# Four particles are situated at the corners of a square of side ‘a’ and move at a constant speed v.

Each particle maintains a direction towards the particle at the next corner. Time when particles will meet each other will be $$\frac{2a}{\sqrt n v}$$. Find the value of n

MY SOLUTION

The particles at adjacent corners will move with speed Vi and Vj. Therefore, relative velocity of first particle wrt second will be $$V_r=\sqrt{V^2+V^2}$$ $$=\sqrt 2 V$$ Distance to be covered is ‘a’

Time=$$\frac{a}{\sqrt 2 V}$$

Equating this with the original given expression $$\frac{a}{\sqrt 2 v}=\frac{2a}{\sqrt n V}$$$$n=8$$ Answer given is 4, what have I don’t wrong?

• The relative velocity of one wrt the other will not in thee direction of the line joining them. You need to take the velocity along the line joining them. Commented Mar 24, 2021 at 13:56
• Does this answer your question? Case of the 'mice problem' for $n=3$ Commented Jul 5, 2021 at 15:04

Consider what has happened after a small interval of time, $$\delta t$$. The first particle will have travelled a distance of $$v \delta t$$ towards the second particle whereas the second particle has travelled a distance of $$v \delta t$$ perpendicular to the line joining them.
The distance between them is the hypotenuse of a right-angled triangle with the other sides having lengths $$a-v \delta t$$ and $$v \delta t$$. Ignoring distances as small as $$(\delta t)^2$$, the distance between the two particles is just $$a-v \delta t$$.
Therefore the distance between two particles will have decreased by $$v \delta t$$. This will be the same throughout the motion and so the particles will meet when $$a=vt$$, i.e. after time $$\frac {a}{v}$$.
In your solution, the relative velocity should just be $$v$$ along the line joining the particles since the second particle is moving at right angles to the first at every instant.