A particle moves in a circle of radius $R$ with a constant speed $v$. The change in velocity after the particle has covered three-quarters of the circle is:

Solution

1. Let the initial velocity be ${\vec{v}}_i=v\hat{i}$.
2. After three-quarters of a circle, the particle is at a position corresponding to a $270^{\circ }$ rotation, so ${\vec{v}}_f=v\hat{j}$.
3. The change in velocity is $$\begin{array}{rl}\Delta \vec{v}& ={\vec{v}}_f-{\vec{v}}_i\\ & =v\hat{j}-v\hat{i}\end{array}$$.
4. The magnitude is $$\begin{array}{rl}|\Delta \vec{v}|& =\sqrt{v^2+(-v{)}^2}\\ & =v\sqrt{2}\end{array}$$.
5. Hence the answer is (A).