Question

A police inspector in a jeep chasing a pick pocket on a straight road. The jeep going at its maximum speed v (assumed uniform). The pick pocket rides on a motor-cycle of a waiting friend when the jeep is at a distance d away and the motorcycle starts with a constant acceleration a. The pick pocket will be caught if
A. $v \geqslant \sqrt {2ad}$
B. ${v^2} \geqslant \sqrt {2ad}$
C. $v \geqslant \sqrt {3ad}$
D. $v \geqslant \sqrt {2a{d^2}}$

Answer 1

The question talks about the application of newton’s third law of rectilinear motion which is represented by ${v^2} = {u^2} + 2ad$, where a is acceleration, d is distance, v is the final velocity and u is the initial velocity. Read the question carefully and find the value of initial velocity.

Complete step by step answer:
We are given a police inspector in a jeep chasing a pick pocket on a straight road. The jeep going at its maximum speed v and the pick pocket rides on a motor-cycle when the jeep is at a distance away and the acceleration of motorcycle is a.
We have to find the velocity of the jeep required to catch the pickpocket.
Since the jeep is moving and not from rest, it is then moving with a uniform velocity, where the initial velocity is u but attain a final or maximum velocity of v. and also the motorcycle is having at rest and u is equal to zero.
Substituting the value of u in the equation ${v^2} = {u^2} + 2ad$, we get
${v^2} = {0^2} + 2ad$
The equation then becomes ${v^2} = 2ad$
Since the motion is at its peak, the equality sign (=) now changes to greater than or equals to sign ($\geqslant$).
Then the equation then becomes
${v^2} \geqslant 2ad \\\ \implies v \geqslant \sqrt {2ad} \\\$.

So, the correct answer is “Option A”.

Note:
We can also see this in a car or train, where initially it was at rest or stationary position, but immediately it is ignited , it then moves with a speed which might initially be uniform but as time goes on the speed increases.