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 ,
[tex] v_A = v_A cos60^o + v_A sin60^o [/tex]
Now the component , [tex]v_A cos60^o [/tex] 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 ,
[tex] v_{CA}= v_C - v_A [/tex]
Substitute ,
[tex] v_{CA}= v - ( - v cos 60^o) [/tex]
Value of cos 60° is 1/2 ,
[tex] v_{CA}= v + v\bigg(\dfrac{1}{2}\bigg)=v +\dfrac{v}{2}[/tex]
Add the terms,
[tex] v_{CA}= \dfrac{2v + v}{2}=\dfrac{3v}{2}[/tex]
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 ;
[tex]s = v t [/tex]
Substitute ,
[tex]d = \dfrac{3v}{2} t [/tex]
Multiplying both sides by 2 ,
[tex] 2d = 3vt [/tex]
Dividing both sides by 3v ,
[tex] t = \dfrac{2d}{3v} [/tex]
Similarly , taking any two particles you will get the same time .
Hence the particles meet after 2d/3v time .
I hope this helps .
