Question

The position vector of the particle is $\vec{r} (t) = a \, \cos \omega t \hat{i} + a \, \sin \, \omega t \, \hat{j}$, 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

Answer 1

Answer: (A)

perpendicular to the velocity

Answer 2

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