Question

A speeder is driving down the road at a constant 15 m/s, passes a police officer parked on the roadside. The officer pauses 9 seconds, then pursues the speeder, accelerating at a constant 5 m/s^2.

How much time does it take the police officer to catch the speeder?
How far did the police officer drive before the speeder was caught?

Answer 1

`3+3√(7)` seconds

299.0588 meters

Step-by-step explanation:

Given

• speeder has a constant speed,x=15
• officer starts 9 seconds after speeder crosses him and accelerates at 5 from rest

Let assume S as the distance covered by police before he catches him

let T be the time taken by him to do so

`(distance=\text{initial velocity}* time+(1)/(2) * acceleration* time)`

Therefore S
`=(1)/(2) aT^(2) =(5)/(2) T^(2)`(since initial velocity=0)

This same distance is covered by the speeder in time T+9 as officer starts after pausing 9 seconds

Therefore S
`= (T+9)* 15=15T+ 135`

equating both the equations

`(5)/(2)}T^(2)=15T+ 135\\5T^(2)=30T+270\\T^(2)=6T+54\\T^(2)-6T-54=0`

Solving the quadratic we get

T=3+3
`√(7)` or T=3-3
`√(7)`(not possible as T cannot be less than 0)

So it takes 3+3
`√(7)` seconds for the officer to catch the speeder

⇒Distance covered
`=(5)/(2) T^(2)`=299.0588 m