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 (Fig. 5.132). By symmetery they will meet at the centroid `O` of the triangle.
As the speed of all the particles is equal they will cover equal distance in any given interval of time.
If we join the instantaneous position of the particle at any time, the particles will from an equilateral triangle of same centroid as initial triangle.
Let us consider the motion of any one particle say `A`. At any instant, its velocity makes an angle `30^@` with line `A O`.
The component of the velocity along `A O` is `V cos 30^@`. This component will be equal to the rate of change of distance between `A and O`.
At `t = 0`, distance between `A and O`,
`A O = (2)/(3) A D = (2)/(3) sqrt( l^2 - ((l)/(2))^2) = (l) /(sqrt(3))`
At time `t = T`, the separation between `A and O` is zero.
Hence, time taken for `A O` to become zero,
`T = ((l//sqrt(3)))/(v cos 30^@) = ((l //sqrt(3)))/(v(sqrt(3 //2))) = (2 d)/(3 v)`
After time `t`, let the distance of separation between the insect be `r`. The velocity of approach (component of relative velocity `v_(rel)` between the insects along the line of their separation) is `= v + v cos 60^@ = 3 v//3`
Since `(v_(rel))_II dt = - dr.` substituting `(v_(rel))_II = (3 v)/(2)`, we obtain
`(3 v)/(2) dt = -dr`
As at time of meeting, integrating both sides, we have
`(3 v)/(2) int_0^(tau) dt = - int _l^0 dr`
This gives `t = (2 l)/(3 v)`.
, .