Question

What is the distance travelled by a point during the time t. if it moves in X-Y plane according to relation.
$X=asin\omega t$
$Y=a(1-cos\omega t)$

Answer 1

Given that,

$x=a\sin \omega t.....\left(I\right)$

$y=a(1-\cos \omega t)$

$y=a-a\cos \omega t.....\left(II\right)$

Now, the velocity for x

On differentiate of equation (I)

$v_x=\frac{dx}{dt}$

$v_x=a\omega \cos \omega t$

Now again differentiate of equation (II)

The velocity for y

$v_y=\frac{dy}{dt}$

$v_y=a\omega \sin \omega t$

Now the velocity is

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

$v=\sqrt{a^2{\omega }^2({\sin }^2\omega t+{\cos }^2\omega t)}$

$v=a\omega$

Now, the distance travelled in time t

$D=v\times t$

$D=a\omega t$

Hence, the distance travelled by a point during the time t is $a\omega t$