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

Moderate

A

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

Velocity is perpendicular to $\vec{r}$ and acceleration is directed away from the origin
C

Velocity and acceleration both are perpendicular to $\vec{r}$
D

Velocity and acceleration both are parallel to $\vec{r}$

Solution

$Given, \vec{r} = \cos ω t \hat{x} + \sin ω t \hat{y}$
$∴ \vec{v} = \frac{d \vec{r}}{d t} = - ω \sin ω t \hat{x} + ω \cos ω t \hat{y}$
$\vec{a} = \frac{d \vec{v}}{d t} = - ω^{2} \cos ω t \hat{x} - ω^{2} \sin ω t \hat{y} = - ω^{2} \vec{r}$
$Since position vector (\vec{r}) is directed away from the origin, so,$
$acceleration (- ω^{2} \vec{r}) is directed towards the origin.$
also, $\Rightarrow \vec{r} ⊥ \vec{v}$
$\vec{r} \cdot \vec{v} = (\cos ω t \hat{x} + \sin ω t \hat{y}) \cdot (- ω \sin ω t \hat{x} + ω \cos ω t \hat{y})$
$= - ω \sin ω t \cos ω t + ω \sin ω t \cos ω t = 0$

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