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 se

Question

3 years ago

11

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 se

conds, 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?

Physics

1 answer:

mamaluj [8] · Answer 1 · 2022-03-24T02:20:56+03:00

1) The police officer catches the speeder after a time of 19.93 s (measured from the moment the speeder passes the officer)

2) The officer catches the speeder after a distance of 299 m

Explanation:

1)

The speeder is driving at constant speed, so we represent the position of the speeder at time t as

$x_s(t) = v_s t =15 t$

where

$v_s = 15 m/s$ is the speed of the speeder

t is the time measured from the moment the speeder passes the officer

The police car starts its motion after 9 seconds, driving from rest and accelerating at $5 m/s^2$ , so its position at time t can be written as

$x_p = \frac{1}{2}a (t-9)^2 = \frac{1}{2}(5)(t-9)^2$

where

$a=5 m/s^2$ is the acceleration of the car

The police officer catches the speeder when the two positions are equal, so:

$15t = \frac{5}{2}(t-9)^2$

And solving for t,

$6t = (t-9)^2 = t^2 -18t + 81\\t^2-24t+81=0$

which gives two solutions:

t = 4.06 s

t = 19.93 s

However, the time must be larger than 9 seconds (because we are measuring the time from the instant the speeder passes the police officer, and the officer starts its motion only 9 seconds later), so the correct solution is

t = 19.93 s

2)

To find how far the police officer went before the speeder was caught, we just substitutite t = 19.93 s into its equation of motion.

We find:

$x_p (19.93)=\frac{1}{2}(5)(19.93-9)^2 = 299 m$