Question

The position vector of the particle is $r\left(t\right)=acos \omega t\hat{i}+asin⁡\omega t\hat{j}$ , where $a$ and $\omega $ are real constants of suitable dimensions. The acceleration is

• A. Perpendicular to the velocity
• B. Parallel to the velocity
• C. Directed away from the origin
• D. Perpendicular to the position vector
• E. Towards the origin

Answer 1

Answer: (A)

Given that, $r(t)=a \cos \omega t \hat{i}+a \sin \omega t \hat{j}$
$\because \vec{v}=\frac{d r(t)}{d t}=-a \omega \sin \omega t \hat{i}+a \omega \cos \omega t \hat{i} j$
$\vec{a}=\frac{d \vec{v}}{d t}=-a \omega^2 \cos \omega t \hat{i}-a \omega^2 \cos \omega t \hat{j}$
$\vec{a} \cdot \vec{v}=a^2 \omega^3 \sin \omega t \cos \omega t-a^2 \omega^3 \sin \omega t \cos \omega t$
$\Rightarrow \vec{a} \cdot \vec{v}=0$
Above result implies that acceleration is perpendicular to velocity.