Question

A particle is moving in a plane with a velocity given by $\overrightarrow{v}=\hat{i}{u}_0+\hat{j}a\omega \cos \omega t$ if the particle is at origin at $t=0$. Distance of the particle from origin at time $3\pi /2\omega$ is

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

Answer 1

Answer: (A)

$\sqrt{a^2+(3\pi u_0/2\omega {)}^2}$

Comparing the given equation with $\overrightarrow{v}=\hat{i}{v}_x+\hat{j}{v}_y$, we get
$v_x=u_0$ and $\frac{dy}{dt}=a\omega \cos \omega t$
or $\frac{dx}{dt}=u_0$ and $\frac{dy}{dt}=a\omega \cos \omega t$.
Integrating $x=\int u_{0}dt$ and $y=\int a\omega \cos \omega dt$ or $x=u_{0}t+c_1$ and $y=a\sin \omega t+c_2$
At $t=0,x=0$ and $y=0$ we get,
$c_1=c_2=0$
$\therefore x=u_{0}t$ and $y=a\sin \omega t$ but $t=3\pi /2\omega$
$\therefore x=u_0\left(3\pi /2\omega \right)$ and
$y=-a$
Then distance from origin, $d=\sqrt{x^2+y^2}=\sqrt{a^2+(3\pi u_0/2\omega {)}^2}$