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(B) Given the position vector: $\vec{r} = \cos(\omega t) \hat{i} + \sin(\omega t) \hat{j}$.$1$. Velocity $\vec{v}$ is the derivative of position with

Vedclass · Accepted Answer

Velocity is perpendicular to $\vec{r}$ and acceleration is directed towards the origin.

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(B) Given the position vector: $\vec{r} = \cos(\omega t) \hat{i} + \sin(\omega t) \hat{j}$.$1$. Velocity $\vec{v}$ is the derivative of position with respect to time:$\vec{v} = \frac{d\vec{r}}{dt} = -\omega \sin(\omega t) \hat{i} + \omega \cos(\omega t) \hat{j}$.$2$. Acceleration $\vec{a}$ is the derivative of velocity with respect to time:$\vec{a} = \frac{d\vec{v}}{dt} = -\omega^2 \cos(\omega t) \hat{i} - \omega^2 \sin(\omega t) \hat{j} = -\omega^2 \vec{r}$.$3$. Since $\vec{a} = -\omega^2 \vec{r}$,the acceleration is directed towards the origin (centripetal acceleration).$4$. To check the relationship between $\vec{r}$ and $\vec{v}$,calculate the dot product:$\vec{r} \cdot \vec{v} = (\cos(\omega t))(-\omega \sin(\omega t)) + (\sin(\omega t))(\omega \cos(\omega t)) = -\omega \sin(\omega t) \cos(\omega t) + \omega \sin(\omega t) \cos(\omega t) = 0$.Since the dot product is $0$,the velocity is perpendicular to the position vector $\vec{r}$.

$A$ particle moves so that its position vector is given by $\vec{r} = \cos(\omega t) \hat{i} + \sin(\omega t) \hat{j}$,where $\omega$ is a constant. Which of the following is true?

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