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(A) Let the constant speed of the particle be $v$ and the radius of the circular path be $R$.When the particle turns by an angle of $90^{\circ}$ (or $

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Let the constant speed of the particle be $v$ and the radius of the circular path be $R$.When the particle turns by an angle of $90^{\circ}$ (or $\pi/2$ radians),the distance traveled along the arc is $s = R 	heta = R(\pi/2) = \pi R / 2$.The time taken for this motion is $t = s / v = (\pi R / 2) / v = \pi R / (2v)$.The displacement is the straight-line distance between the initial and final positions,which is the chord length $AB = \sqrt{R^2 + R^2} = R\sqrt{2}$.The average velocity is defined as the total displacement divided by the total time taken:$\langle v angle = \frac{	ext{Displacement}}{	ext{Time}} = \frac{R\sqrt{2}}{\pi R / (2v)} = \frac{R\sqrt{2} \cdot 2v}{\pi R} = \frac{2\sqrt{2}v}{\pi}$.The ratio of instantaneous velocity $v$ to average velocity $\langle v angle$ is:$\frac{v}{\langle v angle} = \frac{v}{2\sqrt{2}v / \pi} = \frac{\pi}{2\sqrt{2}}$.Comparing this with the given ratio $\pi : x\sqrt{2}$,we have $\frac{\pi}{2\sqrt{2}} = \frac{\pi}{x\sqrt{2}}$.Therefore,$x = 2$.

$A$ particle is moving with constant speed in a circular path. When the particle turns by an angle $90^{\circ}$,the ratio of instantaneous velocity to its average velocity is $\pi : x \sqrt{2}$. The value of $x$ will be $.........$

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