Six particles situated at the corners 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.

The correct option is B

The particles will meet at the point o the center of hexagon.At any instance the particles will form a

hexagon with the same center O.

Let's focus on motion of A

At any instance velocity of A is making an angle ${60}^{\circ }$ from the line AO

The component of velocity of A along AO will be $vcos{60}^{\circ }=\frac{v}{2}$

This component is the rate of decrease of the distance AO

AO = a

$\Rightarrow$ time taken for AO to become 0 t = $\frac{a}{\frac{v}{2}}=\frac{2a}{v}$

Alternate solution:

Velocity of A along AB is v

Velocity of B along AB is -v $cos{60}^{\circ }$

= $-\frac{v}{2}$

So the rate at which the distance AB decrease is $v-\frac{v}{2}=\frac{v}{2}$

$\Rightarrow t=\frac{a}{\frac{v}{2}}$

=$\frac{2a}{v}$