Three particles A, B and C are situated at the vertices of an equilateral triangle ABC of side d at time `t=0.` Each of the particles moves with constant speed v. A always has its velocity along AB, B along BC and C along CA. At what time will the particles meet each other?

Explanation

The particles will meet at the centre of triangle.
`(v_(A//B))_(along AB)=v+vcos 60^(@)=(3v)/2`

Time after which the separation between `A` and `B` becomes zero.
`t=d/((3v)/2)=(2d)/(3v)`
Time after which the three particles meet `=(2d)/(3v)`