Six particles situated at the corners of a regular hexagon of side `a` move at a constant speed `v`. Each particle maintains a direction towards the particle at the next corner. Calculate the time the particles will take to meet each other ?

							total time  = a/v + 2a/v + 3a/v + 4a/v + 5a/v
						

							
@ vikas Kumar Yadav... you are not be an IITan... a never mind...
 
now, 
Initial separation between two particles=Side of hexagon=a
Final separation=0
Therefore,
Relative displacement between two particles=a
Particle B has a component v cos 600 along particle BC (a side)
Therefore, relative velocity with which B and C approaches each other=v- vcos60
=v/2
Since, v is constant, thus time taken by these two balls to meet each other is given by
=(Relative displacement) / Relative velocity
=a/(v/2)=2a/v
So, the time taken by the particles to meet each other=2a/v
 
 
						

							
Relative velocity of A wrt B = V - Vsin30 = V/2
Time = distance / speed Time = 2a/V
In reality each particle will follow a curved path and eventually meet at the center of the hexagon.
						

							How to Sketch figure of this problem.Please clear its figure.How to think about this type of problems.
						

							
 along particle BC (a side)Therefore, relative velocity with which B and C approaches each other=v- vcos60=v/2Since, v is constant, thus time taken by these two balls to meet each other is given by=(Relative displacement) / Relative velocity=a/(v/2)=2a/vSo, the time taken by the particles to meet each other=2a/v600nitial separation between two particles=Side of hexagon=aFinal separation=0Therefore,Relative displacement between two particles=aParticle B has a component v cos