# 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 ?

Vikas TU
14149 Points
8 years ago
total time = a/v + 2a/v + 3a/v + 4a/v + 5a/v
vinayaka
10 Points
8 years ago
@ 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

vinayaka
10 Points
8 years ago
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.
Taral Jain
39 Points
6 years ago
How to Sketch figure of this problem.Please clear its figure.How to think about this type of problems.
ankit singh
askIITians Faculty 614 Points
2 years ago
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