Question

A particle is moving on a circular path of radius R with constant speed $v_0$. What is the change in magnitude of velocity when it completes $\frac{3^{th}}{4}$ of the complete circular path ?

• A. $\sqrt{2}{v}_0$
• B. $v_0$
• C. Zero
• D. $2v_0$

Answer 1

The correct option is C Zero

The magnitude of velocity is the speed.

As the particle is moving in the circular path with the constant speed. Thus the magnitude of velocity will not change whatsoever. Only the direction will be different at different positions on the track.