Particle A starts from rest at x=0 with constant acceleration a. At the same instant particle B is at x=L and moves in the +x direction with constant speed u. Before A catches B, their separation first increases and its maximum value is 2L. The time at which A catches B is:
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Question

Particle A starts from rest at x=0 with constant acceleration a. At the same instant particle B is at x=L and moves in the +x direction with constant speed u. Before A catches B, their separation first increases and its maximum value is 2L. The time at which A catches B is:

Options

G1 Physics · Accepted Answer

Positions: position of A = at²/2, position of B = L+ut. Separation: S(t)=L+ut−at²/2. Maximum separation occurs when dS/dt = u−at = 0 ⇒ t=u/a. At that instant, Sₘₐₓ = L + u²/a − u²/(2a) = L + u²/(2a). Given Sₘₐₓ=2L, u²/(2a)=L ⇒ u=√(2aL). A catches B when at²/2 = L+ut. Thus t = [u+√(u²+2aL)]/a. Using u²=2aL, t = [√(2aL)+2√(aL)]/a = (2+√2)√(L/a). Therefore option B is correct.