Physics - Circular Motion Question with Solution | TestHub

In time $t$ the particle has rotated through an angle $\theta =\omega t$

This means the magnitude of the displacement of the particle is
$s=\mathit{\text{PQ}}=\sqrt{{\mathit{\text{QR}}}^2+{\mathit{\text{PR}}}^2}$

$=\sqrt{{\left(\mathit{\text{a sin}}\omega t\right)}^2+{\left(a-\text{a cos}\omega t\right)}^2}$

$s=2a \text{sin}\frac{\omega t}{2}$