A particle is moving in a plane with a velocity given by, $\overrightarrow{u}={\text{\hspace{0.17em}u}}_0\widehat{i} +\left(\omega a\text{\hspace{0.17em}cos\hspace{0.17em}}\omega \text{t}\right)\widehat{j}$ , where $\widehat{i}$ and  $\widehat{j}$ are unit vectors along x and y-axes respectively. If the particle is at the origin at t = 0, then its displacement from the origin at time $\text{t}=\frac{\text{3π}}{\text{2}\omega }$ will be-

Let ${\text{u}}_X$ and $u_y$ be the components of the velocity of the particle along the

x- and y-directions. Then

$$\begin{gathered} {\text{u}}_{\text{x}}\text{= }\frac{\text{dx}}{dt}{\text{= u}}_0 \\ x=u_{0}t ; substitute given value of t=\frac{3\mathrm{\pi }}{2\omega } \\ x=u_0\frac{3\mathrm{\pi }}{2\omega }---\left(1\right) \\ {\text{ u}}_{\text{y}}\text{= }\frac{\text{dy}}{dt}=\omega \text{a cos }\omega \text{t} \\ dy=(\omega \text{a cos }\omega \text{t)dt} \\ integrating \\ \int dy=\int (\omega \text{a cos }\omega \text{t)dt} \\ y=wa\left(\frac{\sin \omega t}{\omega }\right) \\ y=a\sin \omega t \\ y=a\sin \omega \frac{3\mathrm{\pi }}{2\omega } \\ y=-a---\left(2\right) \end{gathered}$$

$\therefore$ The displacement of the particle from the

origin is $\sqrt{x^2+y^2}=\sqrt{\left[{\left(\frac{3\pi u_0}{2\omega }\right)}^2+a^2\right]}$

Hence correct answer is (A)