Question

A particle is moving in $x-y$ plane. The positive vector of the particle at time $t$ is $r=b(1-\cos \omega t)\hat{i}+b\sin \omega t\hat{j}$, where $b$ and $\omega$ are positive constants. The velocity of the particle at instant $t$ is

• A. $v=\left(b\omega \sin \omega t\right)\hat{i}-\left(b\omega \cos \omega t\right)\hat{j}$
• B. $v=\left(b\omega \cos \omega t\right)\hat{i}+\left(b\omega \sin \omega t\right)\hat{j}$
• C. $v=\left(b\omega \sin \omega t\right)\hat{i}+\left(b\omega \cos \omega t\right)\hat{j}$
• D. None of these

Answer 1

The correct option is C $v=\left(b\omega \sin \omega t\right)\hat{i}+\left(b\omega \cos \omega t\right)\hat{j}$

Given, displacement
$r=b(1-\cos \omega t)\hat{i}+b\sin \omega t\hat{j}$
Velocity,
$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=b\omega \sin \omega t\hat{i}+b\omega \cos \omega t\hat{j}$