Question

A particle moves such that the 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?
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

In the above question we basically first have to determine the velocity and the acceleration of the above-given particles. The velocity of any particle is given as the derivative of its position vector and acceleration is given as the double derivative of the position vector. Once we obtain them, we will verify their direction with respect to their position vector and accordingly draw the necessary conclusion.

Formula used:
$v=\dfrac{dr}{dt}$
$a=\dfrac{{{d}^{2}}r}{d{{t}^{2}}}$
$A\cdot B=0\Rightarrow A\bot B$

Complete step-by-step solution:
In the above question the position vector of a particle with respect to origin is given to us as,
$\overrightarrow{r}=\cos \omega t\widehat{x}+\sin \omega t\widehat{y}$
The first derivative of the above position vector will give us the velocity ‘v’ of the particle and that the second derivative will give the acceleration ‘a’ of the particle. Taking the first derivative and the second derivative of the above position vector we get,
$\begin{aligned} & \overrightarrow{r}=\cos \omega t\widehat{x}+\sin \omega t\widehat{y} \\\ & \dfrac{d\overrightarrow{r}}{dt}=\overrightarrow{v}=\dfrac{d(\cos \omega t\widehat{x}+\sin \omega t\widehat{y})}{dt} \\\ & \Rightarrow \overrightarrow{v}=\dfrac{d(\cos \omega t)}{dt}\widehat{x}+\dfrac{d(\sin \omega t)}{dt}\widehat{y}\text{, }\because \dfrac{d(\cos \omega t)}{dt}=-\sin \omega t\text{ and }\dfrac{d(\sin \omega t)}{dt}=\cos \omega t \\\ & \Rightarrow \overrightarrow{v}=-\sin \omega t\widehat{x}+\cos \omega t\widehat{y} \\\ & \dfrac{{{d}^{2}}\overrightarrow{r}}{d{{t}^{2}}}=\overrightarrow{a}=\dfrac{d\overrightarrow{v}}{dt} \\\ & \Rightarrow \overrightarrow{a}=\dfrac{dv}{dt}=\dfrac{d(-\sin \omega t\widehat{x}+\cos \omega t\widehat{y})}{dt} \\\ & \Rightarrow \overrightarrow{a}=\dfrac{d(-\sin \omega t)}{dt}\widehat{x}+\dfrac{d(\cos \omega t)}{dt}\widehat{y} \\\ & \Rightarrow \overrightarrow{a}=-\cos \omega t\widehat{x}-\sin \omega t\widehat{y} \\\ & \Rightarrow \overrightarrow{a}=-(\cos \omega t\widehat{x}+\sin \omega t\widehat{y})=-\overrightarrow{r} \\\ \end{aligned}$
In the above result obtained if we see the acceleration of the particle it is equal to the negative of the position vector of the particle. The position vector of the particle is directed away from the origin, therefore we can conclude that the acceleration of the above particle will be directed inwards.
Let us say there are two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$. If their dot product is zero, we can say that they are perpendicular. Mathematically this can be written as, $A\cdot B=0\Rightarrow A\bot B$
If we take the dot product of the position vector and velocity vector of the above particle we get,
$\begin{aligned} & \overrightarrow{r}\cdot \overrightarrow{v} \\\ & \Rightarrow (\cos \omega t\widehat{x}+\sin \omega t\widehat{y})\cdot (-\sin \omega t\widehat{x}+\cos \omega t\widehat{y}) \\\ & \Rightarrow -\sin \omega t\cos \omega t+\sin \omega t\cos \omega t=0 \\\ \end{aligned}$

Hence we can say that the velocity and the position vector of the particle are perpendicular to each other.nce the correct answer of the above question is option (c).

Note: While taking the dot product, it is to be noted that we basically multiply the components of the same unit vector to each other. If we observe the above equation of position vector, it basically represents the locus of points which remains fixed from the origin. Hence we can say that the particle moves in a circle. For a particle moving in a circular path, we can say that the velocity is always perpendicular to the radius, and acceleration is always directed towards the center.