Question

A particle moves in the $X-Y$ plane. The position vector of the particle at any time it' is $\overset{ \rightarrow }{r}=\left(2 t^{2} \hat{i} + 3 t^{3} \hat{j}\right)m$ . 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. $\frac{9}{656}rads^{- 1}$
• B. $\frac{9}{164}rads^{- 1}$
• C. $\frac{9}{328}rads^{- 1}$
• D. $\frac{9}{82}rads^{- 1}$

Answer 1

Answer: (C)

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