Question

A car traveling a constant speed of 24 m/s passes a trooper hidden behind a billboard. One second after the speeding car passes the billboard, the trooper sets off in chase with a constant acceleration of 3.00 m/s^s. (a) How long does it take the trooper to overtake the speeding car? (b) How fast is the trooper going at that time?

Answer 1

(a) The trooper overtakes the speeding car in 16.94 seconds

(b) The trooper is going with a velocity of 50.83 m/s

Explanation:

Speed of the car = 24 m/s

acceleration of the trooper, a = 3 m/s²

Let the trooper catches the car in t seconds

Distance travelled by the trooper in t seconds

Using the second equation of motion

$s=ut+\frac{1}{2}at^2[/tex[</p><p>[tex]s=0\times t+\frac{1}{2}\times 3\times t^2$

$\implies s=\frac{3t^2}{2}$

Car will travel the same distance in (t+1) seconds

Thus,

$s=24\times (t+1)$

or, $\frac{3t^2}{2}=24(t+1)$

or, $3t^2=48(t+1)$

or, $3t^2-48t-48=0$

or, $t=\frac{-(-48)\pm\sqrt{48^2-4\times 3\times (-48)}}{2\times 3}$

or, $t=\frac{48\pm\sqrt{2880}}{6}$

or, $t=\frac{48\pm53.67}{6}$

or, $t=16.94$ seconds (since the other value is negative)

Therefore, the trooper overtakes the car in about 17 seconds

velocity of the trooper at that time:

Using the first equation of motion

$v=u+at$

$v=0+3\times 16.94$

$\implies v=50.83$ m/s

Hope this answer is helpful.

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