Question

A particle starts moving from rest at $t=0$ with a tangential acceleration of constant magnitude of $\pi {\text{m/sec}}^2$ along a circle of radius $6\text{m}$. The values of average velocity and average speed during the first $2\sqrt{3}\text{seconds}$ of motion are

• A. Average velocity $=2\sqrt{3}\text{m/s}$; Average speed $=\sqrt{3}\text{m/s}$
• B. Average velocity $=2\sqrt{3}\text{m/s}$; Average speed $=\sqrt{3}\pi \text{m/s}$
• C. Average velocity $=4\sqrt{3}\text{m/s}$; Average speed $=\sqrt{3}\pi \text{m/s}$
• D. Average velocity $=\sqrt{3}\text{m/s}$; Average speed $=\sqrt{3}\pi \text{m/s}$

Answer 1

The correct option is B Average velocity $=2\sqrt{3}\text{m/s}$; Average speed $=\sqrt{3}\pi \text{m/s}$

From the second equation of motion, we have
Total distance travelled by the particle after time $t=2\sqrt{3}\text{s}$ is
$s=\frac{1}{2}a{t}^2=\frac{1}{2}\times \pi \times (2\sqrt{3}{)}^2=6\pi \text{m}$

Angular displacement after time $t=2\sqrt{3}\text{s}$ is
$\theta =\frac{1}{2}\alpha t^2=\frac{1}{2}\times \frac{a}{r}\times t^2$
$=\frac{1}{2}\times \frac{\pi }{6}\times (2\sqrt{3}{)}^2=\pi \text{rad}$
So, linear displacement is $2r$

Thus, average velocity $=\frac{\text{Total displacement}}{\text{Total time taken}}=\frac{2r}{t}=\frac{2\times 6}{2\sqrt{3}}=2\sqrt{3}\text{m/s}$
and average speed $=\frac{\text{Total distance}}{\text{Total time taken}}=\frac{6\pi }{2\sqrt{3}}=\pi \sqrt{3}\text{m/s}$