Question

A police inspector in a jeep is chasing a thief on a straight road. The police jeep is travelling at its maximum speed $v$ (assumed uniform). If the thief hops onto the motorcycle of a waiting friend and the motorcycle starts to move with a constant acceleration $a$ and the police jeep is about distance $d$ away, the thief would be caught by the police if:

• A. $v\le \sqrt{2ad}$
• B. $v\ge 2\sqrt{ad}$
• C. $v\ge \sqrt{2ad}$
• D. $v\le 2ad$

Answer 1

The correct option is C $v\ge \sqrt{2ad}$

Suppose the thief is caught at a time $t$ after the motorcycle has started. Then the distance travelled by the motorcycle during this interval is:

$S=ut+\frac{1}{2}a{t}^2=\frac{1}{2}a{t}^2---\left(1\right)$ (the motorcycle started from rest and thus $u=0$)

Since the jeep is about distance $d$ away from the motorcycle, the police jeep needs to travel a distance $(S+d)$ in the same time $t$ in order to overtake the motorcycle and thus catch the thief.

The maximum speed of the jeep is $v⟹v=\frac{S+d}{t}$ or $S=vt-d---\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,
$vt-d=\frac{1}{2}a{t}^2$
$2vt-2d=a{t}^2$
$⟹a{t}^2-2vt+2d=0$
Solving the quadratic for $t$, we get
$t=\frac{2v\pm \sqrt{(2v{)}^2-4a(2d)}}{2a}$
$t=\frac{v\pm \sqrt{v^2-2ad}}{a}$

The thief could be caught only when $t$ is not imaginary, i.e., $t$ should be real and positive.

$⟹\sqrt{v^2-2ad}\ge 0$
$v^2-2ad\ge 0$ and $v^2\ge 2ad$
$⟹v\ge \sqrt{2ad}$

The police would be able to catch the thief only when the maximum speed of the jeep, $v\ge \sqrt{2ad}$.