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

Answer :

skyluke89

Answer:

24 seconds

Explanation:

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

[tex]x_s(t)=v_s t[/tex]

where

[tex]v_s=50 m/s[/tex] is the velocity of the Speeder

t is the time in seconds

Here [tex]x_s(t)[/tex] is measured relative the the position at which the Speeder overcomes the trooper

The position of the trooper instead is given by

[tex]x_t(t)=v_t t + \frac{1}{2}a_t t^2[/tex]

where:

[tex]v_t=20 m/s[/tex] is the initial velocity of the trooper

[tex]a_t=2.5 m/s^2[/tex] is its acceleration

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

[tex]x_s(t)=x_t(t)\\v_s t = v_t t + \frac{1}{2}a_t t^2[/tex]

Substituting the values, we get:

[tex]50 t = 20 t + \frac{1}{2}(2.5)t^2\\30t-1.25t^2=0\\t(30-1.25t)=0[/tex]

which has two solutions:

t = 0 (initial instant)

[tex]t=\frac{30}{1.25}=24 s[/tex]

So, the trooper reaches the Speeder after 24 seconds.

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