A particle is projected with velocity v 0 along x -axis. The deceleration on the particle is proportional to the square of the distance from the origin i. e. a = - x 2 . The distance at which particle stops is -

Question

A particle is projected with velocity v 0 along x -axis. The deceleration on the particle is proportional to the square of the distance from the origin i. e. a = - x 2 . The distance at which particle stops is -, 3 v 0 2, v 0 2 3 1 3, 2 v 0 2 3 1 3, 3 v 0 2 2 1 3

MARKS App · Accepted Answer

Given initial velocity = v 0 final velocity = 0 a = - x 2 now we know that a = dv dt and v = dx dt thus a = dv dt = dv dt · dx dx = vdv dx vdv dx = - x 2 vdv = - x 2 dx integrating we get ∫ v 0 0 vdv = ∫ 0 x - x 2 dx v 2 2 v 0 0 = - x 3 3 0 x - v 0 2 2 = - x 3 3 ⇒ x = 3 v 0 2 2 1 3

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