I have been pondering over this question for a while and trying to derive a general formula for this.The problem is:

"In an \(N\)-sided polygon ,objects are placed at each of the vertices.They move with a velocity \(v\) such that they always move towards each other. Find the time when they will meet."

I perhaps feel it is an easy question,but still I have been struggling with it for quite some time.

Help would really be appreciated.

Thanks!

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TopNewestIf the side of the polygon is of length \(a\), then: \[T=\frac{a}{v \left (1-\cos \frac{2\pi}{n} \right) } \]

Proof:At any instant, joining the adjacent particles, we again get a regular \(n\) sided polygon.

We now conclude that the relative velocity (along the line joining them) between any two adjacent particles is constant throughout the motion. Also, all the particles meet when any two adjacent particles meet.

Note that the relative velocity along the line joining two adjacent particles is \(v(1-\cos \frac{2\pi}{n})\).

Hence, time required for all the particles to meet is given by the first equation. – Deeparaj Bhat · 9 months, 2 weeks ago

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@Deeparaj Bhat – Shivam Mishra · 9 months, 2 weeks ago

Please can you present the derivation?Log in to reply

– Deeparaj Bhat · 9 months, 2 weeks ago

I've added it now. Tell me if you any problem understanding. :)Log in to reply

– Shivam Mishra · 9 months, 2 weeks ago

Thanks!!!Log in to reply

– Deeparaj Bhat · 9 months, 2 weeks ago

You're welcome. :)Log in to reply

– Aditya Sky · 9 months, 2 weeks ago

Can we somehow figure out equation of trajectory of motion of objects in case polygon under consideration is a square ? I tried to solve this problem last year but was unable to do so.Log in to reply

Let \(r\) be the distance of a particle from the centre of centre of the square (which is invariant wrt time). Let \(\theta\) denote the angle the radius vector makes with the initial line. Then, \[ \frac{d}{dt} (re^{i\theta})=ve^{i(\theta+\frac{3\pi}{4})} \]

Now, solve. :)

Edit:If the side of the square is \(a\), then \[\sqrt2 r=a-vt \\ \theta- \theta_0 = \ln \left( 1- \frac{v}{a}t \right)\] – Deeparaj Bhat · 9 months, 2 weeks ago

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@shivam mishra

Actually, the trajectory can be found explicitly for a n sided polygon. – Deeparaj Bhat · 9 months, 2 weeks ago

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You have to define "always move towards each other". You most likely mean "move towards the object clockwise from them". – Calvin Lin Staff · 9 months, 2 weeks ago

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– Shivam Mishra · 9 months, 2 weeks ago

Actually by that phrase I meant that their velocity vectors are always directed towards the particle that is in front of them,it could be clockwise or anticlockwise.Log in to reply

– Calvin Lin Staff · 9 months, 2 weeks ago

Right, so that should be clarified. "always move towards each other" is an ambiguous phrase.Log in to reply

@Harsh Shrivastava@Rajdeep Dhingra@Prakhar Bindal Please help!!! – Shivam Mishra · 9 months, 2 weeks ago

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– Rajdeep Dhingra · 9 months, 2 weeks ago

Deeparaj bhaiya has already explained. Sorry I'm late.Log in to reply

– Deeparaj Bhat · 9 months, 2 weeks ago

Don't call me 'bhaiya' yaar.Log in to reply

– Shivam Mishra · 9 months, 2 weeks ago

No problem ,well congrats on your ijso selection.Log in to reply

– Rajdeep Dhingra · 9 months, 2 weeks ago

Thanks.Log in to reply