Question

19
Three particles A,B and C are situated at
the vertices of an equilateral triangle ABC
of side d att=0. Each of the particle 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​

Answer 1

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

along BA is vcos60= v/2

. Thus, the separation AB decreases at the rate

v−(−v/2) = 3v/2

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

t= d/(3v/2)

= 2d/3v