Question

A body is moving with uniform speed $v$ on a horizontal circle from A as shown in the figure. Change in the velocity in the first quarter revolution is:

• A. $v^2$
• B. $\sqrt{2}v$ southwest
• C. $\sqrt{2}$ northwest
• D. $2v$ west

Answer 1

The correct option is B $\sqrt{2}v$ southwest

For quarter revolution
$△\overline{v}=v_2(-\hat{i})-v_1\hat{j}$
or $△\overline{v}=-v\hat{i}-v\hat{j}$

On calculating the magnitude
$△v=\sqrt{v^2+v^2}=\sqrt{2}v$
Also $\alpha ={\tan }^{-1}\frac{v}{v}={45}^{\circ }$
$△\overline{v}=\sqrt{2}v$ southwest.