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 distance of the particle at instant $t=\frac{\pi }{3}$ is

• A. $2\pi b$
• B. $3\pi b$
• C. None of these
• D. $\pi b$

Answer 1

The correct option is D $\pi b$

Given, displacement
$r=b(1-\cos \omega t)\hat{i}+b\sin \omega t\hat{j}$
Velocity,
$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=b\omega \sin \omega t\hat{i}+b\omega \cos \omega t\hat{j}$
At $t=\frac{\pi }{3}$
$v_x=b\omega \sin \pi =0$
$v_y=b\omega \cos \pi =-b\omega$
Net speed,
$\therefore v=\sqrt{v_x^2+v_y^2}=\sqrt{0^2+(-b\omega {)}^2}=b\omega )$
Distance $s={\int }_0^{t}vdt$
$s={\int }_0^{\pi /\omega }b\omega dt$
$s={\left[b\omega t\right]}_0^{\pi /\omega }$
$s=b\pi$