NEET UG 2016 — Physics Mechanics

2016mcqmedium

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?

Official previous-year question

Held on 30 Apr 2016 · Verified 9 Jul 2026.

Options

A
Velocity and acceleration both are perpendicular to $\vec{r}$ .
B
Velocity and acceleration both are parallel to $\vec{r}$ .
C
Velocity is perpendicular to $\vec{r}$ and acceleration is directed towards the origin.
D
Velocity is perpendicular to $\vec{r}$ and acceleration is directed away from the origin.

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Solution

Position vector is, $\vec{r}=\mathrm{cos}⁡\omega t\hat{x}+\mathrm{sin}⁡\omega t\hat{y}$

Velocity of particle is $\vec{v}=\frac{d\vec{r}}{dt}$

$\vec{v}=-sin\omega t\times \omega \hat{x}+cos\omega t\times \omega \hat{y}$

$\Rightarrow \vec{v}=\omega (-sin\omega t\hat{x}+cos\omega t\hat{y})$

Acceleration of the particle is

$\vec{a}=\frac{d\vec{v}}{dt}$

$\Rightarrow \vec{a}=-{\omega }^{2}(cos\omega t\hat{x}+sin\omega t\hat{y})$

$\Rightarrow \vec{a}=-{\omega }^{2}\vec{r}$ ,

So direction of $\vec{r}$ and $\vec{a}$ are opposite.

As $\vec{v}\cdot \vec{a}=0\Rightarrow \vec{v}\perp \vec{a}$ and

$\vec{v}\cdot \vec{r}=0\Rightarrow \vec{v}\perp \vec{r}$ ,

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

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Options

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

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

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

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

Solution

Position vector is, $\vec{r}=\mathrm{cos}⁡\omega t\hat{x}+\mathrm{sin}⁡\omega t\hat{y}$

Velocity of particle is $\vec{v}=\frac{d\vec{r}}{dt}$

$\vec{v}=-sin\omega t\times \omega \hat{x}+cos\omega t\times \omega \hat{y}$

$\Rightarrow \vec{v}=\omega (-sin\omega t\hat{x}+cos\omega t\hat{y})$

Acceleration of the particle is

$\vec{a}=\frac{d\vec{v}}{dt}$

$\Rightarrow \vec{a}=-{\omega }^{2}(cos\omega t\hat{x}+sin\omega t\hat{y})$

$\Rightarrow \vec{a}=-{\omega }^{2}\vec{r}$ ,

So direction of $\vec{r}$ and $\vec{a}$ are opposite.

As $\vec{v}\cdot \vec{a}=0\Rightarrow \vec{v}\perp \vec{a}$ and

$\vec{v}\cdot \vec{r}=0\Rightarrow \vec{v}\perp \vec{r}$ ,

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

NEET UG 2016 — Physics Mechanics

Options

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Solution

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NEET UG 2016 — Physics Mechanics

Options

Did you get this right?

Solution

Your note

NEET UG Mechanics in other years

More NEET UG Mechanics questions

Other Physics topics for NEET UG

Electromagnetism

Modern Physics

Thermodynamics

Optics

Waves & Oscillations

Keep practicing Mechanics