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

Answers:

a) 19.937 s

b) 299 m

Explanation:

a) Firstly, we have to establish the equations of motion for both the speeder and the police officer:

For the speeder, its velocity $V_{s}$ is:

$V_{s}=\frac{X_{s}}{t}$

So, its position $X_{s}$ is:

$X_{s}=V_{s} t$ (1)

Where $V_{s}=15 m/s$ and $t$ the time

For the police officer, its position $X_{off}$ is:

$X_{off}=\frac{1}{2}a_{off}(t-9s)^{2}$ (2) Because the officer began to drive after 9 seconds

Where $a_{off}=5 m/s^{2}$

Now, when the position of both the officer and the speeder is the same ($X_{s}=X_{off}$) is when the officer catches the speeder, hence:

$V_{s} t=\frac{1}{2}a_{off}(t-9s)^{2}$ (3)

$15 m/s t=\frac{1}{2}5m/s^{2}(t-9s)^{2}$ (4)

$5m/s^{2} t^{2}-120 m/s t +405 m=0$ (5)

Applying common factor 5:

$5(1m/s^{2} t^{2}-24 m/s t +81 m)=0$ (6)

This can be rewritten as:

$t^{2}-24 t +81=0$ (7)

Solving for $t$ we have two results:

$t_{1}=19.937 s$

$t_{2}=4.063 s$

We choose $t_{1}=19.937 s$, remembering the police officer began to drive $9 s$ after the other car passed by its side. So, the time is 19.937 s.

b) Now, if we want to know the distane the police officer drove before catching the speeder, we have to input the calculated time in the officer's equation (2):

$X_{off}=\frac{1}{2}5m/s^{2}(19.937 s-9s)^{2}$ (8)

$X_{off}=299.04 m \approx 299 m$ (9)

If we want to prove this is the same distance the other car drove, we have to input also this time (1):

$X_{s}=(15 m/s)(19.937 s)$ (10)

$X_{s}=299.05 m \approx 299 m$ (11)