A particle moves such that its position vector $\vec{r}(t) = \cos\omega t\,\hat{i} + \sin\omega t\,\hat{j}$ where $\omega$ is a constant and $t$ is time. Then which of the following statements is true for the velocity $\vec{v}(t)$ and acceleration $\vec{a}(t)$ of the particle:
Answer: (D) $\vec{v}$ is perpendicular to $\vec{r}$ and $\vec{a}$ is directed towards the origin
$$\vec{v} = \frac{d\vec{r}}{dt} = \omega(-\sin\omega t\,\hat{i} + \cos\omega t\,\hat{j})$$
$\vec{v}\cdot\vec{r} = \omega(-\sin\omega t\cos\omega t + \cos\omega t\sin\omega t) = 0$, so $\vec{v} \perp \vec{r}$.
$$\vec{a} = \frac{d\vec{v}}{dt} = -\omega^2(\cos\omega t\,\hat{i} + \sin\omega t\,\hat{j}) = -\omega^2\vec{r}$$
So $\vec{a}$ points opposite to $\vec{r}$, i.e. towards the origin (uniform circular motion).
Solution by Sreeraj P, M.Sc Physics