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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 20 m. The acceleration of ‘P’ when $t = 2$ s is nearly.

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13 $m / s^{2}$

12 $m / s^{2}$

7.2 $m / s^{2}$

14 $m / s^{2}$

answer is D.

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Detailed Solution

$s = t^{3} + 5 \Rightarrow v e l o c i t y, v = \frac{d s}{d t} = 3 t^{2}$

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

Radial acceleration $a_{c} = \frac{v^{2}}{R} = \frac{9 t^{4}}{R}$

At $t = 2 s, a_{1} = 6 \times 2 = 12 m / s^{2}$

$a_{c} = \frac{9 \times 16}{20} = 7.2 m / s^{2}$

$∴$ Resultant acceleration $= \sqrt{a_{1}^{2} + a_{c}^{2}} = \sqrt{{(12)}^{2} + {(7.2)}^{2}} = \sqrt{144 + 51.84} = \sqrt{195.84} = 14 m / s^{2}$

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