ut+(1)/(2)at^2`

`s=0* 12.94+(1)/(2)* 2.17* 12.94^2=181.9377m`

So speeder will travel 181.9377 m before coming to rest