Question

A point moves along a circle with velocity $v=at,$ where $a=0.5{\text{m/s}}^2$. Then the total acceleration of the point at the moment, when it has covered ${\left(\frac{1}{10}\right)}^{th}$ of the circle after the beginning of motion is

• A. $0.5{\text{m/sec}}^2$
• B. $0.6{\text{m/sec}}^2$
• C. $0.7{\text{m/sec}}^2$
• D. $0.8{\text{m/sec}}^2$

Answer 1

The correct option is D $0.8{\text{m/sec}}^2$

Given : $v=at$
and $a=0.5{\text{m/sec}}^2$
now tangential acceleration $a_T=\frac{dv}{dt}$
$a_T=a=0.5{\text{m/sec}}^2$
Centripetal acceleration, $a_c=\frac{v^2}{r}=\frac{(at{)}^2}{r}=0.5\times 0.5\times \frac{t^2}{r}....\left(1\right)$
at $t=0,{\omega }_0=0$ and $a_T=\alpha r$
Now, $\theta ={\omega }_{0}t+\alpha t^2$
$(2\pi \times \frac{1}{10})=\frac{1}{2}\left(\frac{a_T}{r}\right).t^2=\frac{1}{2}\times 0.5\frac{t^2}{r}$
$\frac{t^2}{r}=\frac{2\pi }{5\times 0.5}$
from equation (1)
$a_c=0.5\times 0.5\times \frac{2\pi }{5\times 0.5}=\frac{\pi }{5}$
Net acceleration $a=\sqrt{a_T^2+a_c^2}$
$=\sqrt{(0.5{)}^2+{\left(\frac{\pi }{5}\right)}^2}=0.8{\text{m/sec}}^2$