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

Let at any instant the velocity od `B` makes an angle `prop` with that of x - axis and the time to collide is `T`.
`V_(a pp) = V - u cos prop`
`l = int_0^T V_(app) d t = int_0^T (V - u cos prop) dt` …(i)
Now equating the displacement of `A and B` along x direction, we get
`uT = int V cos prop dt` ....(ii)
Now from (i) and (ii), we get
`l = VT - int_0^T u cos prop dt`
=`VT -(u)/(V) int_0^T V cos prop dt = VT - (v)/(V). uT rArr T = T = (l V)/(v^2 - u^2)`
Now distance travelled by `A and B`
=`u xx (lV)/(v^2 - u^2) and v xx (lV)/(v^2 - u^2)`
=`(u V l)/(V^2 - u^2) and (V^2 l)/(V^2 - u^2)`.
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