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

Answer:

$3+3\sqrt{7}$ seconds

299.0588 meters

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

$(distance=\text{initial velocity}\times time+\frac{1}{2} \times acceleration\times time)$

Therefore S$=\frac{1}{2} aT^{2} =\frac{5}{2} T^{2}$(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$= (T+9)\times 15=15T+ 135$

equating both the equations

$\frac{5}{2}}T^{2}=15T+ 135\\5T^{2}=30T+270\\T^{2}=6T+54\\T^{2}-6T-54=0$

Solving the quadratic we get

T=3+3$\sqrt{7}$ or T=3-3$\sqrt{7}$(not possible as T cannot be less than 0)

So it takes 3+3$\sqrt{7}$ seconds for the officer to catch the speeder

⇒Distance covered $=\frac{5}{2} T^{2}$=299.0588 m