A point $P$ moves in counterclockwise direction on a circular path as shown in the figure. The movem

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A point $P$ moves in counterclockwise 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 is nearly\

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(a) : $s=t^{3}+5$\ $	herefore v=\frac{d s}{d t}=\frac{d}{d t}\left(t^{3}+5ight)=3 t^{2}$\ Tangential acceleration,\ $a_{t}=\frac{d v}{d t}=\frac{d}{d t}\left(3 t^{2}ight)=6 t$\ At $t=2 \mathrm{~s}$,\ $v=3(2)^{2}=12 \mathrm{~m} / \mathrm{s}, a_{t}=6(2)=12 \mathrm{~m} / \mathrm{s}^{2}$\ Centripetal acceleration,\ $a_{c}=\frac{v^{2}}{R}=\frac{(12)^{2}}{20}=\frac{144}{20}=7.2 \mathrm{~m} / \mathrm{s}^{2}$\ Net acceleration,\ $a=\sqrt{\left(a_{c}ight)^{2}+\left(a_{t}ight)^{2}}$\ $=\sqrt{(7.2)^{2}+(12)^{2}} \approx 14 \mathrm{~m} \mathrm{~s}^{-2}$\