Six particles positioned at the corners of a regular hexagon of...

The particles are located at the vertices of a regular hexagon. The velocity of each particle is directed towards the adjacent particle.

The angle between the line joining two adjacent particles and the velocity of the second particle is the exterior angle of the regular hexagon, which is $60^{{}^{{}^{\circ}}}$.

The relative velocity of approach between any two adjacent particles is:

$v_{app} = v - v \cos(60^{{}^{{}^{\circ}}}) = v - \dfrac{v}{2} = \dfrac{v}{2}$

The initial distance between the adjacent particles is $a$. The time taken for the particles to meet is the ratio of the initial distance to the relative velocity of approach:

$t = \dfrac{a}{v_{app}} = \dfrac{a}{\dfrac{v}{2}} = \dfrac{2a}{v}$

Alternatively, the distance of each particle from the center of the hexagon is $a$. The angle between the velocity vector and the position vector from the center is $60^{{}^{{}^{\circ}}}$.

The component of velocity directed towards the center is:

$v_{r} = v \cos(60^{{}^{{}^{\circ}}}) = \dfrac{v}{2}$

Since the particles meet at the center, the time taken is:

$t = \dfrac{a}{v_{r}} = \dfrac{2a}{v}$