Question

Three particles a b c are situated at the vertices of an equilateral triangle

Answer 1

Answer:

distance, AO(=23AD=23a2−a2/4−−−−−−−−√=a3–√

Therefore, the time taken by particle at (A) to go from (A) to (O)

AOvcos30∘=a/3–√v3–√/2=2a3v.

Explanation:

Figure above shows the initial direction of velocities of A, B and C i.e. when A, B,C are at the ends of an equilateral triangle of side 'a' each.

The velocity of A is along AB. So, as B moves, then the direction of velocity of A changes continuously towards the direction of B.

Similarly, the velocity of B is along BC and velocity of C is along CA.

The component of velocity of B along BA = v cos60o =  

So, the effective speed with which A and B approach each other  

Time taken in reducing the separation AB from 'a' to zero is  

Since all three particles A, B and C move with same constant speed v, hence C will also meet A and B at the same time