Question

A state trooper is traveling down the interstate at 20 m/s. He sees a speeder traveling at 50 m/s approaching from behind. At the moment the speeder passes the trooper, the trooper hits the gas and gives chase at a constant acceleration of 2.5 m/s^2.
Required:
a. Assuming that the speeder continues at 60 m/s , how long will it take the trooper to catch up to the speeder?
b. How far down the highway will the trooper travel before catching up to the speeder?

Answer 1

Answer:

Explanation:

From the given information;

Let assume that the distance travelled by the speeder prior to the time the trooper catches with it to be = d

the time interval to be = t

Then, the speeder speed = distance/time

Making distance the subject; then:

distance (d) = speed × time

d = (50 m/s)t

d = 50 t --- (1)

Now, for the trooper; Using the equation of motion:

$d = ut + \dfrac{1}{2}at^2$

$d = (20)t+\dfrac{1}{2}(2.5)t^2$

d = 20t + 1.25t²

Replace the value of d in (1) to the above equation, we have:

50 t = 20 t + 1.25t²

50t - 20t = 1.25t²

30t = 1.25t²

30 = 1.25t

$t = \dfrac{30}{1.25}$

t = 24 seconds

From (1), the distance far down the highway the trooper will travel prior to the time it catches up with the speeder is:

= 50t

= 50(24)

= 1200 seconds