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?

Answer

`t=2a//3v`

Explanation

From the symmetry of the problem all the three points are always located at the vertices of equilateral triangle of varying side length and finally meet at the centroid of the initial equilateral triangle whose side length is a, in the sought time interval (say t).


Let us consider an arbitrary equilateral triangle of edge length `l` (say).
Then the rate by which 1 approaches 2, 2 approches 3, and 3 approches 1, becomes:
`(-dl)/(dt)=v-vcos((2pi)/(3))`
On integrating: `-undersetaoverset0intdl=(3v)/(2)underset0oversettintdt`
`a=3/2vt` so `t=(2a)/(3v)`