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?
What is the magnitude of acceleration of any particle at beginning at t=0?
Find the radius of curvature of the path followed just at the beginning?
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Hii
Explanation:
Velocity of A is v along AB. The velocity of B is along BC.Its component
along BA is vcos60= 2v
Thus, the separation AB decreases at the rate
v−(− 2v)= 23v
Since, this rate is constant, the time taken in reducing the separation AB from d to zero is
t= (3v/2)
d = 3v2d
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