Question

A point $P$ moves in counter clockwise direction on a circular path a shown in figure. The movement of$P$ is such that it sweeps out a length $s=t^2+5$, where $s$ is in metre and $t$ in second. The radius of the path is $20m$. The acceleration of $P$ when $t=2s$ is nearly

• A. $2.1m/s^2$
• B. $12m/s^2$
• C. $7.2m/s^2$
• D. $14m/s^2$

Answer 1

The correct option is A $2.1m/s^2$

given, distance, $s=t²+5$

differentiate with respect to time,

speed $=v=\frac{ds}{dt}⟹v=2t$

at $t=2sec$ speed ,$v=2\times 2=4m/s$

so, centripetal acceleration, $=a_c=\frac{v^2}{R}$ where R is the radius of the path and $R=20m$. so, $a_c=\frac{4^2}{20}=\frac{16}{20}=0.8m/s²$

now rate of change of speed, $=\frac{dv}{dt}=2$

e.g., tangential acceleration $a_t=2m/s²$

now, net acceleration$=a=\sqrt{a_t^2+a_c^2}=\sqrt{4+0.64}$

$=2.154m/s²$