Question

A speeding motorist traveling with velocity Vm is spotted by a police car. The police car is initially at rest, but the instant the motorist passes, accelerates with constant acceleration ap. Assume the police car maintains this acceleration and the motorist maintains velocity Vm. Find: a. The velocity of the police car when it catches up to the motorist. b. The time it takes for the police car to catch the motorist. c. How far does the police car travel to catch the motorist?

Answer 1

Step-by-step explanation:

Given

Let us suppose police car and motorist travel in straight line and police car catches motorist after s distance

Distance travel by motorist

`s=v_mt`----1

Distance traveled by Police car

`s=ut+(at^2)/(2)`

`s=0+(a_pt^2)/(2)`

`s=(a_pt^2)/(2)`----2

from 1 & 2 we get

`t=(2v_m)/(a_p)`

(a)Velocity of Police car after t sec

`v=u+a_pt`

`v=0+a_p* (2v_m)/(a_p)`

`v=2v_m`

(b)time taken by police car is

`t=(2v_m)/(a_p)`

(c)Distance travel by police car
`=(2v_m^2)/(a_p)`