Question

A point $P$ moves in counter-clockwise direction on a circular path as shown in the figure. The movement of ${}^{\prime }{P}^{\prime }$ 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

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

Answer 1

Answer: (D)

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