Question

A particle moving in a plane with velocity given by $\overrightarrow{u}=u_0\hat{i}+(a\omega \cos \omega t)\hat{j}$, where $\hat{i}$ and $\hat{j}$ are unit vectors along $x$ and $y$ axis, respectively. If the particle is at origin at $t=0$, The distance from origin at time $3\pi /2\omega$ is

• A. $a^2+{\omega }^2$
• B. ${[{\left(\frac{3\pi u_0}{2\omega }\right)}^2+a^2]}^{1/2}$
• C. $\sqrt{a^2+{\left(\frac{2}{3}\pi u_0\right)}^2}$
• D. $\sqrt{a^2+{\left(\frac{\pi u_0}{3}\right)}^2}$

Answer 1

The correct option is B ${[{\left(\frac{3\pi u_0}{2\omega }\right)}^2+a^2]}^{1/2}$

$Using,\overrightarrow{u_0}=\frac{d\overrightarrow{r}}{dt},\overrightarrow{r}=\int \overrightarrow{u_0}dt$
$\overrightarrow{r}=u_{0}t\hat{i}+a\sin wt\hat{j}+constant$
Sine $r=0att=0$
Therefore $constant=0$
$\overrightarrow{r}=u_{0}t\hat{i}+a\sin wt\hat{j}$

$att=\frac{3\pi }{2w}$

$\overrightarrow{r}=\frac{3u_0\pi }{2w}\hat{i}+a\sin \frac{3\pi }{2}\hat{j}$

$\overrightarrow{r}=\frac{3u_0\pi }{2w}\hat{i}+a\hat{j}$
$\left|\overrightarrow{r}\right|=\sqrt{{\left(\frac{3u_0\pi }{2w}\right)}^2+a^2}$
$\text{hence option B}$