Question

A point moves on the $x-y$ plane according to the law $x=a\sin \omega t$ and $y=a(1-\cos \omega t)$ where $a$ and $\omega$ are positive constants and $t$ is in seconds. Find the distance covered in time $t_0$.

• A. $a\omega t_0$
• B. $\sqrt{2a^2+2a^2\cos \omega t_0}$
• C. $2a\sin \frac{\omega t_0}{2}$
• D. $2a\cos \frac{\omega t_0}{2}$

Answer 1

The correct option is A $a\omega t_0$

Position vector of the point $\overrightarrow{r}=a\sin wt\hat{i}+a(1-\cos wt)\hat{j}$

Velocity vector $\overrightarrow{v}=\frac{\overrightarrow{dr}}{dt}=aw\cos wt\hat{i}+aw\sin wt\hat{j}$

Distance covered in time $t_o$ by the point is the integral of speed or magnitude of velocity vector

$={\int }_0^{t_0}\left|\overrightarrow{v}\right|dt={\int }_0^{t_0}\sqrt{{\left(aw\cos wt\right)}^2+{\left(aw\sin wt\right)}^2}dt={\int }_0^{t_0}awdt=aw{t}_o$