Question

A point moves in the plane $xy$ according to the law $x=\alpha \sin \omega t$, $y=\alpha (1-\cos \omega t)$, where $\alpha$ and $\omega$ are positive constants and $t$ is time. The distance traversed by the point in time $t_0$ is $\frac{4\alpha \omega t_0}{x}$. Find $x$.

Answer 1

Given, $x=\alpha \sin \omega t$
$v_x=\frac{dx}{dt}$
$=\alpha \cos \omega t\cdot \omega$
$=\alpha \omega \cos \omega t$

$y=\alpha (1-\cos \omega t)$
$v_y=\frac{dy}{dt}=\alpha [0-(-\sin \omega t\left)\omega \right]=\alpha \omega \sin \omega t$

$v=\sqrt{v_x^2+v_y^2}=\alpha \omega =$constant

Distance in time $t_0$
$s=v{t}_0$
$=\alpha \omega t_0$