A) \[\overrightarrow{v}\]and \[\overrightarrow{a}\]both are perpendicular to \[\overrightarrow{r}\]
B) \[\overrightarrow{v}\] is perpendicular to \[\overrightarrow{r}\] and \[\overrightarrow{a}\] is directed towards the origin
C) \[\overrightarrow{v}\] and \[\overrightarrow{a}\] both are parallel to \[\overrightarrow{r}\]
D) None of these
Correct Answer: B
Solution :
| [b] \[\vec{r}=cos\omega t\text{ }\hat{i}+sin\omega t\,\,\hat{j}\] |
| \[\vec{v}=\frac{d\vec{r}}{dt}=\omega \left( -\sin \omega t\,\,\hat{i}+\cos \omega t\,\,\hat{j} \right)\] |
| \[\vec{a}=\frac{d\vec{v}}{dt}=-{{\omega }^{2}}\left( \cos \omega t\,\,\hat{i}+\sin \omega t\,\,\hat{j} \right)\] |
| \[\vec{a}=-{{\omega }^{2}}\vec{r}\] \[\therefore \,\,\vec{a}\,\,is\,\,antiparallel\,\,to\,\,\vec{r}\] |
| Also \[\vec{v}\cdot \vec{r}=0\] \[\therefore \,\,\,\,\,\vec{v}\bot \vec{r}\] |
| Actually particle is in state of uniform circular motion. |
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