A point $\mathrm{P}$ moves in counter-clockwise direction on a circular path as shown in the figure. The…
A point $\mathrm{P}$ moves in counter-clockwise direction on a circular path as shown in the figure. The movement of ' $\mathrm{P}$ ' 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 $20 \mathrm{~m}$. The acceleration of ' $\mathrm{P}$ ' when $t=2 \mathrm{~s}$ is nearly
$13 \mathrm{~m} / \mathrm{s}^2$
$12 \mathrm{~m} / \mathrm{s}^2$
$7.2 \mathrm{~m} / \mathrm{s}^2$
$14 \mathrm{~m} / \mathrm{s}^2$
Solution
$
S=t^3+5
$
$\therefore \quad$ speed, $\mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}=3 \mathrm{t}^2$
and rate of change of speed $=\frac{\mathrm{dv}}{\mathrm{dt}}=6 \mathrm{t}$
$\therefore \quad$ tangential acceleration at $\mathrm{t}=2 \mathrm{~s}, \mathrm{a}_{\mathrm{t}}=6 \times 2=12 \mathrm{~m} / \mathrm{s}^2$
at $t=2 \mathrm{~s}, \mathrm{v}=3(2)^2=12 \mathrm{~m} / \mathrm{s}$
$\therefore \quad$ centripetal acceleration, $\quad a_c=\frac{v^2}{R}=\frac{144}{20} \mathrm{~m} / \mathrm{s}^2$
$\therefore \quad$ net acceleration $=\sqrt{\mathrm{a}_t^2+\mathrm{a}_{\mathrm{i}}^2}$ $\approx 14 \mathrm{~m} / \mathrm{s}^2$
