JEE MAIN - Physics (2020 - 8th January Evening Slot - No. 3)
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 :
$$\overrightarrow v $$ and $$\overrightarrow a $$ both are perpendicular to $$\overrightarrow r $$
$$\overrightarrow v $$ and $$\overrightarrow a $$ both are parallel to $$\overrightarrow r $$
$$\overrightarrow v $$ is perpendicular to $$\overrightarrow r $$ and $$\overrightarrow a $$ is directed
towards the origin
$$\overrightarrow v $$ is perpendicular to $$\overrightarrow r $$ and $$\overrightarrow a $$ is directed
away from the origin
Explanation
$$\overrightarrow r \left( t \right) = \cos \omega t\widehat i + \sin \omega t\widehat j$$
$$\overrightarrow v = {{d\overrightarrow r } \over {dt}}$$ = $$ - \omega \sin \omega t\,\widehat i + \omega \cos \omega t\widehat j$$
$$\overrightarrow a = {{d\overrightarrow v } \over {dt}}$$ = $$ - {\omega ^2}\cos \omega t\,\widehat i - {\omega ^2}\sin \omega t\widehat j$$
= $$ - {\omega ^2}\left( {\cos \omega t\,\widehat i + \sin \omega t\widehat j} \right)$$
= $$ - {\omega ^2}\overrightarrow r $$
$$ \therefore $$ $$\overrightarrow a $$ is antiparallel to $$\overrightarrow r $$ and it's direction towards the origin.
$$\overrightarrow v .\overrightarrow r = $$ $$\omega \left( { - \sin \omega t\cos \omega t + \cos \omega t\sin \omega t} \right)$$ = 0
So $$\overrightarrow v \bot \overrightarrow r $$.
$$\overrightarrow v = {{d\overrightarrow r } \over {dt}}$$ = $$ - \omega \sin \omega t\,\widehat i + \omega \cos \omega t\widehat j$$
$$\overrightarrow a = {{d\overrightarrow v } \over {dt}}$$ = $$ - {\omega ^2}\cos \omega t\,\widehat i - {\omega ^2}\sin \omega t\widehat j$$
= $$ - {\omega ^2}\left( {\cos \omega t\,\widehat i + \sin \omega t\widehat j} \right)$$
= $$ - {\omega ^2}\overrightarrow r $$
$$ \therefore $$ $$\overrightarrow a $$ is antiparallel to $$\overrightarrow r $$ and it's direction towards the origin.
$$\overrightarrow v .\overrightarrow r = $$ $$\omega \left( { - \sin \omega t\cos \omega t + \cos \omega t\sin \omega t} \right)$$ = 0
So $$\overrightarrow v \bot \overrightarrow r $$.
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