Question

A police officer at rest at the side of the highway notices a speeder moving at 62 km/h along a straight level road near an elementary school. When the speeder passes. the officer accelerates at 3.0 m/s^2 in pursuit. The speeder does not notice until the police officer catches up.

(a) How long will it take for the officer to catch the speeder?

Answer 1

62 km/h * 1000 m/km * 1 h/3600 s = 17.22 m/s 
then 
a 
the d for both is equal so 
V t = Vit + 1/2 a t^2 
17.22 t = 0 + 1/2 (3.0) t^3 
1.5 t = 17.22 
t = 11.5 s 
b 
d = 17.22 * 11.5 = 198 m 
or 
d = 0 + 1/2 (3.0) (11.5^2) = 198 m 
c 
Vf = a t = 3.0 * 11.5 = 34.5 m/s or 124 km/h 
Hope it helps....

Answer 2

Given:

A police officer at rest at the side of the highway notices a speeder moving at 62 km/h along a straight level road near an elementary school.

When the speeder passes. the officer accelerates at 3.0 m/s² in pursuit.

To Find:

The time taken by the officer to catch the speeder

Solution:

The time taken by the officer to catch the speeder is 11.48 seconds.

When the officer catches the speeder, the distance traveled by the speeder and the officer must be the same.

Let this distance be s and the time taken by the officer to catch the speeder be t.

For the speeder:

s =  62 km/h X t

or s = 17.22 m/s X t    - (1)

For the officer,

Using the second equation of motion,

s = ut + 1/2 at²

Since the officer starts from rest, t = 0 m/s

or s = 1/2 X 3 X t²   - (2)

Equating (1) and (2) we get:

17.22t = 1.5t²

or t = 17.22 / 1.5

= 11.48 s