Easy MCQ +4 / -1 PYQ · JEE Mains 2020

A particle moves such that its position vector $$\overrightarrow r \left( t \right) = \cos \omega t\widehat i + \sin \omega t\widehat j$$ where $\omega$ is a constant and t is time. Then which of the following statements is true for the velocity $\overrightarrow v \left( t \right)$ and acceleration $\overrightarrow a \left( t \right)$ of the particle :

  1. A $\overrightarrow v$ and $\overrightarrow a$ both are perpendicular to $\overrightarrow r$
  2. B $\overrightarrow v$ and $\overrightarrow a$ both are parallel to $\overrightarrow r$
  3. C $\overrightarrow v$ is perpendicular to $\overrightarrow r$ and $\overrightarrow a$ is directed towards the origin Correct answer
  4. D $\overrightarrow v$ is perpendicular to $\overrightarrow r$ and $\overrightarrow a$ is directed away from the origin

Solution

$$\overrightarrow r \left( t \right) = \cos \omega t\widehat i + \sin \omega t\widehat j$$ <br><br>$\overrightarrow v = {{d\overrightarrow r } \over {dt}}$ = $- \omega \sin \omega t\,\widehat i + \omega \cos \omega t\widehat j$ <br><br>$\overrightarrow a = {{d\overrightarrow v } \over {dt}}$ = $- {\omega ^2}\cos \omega t\,\widehat i - {\omega ^2}\sin \omega t\widehat j$ <br><br>= $$ - {\omega ^2}\left( {\cos \omega t\,\widehat i + \sin \omega t\widehat j} \right)$$ <br><br>= $- {\omega ^2}\overrightarrow r$ <br><br>$\therefore$ $\overrightarrow a$ is antiparallel to $\overrightarrow r$ and it's direction towards the origin. <br><br>$\overrightarrow v .\overrightarrow r =$ $$\omega \left( { - \sin \omega t\cos \omega t + \cos \omega t\sin \omega t} \right)$$ = 0 <br><br>So $\overrightarrow v \bot \overrightarrow r$.

About this question

Subject: Physics · Chapter: Kinematics · Topic: Motion in a Straight Line

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