three particles A,B and C are situated at the vertices of an equilateral triangle ABC of side d at 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 ?

if they are moving with same velocities then i think they will never meet 

They will meet after 3v/2 seconds .... where "v" is the constant speed of all the three!

the particle will meet at the centroid of the equilateral triangle whose distace fro any of the vertices is equal to d=a/(root 3).(a+ side of triangle. hence time=d/v.

velocity of A is v along AB 

the velocity of B is along BC . its component along BA is vcos60 = v/2. thus the separation AB decreases at the rate

v+v/2=3v/2

sice this rate is constt. the time taken in reducing the separation AB from d to 0 is 

t=d/(3v/2) =2d/3v

this question can be solved by two methods.

first note that the particles collide at the centroid of the triangle.

for this the particles have to cover a distance of d/√3.

and at any instant of time the speeds of the particles is same as v and they preserve the equilateral triangle shape till they collide.

the component of the speed of the particle along this distance is (√3/2)*v

so the time taken to collide is 2d/3v.

in the second method we look at one of the particles from another one. the relative speed is constant and equals √3v/2.

this is inclined at 30º to the line joining the particles always.

the component of this speed along this line is 3v/2.

in order to collide the distance covered by one particle relative to the other is d and this is along the line joining them.

so the time taken is 2d/3v.