A particle starting from rest rotates in a circle of radius $R=\sqrt{2}\text{m}$ with an angular acceleration $\alpha =\frac{\pi }{4}{\text{rad/sec}}^2$. The magnitude of average velocity of the particle over the time in rotating a quarter of a circle is

The correct option is A $1\text{m/s}$
Given, particle starts from rest
So, tangential acceleration $a_t=R\alpha =\frac{\sqrt{2}\pi }{4}{\text{m/sec}}^2$
$\therefore$ Distance travelled in time ${}^{\prime }{t}^{\prime }$ by the particle is
$s=\frac{1}{2}{a}_t{t}^2=\frac{\sqrt{2}\pi t^2}{8}$.....(i)

For one quarter circle, distance travelled is
$s=\frac{\pi R}{2}=\frac{\sqrt{2}\pi }{2}$......(ii)

From eq. (i) and (ii), we have
$\frac{\sqrt{2}\pi }{2}=\frac{\sqrt{2}\pi t^2}{8}$
$\Rightarrow t=2s$ is the time taken to rotate one quarter circle.

Now, the particle motion can be shown as

$\Rightarrow \text{Average velocity}=\frac{\text{Total Displacement}}{\text{Total Time}}$
$=\frac{\sqrt{2}R}{t}=\frac{\sqrt{2}\times \sqrt{2}}{2}=1\text{m/s}$