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(A) Given the position vector: $\overrightarrow{r}(t) = \cos \omega t \hat{i} + \sin \omega t \hat{j}$.To find the velocity $\overrightarrow{v}(t)$,we

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$\overrightarrow{v}$ is perpendicular to $\overrightarrow{r}$ and $\overrightarrow{a}$ is directed towards the origin.

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(A) Given the position vector: $\overrightarrow{r}(t) = \cos \omega t \hat{i} + \sin \omega t \hat{j}$.To find the velocity $\overrightarrow{v}(t)$,we differentiate $\overrightarrow{r}(t)$ with respect to time $t$:$\overrightarrow{v}(t) = \frac{d\overrightarrow{r}}{dt} = -\omega \sin \omega t \hat{i} + \omega \cos \omega t \hat{j}$.To find the acceleration $\overrightarrow{a}(t)$,we differentiate $\overrightarrow{v}(t)$ with respect to time $t$:$\overrightarrow{a}(t) = \frac{d\overrightarrow{v}}{dt} = -\omega^2 \cos \omega t \hat{i} - \omega^2 \sin \omega t \hat{j} = -\omega^2 (\cos \omega t \hat{i} + \sin \omega t \hat{j}) = -\omega^2 \overrightarrow{r}$.Now,check the dot product of $\overrightarrow{v}$ and $\overrightarrow{r}$:$\overrightarrow{v} \cdot \overrightarrow{r} = (-\omega \sin \omega t)(\cos \omega t) + (\omega \cos \omega t)(\sin \omega t) = 0$.Since the dot product is $0$,$\overrightarrow{v}$ is perpendicular to $\overrightarrow{r}$.Since $\overrightarrow{a} = -\omega^2 \overrightarrow{r}$,the acceleration vector is in the opposite direction of the position vector $\overrightarrow{r}$,meaning it is directed towards the origin.

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