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 $t = 2s$ is nearly [image:4_image45.emf]

Question

Dr. Abubacker · Accepted Answer

(a) As $S = t^{3} + 5$
$\frac{ds}{dt} = 3t^{2} = v$
$	herefore a_{t} = \frac{dv}{dt} = 6t$
at $t = 2\ sec$
$\left| \overrightarrow{a} ight| = \sqrt{a_{c}^{2} + a_{t}^{2}}$
$= \sqrt{\left( \frac{v^{2}}{R} ight)^{2} + a_{t}^{2}} = \sqrt{\left( \frac{4t^{4}}{R} ight)^{2} + \left( \frac{dv}{dt} ight)^{2}}$
$= \sqrt{(7.2)^{2} + 144}$
$\left| \overrightarrow{a} ight| = 14m/s^{2}$

Explanation

Reviewed by Dr. Abubacker

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