Triangle Convergence Kinematics

The trick here is to understand that at any given time, the particles will be at the vertices of an equilateral triangle (although the triangle would be rotating). Due to symmetry, the component of any particle's velocity towards the center of the triangle will always remain the same.

Displacement of the particles $=$ Distance from vertex to the center of the triangle $=$ $\frac{6}{\sqrt3}$

Component of velocity towards center of the triangle $=vcos(\frac{60}{2})=vcos(30)=v\frac{\sqrt3}{2}$ .

Therefore, time taken for convergence $=\frac{Displacement}{Velocity}=\frac{6\times2}{\sqrt3\times2\times\sqrt3}=\boxed{2.000sec}$

The relative velocity between two of the particles is $v_r=v+v\cos \theta=v(1+\cos \theta)$ . Now at time $t$ the distance traveled by the relative velocity must be equal to the length of the triangle which is $d$ . Thus $v_r \cdot t = d \Rightarrow t=\dfrac{d}{v_r}$ . In the problem $\cos \theta = 1/2$ , $d=6m$ and $v=2ms^{-1}$ , substituting these values in our equation, we get $t=2s$ .

PS: I didn't know how to include an image without which the proof is incomplete, sorry about that.

This is a very famous problem named as Man Chasing Problem(name given by me). First of all lets understand the problem.Firstly all the 3 points were on the vertices of the triangle ABC.Then they start moving.As the one at C starts moving,the one at B start chasing him and leave the path which it was following i.e the side of the triangle,then the one who started from C start chasing the one at A and the one at A start chasing the one who started from B and so on. So all of them left their initial paths i.e the sides of the triangle and started going inside the triangle.The main aim of each point is to chase the other one who is ahead of it.Therefore by symmetry they will meet at the centroid of the triangle.Lets talk about the point at B.Its initial velocity was 2m/s along BC.Consider the journey of B point from vertex B to centroid G.The initial point of journey was vertex B and final point was centroid G, therefore the displacement will be BG and we know that the direction of the velocity is in the direction of displacement therefore velocity along BG is the component of initial velocity that is along BC and also note that the speed is constant,therefore the component of initial velocity along BG will be $vcos30^{\circ}$ and the magnitude of displacement will be BG= $2\sqrt3$ which can be found easily using geometry.Therefore $t=\frac{s}{v} = \frac{2\sqrt{3}}{\sqrt{3}}=\boxed 2.$

I used a numerical solution within Excel. I created columns for the position and velocity of each point at increments of time. Then I calculated x and y velocities for each particle after a small instant of time and used those velocities to calculate a new x and y position. I used a plot of my values to find where the particles intersect and examined my table to find the time where that takes place. The hardest part of my solution for me was correctly writing instructions for excel to calculate the angle between two particles and get the correct sign for the velocity. To create my and find an approximate solution, I used a time increment of .01 seconds, and then to get a solution I was more confident in, I used a time increment of .001 seconds.

Well for such type of motion in any regular polygon

Time Taken = $\frac{a}{2v\cos^{2}\theta/2}$

a - side length

theta - angle

v - velocity

TAKE COMPONENT OF VELOCITY TOWARDS THE CENTROID AND DISPLACEMENT

Let the 3 points be A,B,C

The velocity of A is v=2 m/s is along AB.

The velocity of B is v=2 m/s is along BC.

Its component along BA is v*cos60 = v/2 = 1 m/s

Thus the separation AB decreases at the rate

v'= v + (v/2) = 2+1 = 3 m/s

Since this rate is constant,

The time taken in reducing the separation AB from 6 to 0 is

t = d/v' = 6/3= 2 sec

Firstly, we resolve the components of velocities along the side of the triangle. Let the velocities be v ,then. -vcos60-v= dL/dt. L is side of triangle at any instant formed by 3 particles. Let the side of the original triangle be a. Now -vcos60-v=. -3v/2. Time at which they meet at center is. Side of triangle/speed of approach= 2a/3v= 2seconds

Let the three particles be p,q,r initially at A,B,C points of equilateral triangle respectively..... the veloctiy of P is always directed towards side AB,the velocity of q is directed towards BC nd etc..thus this particles will come come close to each other as relative distance between them decreases..thus they will meet at centroid..now find the distance of centroid from any point and then find the component of velocity of the particle initially situated at that point in the direction towards the line joining centroid and the point..,thus t= d/v