Question

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 :

• A. $13$ $m/s^2$
• B. $12$ $m/s^2$
• C. $7.2$ $m/s^2$
• D. $14$ $m/s^2$

Answer 1

The correct option is D $14$ $m/s^2$

$S=t^3+5$
$\cdot$ Speed, $v=\frac{ds}{dt}=3t^2$
and rate of change of speed $=\frac{dv}{dt}=6t$
$\cdot$. tangential acceleration at $t=2s,a_t=6\times 2=12m/s^2$ at $t=2s,v=3(2{)}^2=12m/s$
$\cdot$. Centripetal acceleration, $a_c=\frac{v^2}{R}=\frac{144}{20}m/s^2$
$\cdot$ Net acceleration $=\sqrt{a_t^2+a_i^2}$
$\approx 14m/s^2$