Question

Six particles situated at the corner of a regular hexagon of side a move at a constant speed v. Each particle maintains a direction towards the particle at the next corner. Calculate the time the particles will take to meet each other.  

Answer 1

A regular hexagon has a side a. Six particles situated at the corners of the hexagon are moving with a constant speed v.
As per the question, each particle maintains a direction towards the particle at the next corner. So, particles will meet at centroid O of triangle PQR. Now, at any instant, the particles will form an equilateral triangle PQR with the same centroid O.
We know that P approaches Q, Q approaches R and so on.
Now, we will consider the motion of particle P. Its velocity makes an angle of 60˚.
This component is the rate of decrease of distance PO. 

Relative velocity between P and Q:

\[\vec{\text{ v } }_{\text{PQ }} = \vec{\text{v }}_P - \vec{\text{v }}_Q = \vec{\text{v}} - \vec{\text{v}} \cos 60^\circ\]

\[ = \vec{\text{v}} - \frac{\vec{\text{v}}}{2} = \frac{\vec{\text{v}}}{2}\]

\[\text{ Time } , t = \frac{\text{ Displacement }}{\text{ Velocity } }\]

\[ = \frac{a}{\text{ v }/2} = \frac{2\text{a} }{\text{v }}\]

Hence, the time taken by the particles to meet each other is \[\frac{2\text{a }}{\text{v } }\] .