Question

A police inspector in a jeep is chasing a pickpocket on a straight road. The jeep is going at a maximum speed $v$ (assume uniform). The pickpocket rides on the motorcycle of a waiting friend when the jeep is at a distance $d$ away, and the motorcycle starts with a constant acceleration $a$. What should be the speed of the jeep so that the pickpocket will be caught?

• A. $v\ge \sqrt{\frac{3}{2}ad}$
• B. $v\ge \sqrt{2ad}$
• C. $v\ge \sqrt{6ad}$
• D. $v\ge 2ad$

Answer 1

Answer: (B)

$v\ge \sqrt{2ad}$

Consider the motion with respect to the jeep. The jeep moves with initial speed $v$ towards direction of motorcycle.

Now an deceleration of $a$ occurs to the jeep.

Under this deceleration too, the jeep must be able to reach a distance $d$ in direction of motorcyclist.

Therefore distance traveled by jeep=$\frac{v^2}{2a}\ge d$

$⟹v\ge \sqrt{2ad}$