Question

A point $P$ moves in clockwise direction on a circular path as shown in figure. The movement of $P$ is such that it sweeps a distance $s=2t^3+3t^2+6$, where $s$ is in metres and $t$ is in seconds. The radius of the path is $36\text{m}$. The acceleration of ${}^{\prime }{P}^{\prime }$ when $t=1\text{s}$ is

• A. $18{\text{m/s}}^2$
• B. $4{\text{m/s}}^2$
• C. $\sqrt{350}{\text{m/s}}^2$
• D. $\sqrt{340}{\text{m/s}}^2$

Answer 1

The correct option is D $\sqrt{340}{\text{m/s}}^2$

Given, distance covered along circular path
$s=2t^3+3t^2+6$
$\Rightarrow$ Velocity, $v=\frac{ds}{dt}=6t^2+6t$

Tangential acceleration, $a_t=\frac{dv}{dt}=12t+6$
and radial acceleration, $a_c=\frac{v^2}{R}=\frac{(6t^2+6t{)}^2}{R}=\frac{36t^2(t+1{)}^2}{R}$

At $t=1\text{s}$,
$a_t=12\left(1\right)+6=18{\text{m/s}}^2$
$a_c=\frac{36\left(1{)}^2\right(2{)}^2}{36}=4{\text{m/s}}^2$
$\therefore$ Resultant acceleration at $P$ is
$a=\sqrt{a_t^2+a_c^2}=\sqrt{{18}^2+4^2}=\sqrt{340}{\text{m/s}}^2$