A) parallel to position vector
B) perpendicular to position vector
C) along the origin
D) along to opposite of origin
Correct Answer: B
Solution :
\[\overrightarrow{r}=(a\cos \omega t)\hat{i}+(a\sin \omega t)\hat{j}\] \[\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=-(a\omega \sin \omega t)\hat{i}+(a\omega \cos \omega t)\hat{j}\] \[{{m}_{1}}=\frac{a\sin \omega t}{a\cos \omega t}=\tan \omega t\] \[{{m}_{2}}=\frac{a\omega \cos \omega t}{-a\omega \sin \omega t}=-\cot \omega t\] \[\therefore \] \[{{m}_{1}}{{m}_{2}}=-1\] Therefore, velocity of particle will be perpendicular to position vector.
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