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!

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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.

Log in to reply

Please can you present the derivation? @Deeparaj Bhat

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply

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

Ok. I was able to find the trajectory in polar coordinates. We shall consider the trajectory of a single particle.

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 \[\sqrt2 r=a-vt \\ \theta- \theta_0 = \ln \left( 1- \frac{v}{a}t \right)\]

Log in to reply

@shivam mishra

Actually, the trajectory can be found explicitly for a n sided polygon.

Log in to reply

@Harsh Shrivastava@Rajdeep Dhingra@Prakhar Bindal Please help!!!

Log in to reply

Deeparaj bhaiya has already explained. Sorry I'm late.

Log in to reply

No problem ,well congrats on your ijso selection.

Log in to reply

Log in to reply

Don't call me 'bhaiya' yaar.

Log in to reply

You have to define "always move towards each other". You most likely mean "move towards the object clockwise from them".

Log in to reply

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

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

Log in to reply