Physics, 13.02.2022 05:50 mzbugsbunny3029

$\bold{\huge{\underline{ Hola \: Brainlians!!! }}}$ 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 ?

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Answer from: nati5720

It was ans of your questions

Three particles A, B and C are situated at the vertices of an equilateral triangle ABC of side d a

Answer from: oliup567

Given :-

Three particles A, B and C are situated at the vertices of an equilateral triangle ABC.The side of the triangle is d .A always has its velocity along AB, B along BC and C along CA.

To Find :-

Time at which the particles will meet .

Solution :-

See attachment .

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

v_A = v_A cos60^o + v_A sin60^o

Now the component , v_A cos60^o 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 ,

$v_{CA}= v_C - v_A$

Substitute ,

$v_{CA}= v - ( - v cos 60^o)$

Value of cos 60° is 1/2 ,

$v_{CA}= v + v\bigg(\dfrac{1}{2}\bigg)=v +\dfrac{v}{2}$

Add the terms,

$v_{CA}= \dfrac{2v + v}{2}=\dfrac{3v}{2}$

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

Since the speed is constant , we can use distance = speed*time , as ;

s = v t

Substitute ,

$d = \dfrac{3v}{2} t$

Multiplying both sides by 2 ,

2d = 3vt

Dividing both sides by 3v ,

$t = \dfrac{2d}{3v}$

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

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

I hope this helps .

Answer from: Quest

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Answer from: Quest

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