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 vector is rate of change of position vector. 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.You need to login to perform this action.
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