A) Parallel to the position vector
B) Perpendicular to the position vector
C) Directed towards the origin
D) Directed away from the origin
Correct Answer: B
Solution :
\[\vec{r}=(a\cos \omega \,t)\hat{i}+(a\sin \omega \,t)\hat{j}\] \[\vec{v}=\frac{d\vec{r}}{dt}=-a\omega \sin \omega \,t\,\hat{i}+a\omega \cos \omega \,t\,\hat{j}\] As \[\vec{r}.\vec{v}=0\] therefore velocity of the particle is perpendicular to the position vector.
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