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 meters and t is in seconds. The radius of the path is 20 m. The acceleration of 'P' when t = 2 s is nearly

• A. 13 m/s²
• B. 12 m/s²
• C. 7.2 m/s²
• D. 14 m/s²

Answer 1

The correct option is D

14 m/s²

$S=t^3+5$
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/$s^2$
$Att=2s,v=3(2{)}^2=12m/s^2$
$\therefore$ centripetal acceleration,
=$\frac{v^2}{R}=\frac{144}{20}m/s^2$
$\therefore$ net acceleration=$\sqrt{a_t^2+a_i^2}\approx 14m/s^2$