A) parallel to position vector
B) perpendicular to position vector
C) directed towards the origin
D) directed away from the origin
Correct Answer: B
Solution :
Key Idea: Velocity is the rate of change of position vector, \[\vec{r}=a(\cos \omega t)\hat{i}+(a\sin \omega t)\hat{j}\] Velocity vector, \[\vec{v}=\frac{d\vec{r}}{dt}\] \[=\frac{d}{dt}[(a\cos \omega t)]\hat{i}+(a\sin \omega t)\hat{j}]\] \[=(-a\sin \omega t)\hat{i}+(a\cos \omega t)\hat{j}\] Now, \[\vec{v}.\vec{r}=[(-a\sin \omega t)\hat{i}+(a\cos \omega t)\hat{j}]\] \[[(a\cos \omega t)\hat{i}+(a\sin \omega t)\hat{j}]\] \[=(-a\sin \omega t)(a\cos \omega t)+(a\cos \omega t)(a\sin \,\omega t)\] \[=-{{a}^{2}}\sin \omega t\,\cos \omega t+{{a}^{2}}\cos \omega t\sin \omega t\] \[=0\] As dot product of velocity vector with position vector is zero, so velocity vector of the particle is perpendicular to position vector.
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