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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=aFinal separation=0Therefore,Relative displacement between two particles=aParticle 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/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/v
Relative velocity of A wrt B = V - Vsin30 = V/2Time = distance / speed Time = 2a/VIn 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.
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