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A particle starts from the origin and moves along the X-axis such that the velocity at any instant is given by $4 t^{3} - 2 t,$ where t is in second and velocity is in $m s^{- 1} .$ What is the acceleration of the particle when it is 2 m from the origin?

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$22 m s^{- 2}$

$12 m s^{- 2}$

$28 m s^{- 2}$

$10 m s^{- 2}$

answer is C.

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Detailed Solution

$\frac{d x}{d t} = 4 t^{3} - 2 t$

$o r d x = 4 t^{3} d t - 2 t d t$

Integrating, $x = \frac{4 t^{4}}{4} - \frac{2 t^{2}}{2} = t^{4} - t^{2}$

When x=2,

$∴ t^{4} - t^{2} - 2 = 0$

$t^{2} = \frac{- (- 1) \pm \sqrt{1 + 8}}{2}$

$o r t^{2} = \frac{1 \pm 3}{2} = 2$ (ignoring –ve sign)

Again, $\frac{d^{2} x}{d t^{2}} = 12 t^{2} - 2$

When $t^{2} - 2, a c c e l e r a t i o n = 12 \times 2 - 2 = 22 m s^{- 2}$

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