Question

A particle moves in the $X-Y$ plane. The position vector of the particle at any time it' is $\overrightarrow{r}=\left(2t^2\hat{i}+3t^3\hat{j}\right)m$ . The rate of change of $\theta$ at time $t=4sec$ (where $\theta$ is the angle which its velocity vector makes with positive $X$ -axis) is

• A. $\frac{9}{656}rad{s}^{-1}$
• B. $\frac{9}{164}rad{s}^{-1}$
• C. $\frac{9}{328}rad{s}^{-1}$
• D. $\frac{9}{82}rad{s}^{-1}$

Answer 1

Answer: (C)

Component of displacement vector along axes will be,
$x=2t^2,y=3t^3$
Then,
$v_x=\frac{dx}{dt}=4t\&v_y=\frac{dy}{dt}=9t^2$
$tan\theta =\frac{v_y}{v_x}=\frac{9}{4}t$
Upon differentiation with respect to time we get,
${\sec }^2\theta \left(\frac{d\theta }{dt}\right)=\frac{9}{4}$
$\frac{d\theta }{dt}=\frac{9/4}{1+ta{n}^2\theta }=\frac{9/4}{1+\frac{81}{16}{t}^2}$
at $t=4,\frac{d\theta }{dt}=\frac{9}{4}\times \frac{1}{82}=\frac{9}{328}rad{s}^{-1}$