A particle moves a distance 'S' in time 't' where S=t The acceleration at t=4sec is

s=t v=12t=12. t−1/2 a=−14. t−3/2 =−132

A particle moves a distance 'S' in time 't' where S=t The acceleration at t=4sec is

s=t v=12t=12. t−1/2 a=−14. t−3/2 =−132

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A particle moves a distance $' S'$ in time $' t'$ where $S = \sqrt{t}$ The acceleration at $t = 4 \sec$ is

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$2 units / \sec^{2}$

$\frac{1}{4} units / \sec^{2}$

$\frac{- 2}{17} units / \sec^{2}$

$\frac{- 1}{32} units / \sec^{2}$

answer is D.

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Detailed Solution

$s = \sqrt{t}$

$v = \frac{1}{2 \sqrt{t}} = \frac{1}{2} . t^{- 1 / 2}$

$a = \frac{- 1}{4} . t^{- 3 / 2}$ $= \frac{- 1}{32}$

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A particle moves a distance 'S' in time 't' where S=t The acceleration at t=4sec is

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A particle moves a distance $' S'$ in time $' t'$ where $S = \sqrt{t}$ The acceleration at $t = 4 \sec$ is