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A point P moves in 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 metres and t is in seconds. The radius of the path is 20m. The acceleration of ‘P’ when t = 2s is nearly.

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_{$14 m / s^{2}$}

_{$13 m / s^{2}$}

_{$12 m / s^{2}$}

_{$7.2 m / s^{2}$}

answer is A.

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Detailed Solution

$s = t^{3} + 5$

$∴$ Speed $v = \frac{d s}{d t} = 3 t^{2}$

Tangential acceleration $(a_{t}) = \frac{d v}{d t} = 6 t$

At t = 2s, $v = 3 \times {(2)}^{2} = 12 m s^{- 1}$ and $a_{t} = 6 \times 2 = 12 m s^{- 2}$

Centripetal acceleration $(a_{c}) = \frac{v^{2}}{R} = \frac{{(12)}^{2}}{20} = \frac{144}{20} m s^{- 2}$

$∴$ Acceleration of P is

$a = \sqrt{a_{t}^{2} + a_{c}^{2}}$

Substituting the values of $a_{t}$ and $a_{c}$ , we get $a = 14 m s^{- 2}$ ,

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