Question

A particle starting from rest moves in a circle with a velocity of $v=5\sin 100\pi t$. The magnitude of component of acceleration along the direction of velocity at $t=\frac{1}{2}\text{seconds}$ is (in ${\text{m/s}}^2$)

• A. Zero
• B. $500\pi$
• C. $1000\pi$
• D. $\frac{1}{20\pi }$

Answer 1

The correct option is B $500\pi$

Given, $v=5\sin 100\pi t$
Since $v$ depends on time, it is non uniform circular motion.
And the component of acceleration along $v$ is tangential acceleration.
$\therefore a_t=\frac{dv}{dt}=\frac{d}{dt}\left(5\sin 100\pi t\right)$
$=500\pi \cos \left(100\pi t\right)$
At $t=\frac{1}{2}\text{seconds}$,
$a_t=500\pi \cos \left(50\pi \right)=500\pi {\text{m/s}}^2$