Question 114: Motion in 2D & Vectors

Question

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 :

Accepted Answer

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

Answer

$\overrightarrow v $ and $\overrightarrow a $ both are perpendicular to $\overrightarrow r $

Answer

$\overrightarrow v $ and $\overrightarrow a $ both are parallel to $\overrightarrow r $

Answer

$\overrightarrow v $ is perpendicular to  $\overrightarrow r $ and $\overrightarrow a $ is directed
towards the origin

Answer

$\overrightarrow v $ is perpendicular to $\overrightarrow r $ and $\overrightarrow a $ is directed
away from the origin

Motion in 2D & Vectors

Choose the correct answer: