Question

9. A speeder traveling at 42 m/s passes a motorcycle policeman at rest at the side of the road. The policeman accelerates at 3.23 m/s2. To the nearest tenth of a second how long does it take the policeman to catch the speeder? 10. How far to the nearest tenth of a meter can a runner running at 12 m/s run in the time it takes a rock to fall from rest 56 meters?

Answer 1

9)

The speeder is moving at a constant velocity of 42m/s without acceleration.

The policeman starts driving from rest with a uniform acceleration of 3.23m/s^2.

Let t be the time at which the police catches the speeder.

The distance covered by both of them in time t should be equal, so that the police can catch the speeder.

Using eqn of motion on the speeder,

$s= ut +\frac{gt^2}{2}$

$s= 42t +\frac{0t^2}{2}$ since the acceleration is 0.

For the policeman

$s= 0t +\frac{3.23t^2}{2}$

By equation the above 2 eqns​​​​, we get

$42t =\frac{3.23t^2}{2}$

$t=\sqrt{\frac{42\times 2}{3.23}}$

$t=5.099seconds$

10)

Time taken by the rock to fall 56meters, is obtained from the eqn of motion

$s=ut + \frac{gt^2}{2}$

$56=0t + \frac{9.81t^2}{2}$

$t =\sqrt{ \frac{56\times 2}{9.81}}$

$t =3.3788seconds$

The distance covered by a runner running at 12m/s within this time of 3.3788seconds, will be

$d= v\times t$

$d= 12\times 3.3788$

$d= 40.54meters$