Three particles A, B and C are situated at the vertices of an equilateral triangle ABC of side d at time t = 0. Each of the particles moves with constant speed v. A always has its velocity along AB, B along BC and C along CA. At what time will the particles meet each other ?

- Don't want spam or copied answers
- Don't be greedy for points
- Best answer will be awarded with brainliest.

It was ans of your questions

Given :-

To Find :-

Solution :-

See attachment .

Firstly break the velocity of A , into it's vertical components . That will be ,

Now the component , coincides on side AC and is directed from A to C . Along CA, the velocity of particle C is directed . So both are in opposite directions.

So , the relative velocity of C with respect to A will be ,

Substitute ,

Value of cos 60° is 1/2 ,

Add the terms,

Hence , now we can find the time taken by A and C ,to meet each other will be ,

Substitute ,

Multiplying both sides by 2 ,

Dividing both sides by 3v ,

Similarly , taking any two particles you will get the same time .

Hence the particles meet after 2d/3v time .

I hope this helps .

explanation:

an enlarged virtual image is the answer