Question 178: Motion in 2D & Vectors

Question

A particle moves so that its position vector is given by $$\overrightarrow r  = \cos \omega t\,\widehat x + \sin \,\omega t\,\widehat y,$$ where $$\omega $$ is a constant.
Which of the following is true?

Accepted Answer

$$\overrightarrow r  = \cos \omega t\,\widehat x + \sin \,\omega t\,\widehat y,$$
$$ 	herefore $$ $$\overrightarrow v  =  - \omega \sin \omega t\widehat x + \omega \cos \omega t\widehat y$$
and $$\overrightarrow a  =  - {\omega ^2}\cos \omega t\widehat x - {\omega ^2}\sin \omega t\widehat y$$ = $$ - {\omega ^2}\overrightarrow r $$
$$\overrightarrow r .\overrightarrow v $$ = 0
$$ \Rightarrow $$ $$\overrightarrow r  \bot \overrightarrow v $$
As position vector $$\left( {\overrightarrow r } ight)$$ is directly away from the origin, so, acceleration ($$ - {\omega ^2}\overrightarrow r $$) is directed towards the origin.

Answer

Velocity is perpendicular to $$\overrightarrow r $$ and acceleration is directed towards the origin.

Answer

Velocity  is perpendicular to $$\overrightarrow r $$ and acceleration is directed away  from the origin.

Answer

Velocity and acceleration both are perpendicular to $$\overrightarrow r $$

Answer

Velocity  and acceleration  both are parallel to $$\overrightarrow r $$

Motion in 2D & Vectors

A particle moves so that its position vector is given by $\overrightarrow r = \cos \omega t\,\widehat x + \sin \,\omega t\,\widehat y, where \omega $ is a constant. Which of the following is true?

Choose the correct answer: