Question

Three psrticles A,Band C are situated at the verices of an equilateral triangle ABC of side a at t=0 each of the particles moves with constant speed v .A always has its velocity along AB,B along BC and C .At what time will these particles meet each other​

Answer 1

Answer:

Velocity of A is v along AB. The velocity of B is along BC.Its component

along BA is vcos60=

2

v

. Thus, the separation AB decreases at the rate

v−(−

2

v

)=

2

3v

Since, this rate is constant, the time taken in reducing the separation AB from d to zero is

t=

(3v/2)

d

=

3v

2d