Question

A police party is moving in a jeep at a constant speed v. They saw a thief at a distance x on a motorcycle which is at rest. The moment the police saw the thief, the thief started at constant acceleration a. Which of the following relations is true if the police is able to catch the thief?
  $A.\;\;{v^2} < ax$
  $B.\;\;{v^2} < 2ax$
  $C.\;\;{v^2} \geqslant 2ax$
  $D.\;\;{v^2} = ax$

Answer 1

- Hint: We may change from the outside observer perspective to the perspective of the police van. Here, the thief seems to be approaching us with an initial speed, just like a stone thrown upwards.

Complete step-by-step solution -
Let us say the police succeeded in catching the thief. This means that the police and the thief reached the same spot at the same time somewhere during the course of the chase. We are given that the police vehicle is moving with a speed $v$ and at a Distance $x$ away from the thief.
Solving this question from an outside frame of reference requires a lot of calculations. So let us go to the perspective of police to get an easier and intuitive understanding.

fig: motion of the two bodies from the police’ perspective
If we were inside the police vehicle, moving with a constant speed $v$, we would be seeing the thief coming towards us with a speed $v$ and when the distance between the two is $x$, the thief starts accelerating away from us.
We can visualize this as a thief having an initial velocity of $ - v$ and accelerating away with acceleration $ + a$.

So if the police are to catch the thief, both should cross each other. That means the distance between them should become at least zero.
The minimum speed required would be the speed for which the thief just reaches the police vehicle. That is the final velocity after traveling $x$ distance is 0. We may compare this with the minimum speed a stone should have to cross a height h.
We know the relation between the final velocity $v$, the initial velocity $u$ and the distance travelled $s$:
${v^2} - {u^2} = 2a\;s$
Here, for the minimum initial velocity v, the final velocity would be 0. Also, since the acceleration is in the opposite direction as that of velocity, a will numerically be negative.
$0 - {v^2} = - 2ax$ or $v = \sqrt {2\;a\;x} $

So, to catch the thief, the speed of the police vehicle should be greater than or equal to this minimum speed.
So the correct option is C ${v^2} \geqslant 2ax$

Note : We should remember that for speeds greater than $\sqrt {2\;a\;x} $ . The police would have to slow down after overtaking the thief. Though this is practically obvious, In questions we have to consider that. If an option ${v^2} = 2a\;x$ was present, that would be more apt because only then, the police and thief would be at the same place at the same time with the same speed.