Question

A particle moves so that its position vector is given by \[\overrightarrow r = \cos \omega t\overrightarrow x + \sin \omega t\overrightarrow y \], where $\omega $ is a constant. Which of the following is true?
A) Velocity and acceleration both are perpendicular to $\overrightarrow r $.
B) Velocity and acceleration both are parallel to $\overrightarrow r $.
C) Velocity is perpendicular to $\overrightarrow r $ and acceleration is directed towards the origin.
D) Velocity is perpendicular to $\overrightarrow r $ and acceleration is directed away from the origin.

Answer 1

Hint: In order to find the solution of the given question, first of all we need to find the value of velocity and acceleration of the particle. After that we need to find the relation of the direction between the acceleration and velocity of the particle. Then finally, we can conclude with the correct solution for the given question.

Complete step by step solution:
The value of position vector in the question is given as, \[\overrightarrow r = \cos \omega t\overrightarrow x + \sin \omega t\overrightarrow y \]
First of all we need to find the velocity of the particle. For that we need to differentiate the position vector with respect to time. So, we can write,
$\Rightarrow v = \dfrac{{dr}}{{dt}} = \dfrac{{d(\cos \omega t\overrightarrow x + \sin \omega t\overrightarrow y )}}{{dt}}$
$\Rightarrow v = ( - \sin \omega t)\omega \overrightarrow x + (\cos \omega t)\omega \overrightarrow y $
$\Rightarrow v = - \omega (\sin \omega t\overrightarrow x - \cos \omega t\overrightarrow y )$
Now, we need to find the acceleration of the particle. So, we can write it as,
$\Rightarrow a = \dfrac{{dv}}{{dt}} = \dfrac{{d( - \omega \sin \omega t\overrightarrow x + \omega \cos \omega t\overrightarrow y )}}{{dt}}$
$ \Rightarrow a = - {\omega ^2}\cos \omega t\overrightarrow x - {\omega ^2}\sin \omega t\overrightarrow y $
$ \Rightarrow a = - {\omega ^2}(\cos \omega t\overrightarrow x - \sin \omega t\overrightarrow y )$
$\therefore a = - {\omega ^2}r = {\omega ^2}( - r)$
So, we can say that the acceleration is directed towards $ - r$ that is towards the origin.
Now, let us find the relation between the velocity vector and the position vector.
$\Rightarrow \overrightarrow v .\overrightarrow r = - \omega (\sin \omega t\overrightarrow x - \cos \omega t\overrightarrow y ).(\cos \omega t\overrightarrow x + \sin \omega t\overrightarrow y )$
$ \Rightarrow \overrightarrow v .\overrightarrow r = - \omega (\sin \omega .\cos \omega t + 0 + 0 - \sin \omega t.\cos \omega t)$
$\therefore \overrightarrow v .\overrightarrow r = - \omega (0) = 0$
Since the product of velocity vector and the position vector is zero, we can conclude that both the velocity vector and the perpendicular vectors are perpendicular to each other.

Note: We know that velocity of a body is the ratio of the distance travelled by that body to the corresponding time taken whereas acceleration is the change in the velocities of the body when it is in motion. Also if the product of acceleration and position vector comes to zero, then we can say that both the acceleration and the position vectors are parallel to each other.