Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

The velocity and acceleration both are parallel to $\vec{r}$ .

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

The velocity is parallel to $\vec{r}$ and acceleration is directed away from the origin.

The velocity and acceleration both are perpendicular to $\vec{r}$ .

Step-by-Step Solution

Velocity Calculation: Velocity is the time derivative of the position vector $\vec{r}$ . $\vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}[\cos(\omega t)\hat{x} + \sin(\omega t)\hat{y}] = -\omega \sin(\omega t)\hat{x} + \omega \cos(\omega t)\hat{y}$
Check Perpendicularity: Calculate the dot product of $\vec{r}$ and $\vec{v}$ . $\vec{r} \cdot \vec{v} = (\cos \omega t)(-\omega \sin \omega t) + (\sin \omega t)(\omega \cos \omega t) = -\omega \sin \omega t \cos \omega t + \omega \sin \omega t \cos \omega t = 0$ Since the dot product is zero, velocity $\vec{v}$ is perpendicular to position vector $\vec{r}$ .
Acceleration Calculation: Acceleration is the time derivative of velocity. $\vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}[-\omega \sin(\omega t)\hat{x} + \omega \cos(\omega t)\hat{y}]$ $\vec{a} = -\omega^2 \cos(\omega t)\hat{x} - \omega^2 \sin(\omega t)\hat{y} = -\omega^2 [\cos(\omega t)\hat{x} + \sin(\omega t)\hat{y}]$ $\vec{a} = -\omega^2 \vec{r}$
Direction of Acceleration: The negative sign indicates that acceleration $\vec{a}$ is directed opposite to the position vector $\vec{r}$ . Since $\vec{r}$ points away from the origin, $\vec{a}$ is directed towards the origin .

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