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A particle moves such that its position vector $\overrightarrow{\mathrm{r}}(\mathrm{t})=\cos \omega \mathrm{t} \hat{\mathrm{i}}+\sin \omega \mathrm{t} \hat{\mathrm{j}}$ where $\omega$ is a constant and $t$ is time. Then which of the following statements is true for the velocity $\overrightarrow{\mathrm{v}}(\mathrm{t})$ and acceleration $\overrightarrow{\mathrm{a}}(\mathrm{t})$ of the particle
$\overrightarrow{\mathrm{v}}$ is perpendicular to $\overrightarrow{\mathrm{r}}$ and $\overrightarrow{\mathrm{a}}$ is directed towards the origin
$\overrightarrow{\mathrm{v}}$ and $\overrightarrow{\mathrm{a}}$ both are parallel to $\overrightarrow{\mathrm{r}}$
$\overrightarrow{\mathrm{v}}$ and $\overrightarrow{\mathrm{a}}$ both are perpendicular to $\overrightarrow{\mathrm{r}}$
$\overrightarrow{\mathrm{v}}$ is perpendicular to $\overrightarrow{\mathrm{r}}$ and $\overrightarrow{\mathrm{a}}$ is directed away from the origin
Solution
$\overrightarrow{\mathrm{r}}(\mathrm{t})=\cos \omega \hat{\mathrm{i}}+\sin \omega \mathrm{t} \hat{\mathrm{j}}$
On diff. we get
${\overrightarrow{\mathrm{v}}=-\omega \sin \omega \mathrm{t} \hat{\mathrm{i}}+\omega \cos \omega \mathrm{t} \hat{\mathrm{j}}}$
${\overrightarrow{\mathrm{a}}=-\omega^{2} \overrightarrow{\mathrm{r}}}$
${\overrightarrow{\mathrm{v}} \cdot \overrightarrow{\mathrm{r}}=0}$
