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A point $P$ moves in a counter-clockwise direction on a circular path as shown in the figure. The movement of $P$ is such that it sweeps out a length $s = t^{3} + 5$ , where $s$ is in metre and $t$ is in second. The radius of the path is $20 m$ . The acceleration of $P$ when $t = 2 s$ is nearly, is

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$14 {ms}^{- 2}$

$12 {ms}^{- 2}$

$13 {ms}^{- 2}$

$7.2 {ms}^{- 2}$

answer is D.

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Detailed Solution

Given, $s = t^{3} + 5$

$∴$ Speed, $v = \frac{ds}{dt} = 3 t^{2}$

and rate of change of speed, $a_{t} = \frac{dv}{dt} = 6 t$

$∴$ Tangential acceleration at $t = 2 s$ ,

$a_{t} = 6 \times 2 = 12 {ms}^{- 2}$

and at $t = 2 s, v = 3 (2)^{2} = 12 {ms}^{- 1}$

$∴$ Centripetal acceleration

$\ a_{c} = \frac{v^{2}}{R} = \frac{144}{20} {ms}^{- 2}$

$∴$ Net acceleration $= \sqrt{a_{1}^{2} + a_{c}^{2}}$

$= \sqrt{(12)^{2} + {(\frac{144}{20})}^{2}} = 14 {ms}^{- 2}$

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