Here is a problem I found in a book. Even after many attempts, I couldn't solve it. None of my friends could solve it! Here it is, directly taken from the book.

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.

It must be based on the concept of Relative Velocity (or Change of Frame) somehow. Can anyone help me? Thanks in advance...

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TopNewestI think that's from H.C. Verma, right? – Dinesh Chavan · 2 years, 10 months ago

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– Ameya Salankar · 2 years, 10 months ago

Yeah! Do you know how it's done?Log in to reply

So, the component of the velocity towards center, you can find it easily, and then the distance between center and any particle, then just get the time= \(\dfrac{distance}{speed}\). not tough at all.....

You may want to see the solution I wrote to your problem (square one)

Note that this will happen if and only if the polygon is

regular polygonand thespeedof each of the particles is the same.@Ameya Salankar , @Dinesh Chavan – Aditya Raut · 2 years, 9 months ago

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@Aditya Raut, yes, quite an easy one! Can you please elaborate your

emotionsof physics? – Ameya Salankar · 2 years, 9 months agoLog in to reply

emotionsof physics is getting something by feeling, intuition in physics – Aditya Raut · 2 years, 9 months agoLog in to reply

– Dinesh Chavan · 2 years, 9 months ago

Its a small Issue in our tution classesLog in to reply

Ants on a tetrahedron – Thaddeus Abiy · 2 years, 10 months ago

I can't help but notice the similarity withLog in to reply

Hey friends i just tried to deduce a formula for such type of problems \(t=\dfrac{a}{2v\times \sin(\dfrac{\pi}{n})\times \sin(\dfrac{\pi}{n})}\) where \(a\) is side length and \(n\) denotes the total sides of the regular polygon – Mehul Chaturvedi · 1 year, 11 months ago

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– Shah Yug · 5 months ago

Can you please show how did you do it ?Log in to reply

– Atul Saswat · 1 year, 1 month ago

Nailed it ! NYC DONE BRO >Log in to reply

The time it takes is exactly 2 times the time it takes for any particle to travel the length of one side of the hexagon. If there were 4 particles in a square, the time is the same as travelling a side. Given any regular polygon, since at any time after pursuit begins, the particles are at the corners of the same regular polygon but a shrinking one, the time it takes is the same as finding the length of logarithmic spirals, which have the property of self-similarity.

Most pursuit problems are harder to analyze. – Michael Mendrin · 2 years, 10 months ago

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– Ameya Salankar · 2 years, 10 months ago

But how do we prove that they will eventually collide with each other? (it seems to defy reasoning)Log in to reply

A shrinking hexagon indicates that the particles are getting closer to each other, because the sides are shrinking.

As a counterexample, if each particle was constrained to move only along a circular path common to all of them, then they'll never get any closer. – Michael Mendrin · 2 years, 10 months ago

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@Michael Mendrin, Thanks for explaining! I got it now. – Ameya Salankar · 2 years, 10 months ago

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ANSWER:

2a/v

Plz admire this image for understanding my answer... and plz consider it as a regular hexagonimg

**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 cos60 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

In reality each particle will follow a curved path and eventually meet at the center of the hexagon.** – Ess Kp · 2 years, 7 months ago

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@ess kp , I have edited your comment so now the image is appearing instead of the link. Please see the changes I've made so that next time you can make your image appear in comments, notes and problems. – Aditya Raut · 2 years, 7 months ago

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