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 motion of the particles is roughly sketched in figure.

By symmetry they will meet at the centroid O of the triangle. At any instant the particles will form an equilateral triangle ABC with the same centroid O. Concentrate on the motion of any one particle say A. At any instant its velocity makes angle `30^0` with AO.
The component of this velocity along AO is `vcos 30^@`. This component is the rate of decrease of the distance AO.
Initially,
`AO=2/3 sqrt(d^2-(d/2)^2)=d/sqrt3`
There the time taken for AO to become zero
`=(dsqrt(3))/(v cos 30^0)=(2d)/(sqrt(3 vxxsqrt3)=(2d)/(3v)`
Alternatively : Velocity of A is v along AB. The velocity of B is along BC. Its component along BA is`v cos 60^0 =v/2`. Thus, the separation AB decreases at the rate
`v+v/2=(3v)/2`
Since this rate is constant, the time taken in reducing the separation AB from d to zero is
t = `(d)xx(2)/(3v) = (2d)/(3v)`.