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
${ \frac{m}{s} }^{2}$
Assuming that the Speeder continues at 50 m/s, how long will it take the trooper to catch up to the speeder?​

Answer 1

Answer:

24 seconds

Explanation:

The speeder moves at constant velocity, so we can write the position of the speed at time t as:

$x_s(t)=v_s t$

where

$v_s=50 m/s$ is the velocity of the Speeder

t is the time in seconds

Here $x_s(t)$ is measured relative the the position at which the Speeder overcomes the trooper

The position of the trooper instead is given by

$x_t(t)=v_t t + \frac{1}{2}a_t t^2$

where:

$v_t=20 m/s$ is the initial velocity of the trooper

$a_t=2.5 m/s^2$ is its acceleration

The trooper reaches the speeder when their positions are the same:

$x_s(t)=x_t(t)\\v_s t = v_t t + \frac{1}{2}a_t t^2$

Substituting the values, we get:

$50 t = 20 t + \frac{1}{2}(2.5)t^2\\30t-1.25t^2=0\\t(30-1.25t)=0$

which has two solutions:

t = 0 (initial instant)

$t=\frac{30}{1.25}=24 s$

So, the trooper reaches the Speeder after 24 seconds.