Question

The position vector of a moving particle in $x-y$ plane can be represented as $\overrightarrow{r}=\left(2t\hat{i}+2t^2\hat{j}\right)m$, where $t$ represents the time instant. The rate of change of $\theta$ at time $t=2s$. (where $\theta$ is the angle which its velocity vector makes with positive $x-$axis)

• A. $\frac{2}{5}rad/s$
• B. $\frac{2}{17}rad/s$
• C. $2rad/s$
• D. $4rad/s$

Answer 1

The correct option is B $\frac{2}{17}rad/s$

Here the position vector is shown in the diagram below

$x=2t\Rightarrow v_x=\frac{dx}{dt}=2y=2t^2\Rightarrow v_y=\frac{dy}{dt}=4t$
$\therefore$ $\tan \left(\theta \right)=\frac{v_y}{v_x}=\frac{4t}{2}=2t$
Diffrentiating with respect to time we get,
$\left(se{c}^2\theta \right)$$\frac{d\theta }{dt}=2$
or $\left(1+{\tan }^2\left(\theta \right)\right)\frac{d\theta }{dt}=2,$or $\left(1+4t^2\right)\frac{d\theta }{dt}=2$
$\therefore \frac{d\theta }{dt}=\frac{2}{1+4(2{)}^2}=\frac{2}{17}rad/s$

For detailed solution watch the next video.