A particle starts from rest at x=0 and moves under acceleration a = a₀(1-2x/L), where a₀, L>0. The maximum speed attained by the particle and the time taken to attain this speed are:
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Question

A particle starts from rest at x=0 and moves under acceleration a = a₀(1−2x/L), where a₀,L>0. The maximum speed attained by the particle and the time taken to attain this speed are:

G1 Physics · Accepted Answer

First obtain v as a function of x: v dv/dx = a₀(1−2x/L). Integrating from x=0, v=0: v²/2 = a₀(x−x²/L). Hence v² = 2a₀x(1−x/L). Maximum speed occurs where a=0: 1−2x/L=0 ⇒ x=L/2. Thus vₘₐₓ² = 2a₀(L/2)(1/2) = a₀L/2, so vₘₐₓ=√(a₀L/2). For time: t = ∫₀→L/2 dx / √[2a₀x(1−x/L)]. Put x=L sin²θ. Then the integral becomes t = √(2L/a₀) ∫₀→π/4 dθ = (π/4)√(2L/a₀). Therefore option B is correct.