A particle moves so that its position vector is given by → r = cos ^ x + sin ^ y , where ω is a constant. Which of the following is true?

Correct answer: C. Velocity is perpendicular to → r and acceleration is directed towards the origin. Position vector is, → r = cos ω t ^ x + sin ω t ^ y Velocity of particle is → v = d → r dt → v = -sin ω t × ω ^ x + cos × ω ^ y ⇒ → v = ω(- sin ^ x + cos ^ y ) Acceleration of the particle is → a = d → v dt ⇒ → a = - ω ^ 2 (cos ^ x + sin ^ y ) ⇒ → a = - ω ^ 2 → r , So direction of → r and → a are opposite. As → v · → a = 0 ⇒ → v ⊥ → a and → v · → r = 0 ⇒ → v ⊥ → r , So, velocity is perpendicular to → r and acceleration is directed towards the origin.

NEET 2016

Exam: NEET
Year: 2016
Subject: Physics
Chapter: Motion in One Dimension

NEET 2016 · Physics · Motion in One Dimension

A particle moves so that its position vector is given by → r = cos...

A particle moves so that its position vector is given by

\overset{\to}{r} = cos\omegat\overset{^}{x} + sin\omegat\overset{^}{y}

, where

\omega

is a constant. Which of the following is true?

Options

AVelocity and acceleration both are perpendicular to $\overset{\to}{r}$ .
BVelocity and acceleration both are parallel to $\overset{\to}{r}$ .
CVelocity is perpendicular to $\overset{\to}{r}$ and acceleration is directed towards the origin.
DVelocity is perpendicular to $\overset{\to}{r}$ and acceleration is directed away from the origin.

Correct answer

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

Step-by-step solution

Position vector is,

\overset{\to}{r} = cos \omega t \overset{^}{x} + sin \omega t \overset{^}{y}

Velocity of particle is

\overset{\to}{v} = \dfrac{d\overset{\to}{r}}{dt}

\overset{\to}{v} = -sin \omega t \times \omega\overset{^}{x} + cos \omegat \times \omega\overset{^}{y}

⇒ \overset{\to}{v} = \omega\left(- sin \omegat\overset{^}{x} + cos \omegat\overset{^}{y}\right)

Acceleration of the particle is

\overset{\to}{a} = \dfrac{d\overset{\to}{v}}{dt}

⇒ \overset{\to}{a} = - {\omega}^{2}\left(cos \omegat\overset{^}{x} + sin \omegat\overset{^}{y}\right)

⇒ \overset{\to}{a} = - {\omega}^{2}\overset{\to}{r}

,

So direction of

\overset{\to}{r}

and

\overset{\to}{a}

are opposite.

As

\overset{\to}{v} \cdot \overset{\to}{a} = 0 ⇒ \overset{\to}{v} ⊥ \overset{\to}{a}

and

\overset{\to}{v} \cdot \overset{\to}{r} = 0 ⇒ \overset{\to}{v} ⊥ \overset{\to}{r}

,

So, velocity is perpendicular to

\overset{\to}{r}

and acceleration is directed towards the origin.

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