four particles a,b,c,d.are situated at four corners of square abcd of side a at t=0 .each of the particles move with constant speed v .a always has its velocity along ab,b along bc,c along cd,and d along da .at what time will these particles meet each other?

If the question is as u hav given, then the reply by sriram.a is correct.

Or else the question should be  "................... a always has its velocity along b,b along c, c along d,and d along a ................. "

If it is like the one i have given the take the solution ... 

(it is a problm on relative velocity)

So there is a gr8 method to solve this type of problems.

Just consder two particles. Say A and B.

Since the arrangement given is a square, there is no component of velocity of B towards A.

So along the side containing A and B, relative velocity of A with respect to B is v - 0. Since velocity of B along that line is 0.

So time taken for meeting of A and B = that for all bodies.

So reqd. time  = (dist. b/w A and B) / (rel. velocity of A wrt B) = a/v.

In other cases of geometrical figures given, there will be an angle b/w the velocity of B and A. let that angle be @. So the component of B along the line containing A and B = v.cos@.

So rel. vel. of A wrt B = v - ( - vcos@ ) = v + vcos@

So time = ( v + vcos@ ) / a                             a = length of side...

And the point of meeting is by symmetry at the centre of the fig. given.

NOTE THAT THIS METHOD IS USEFUL FOR SYMMETRICAL FIGURES ONLY.... WHERE ALL SIDE LENGTHS AND ANGLES BETWEEN THE SIDES ARE EQUAL..