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 $s=t^2+5$, 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 approximately :

• A. $13m/s^2$
• B. $2.15m/s^2$
• C. $7.2m/s^2$
• D. $14m/s^2$

Answer 1

The correct option is B $2.15m/s^2$

$s=t^2+sv=\frac{ds}{dt}=2tatt=2s,v=2\times 2=4\frac{m}{s}\therefore a_{completedacceleration}=\frac{v^2}{r}=\frac{4^2}{20}=\frac{16}{20}=\frac{4}{5}andtangentialacceleration=a_t=\frac{dv}{dt}=2hence,a_{net}=\sqrt{{\left(\frac{4}{5}\right)}^2+2^2}=2.15\frac{m}{s^2}$