A particle moves in 00. After its acceleration changes sign, the magnitude of retardation becomes maximum at some position x. The position and speed there are:

Question

A particle moves in 0<x<L with velocity v = β√x (L-x), where β>0. After its acceleration changes sign, the magnitude of retardation becomes maximum at some position x. The position and speed there are:

Options

G1 Physics · Accepted Answer

Use a = v dv/dx = (1/2)d(v²)/dx. Now v² = β²x(L−x)². Therefore a = (β²/2)(L−x)(L−3x). Acceleration changes sign at x=L/3. For x>L/3, retardation magnitude is R = −a = (β²/2)(L−x)(3x−L). Maximize f(x)=(L−x)(3x−L) = −3x²+4Lx−L². df/dx = −6x+4L=0 ⇒ x=2L/3. The speed there is v = β√(2L/3)(L/3) = β√6 L³ᐟ²/9. Therefore option B is correct.

A particle moves in 0<x<L with velocity v = β√x (L−x), where β>0. After its acceleration changes sign, the magnitude of retardation becomes maximum at some position x. The position and speed there are: