Question

A particle moves in $x-y$ plane. The position vector of the particle at any time $t$ is $\overset{ \rightarrow }{r}=\left\{\left(2 t\right) \hat{i} + \left(2 t^{2}\right) \hat{j}\right\} \, m$ . The rate of change of $\theta $ at time $t=2 \, s$ (where $\theta $ is the angle which its velocity vector makes with positive $x$ -axis ) is

• A. $\frac{2}{17}rads^{- 1}$
• B. $\frac{1}{14}rads^{- 1}$
• C. $\frac{4}{7}rads^{- 1}$
• D. $\frac{6}{5}rads^{- 1}$

Answer 1

Answer: (A)

$x=2t \, \Rightarrow v_{x}=\frac{d x}{d t}=2$
$y=2t^{2} \, \Rightarrow v_{y}=\frac{d y}{d t}=4t$
$\therefore \, tan \theta =\frac{v_{y}}{v_{x}}= \, \frac{4 t}{2}=2t$
Differentiating with respect to time we get,
$\left(\sec ^2 \theta\right) \frac{\mathrm{d} \theta}{\mathrm{d} t}=2$
or $\left(1+\tan ^2 \theta\right) \frac{\mathrm{d} \theta}{\mathrm{d} t}=2$
or $\left(1 + 4 t^{2}\right) \, \frac{d \theta }{d t}=2$
or $\frac{d \theta }{d t}=\frac{2}{1 + 4 t^{2}}$
$\frac{ \, d \theta }{d t}at \, t=2 \, s \, is \, \frac{d \theta }{d t}=\frac{2}{1 + 4 \left(2\right)^{2}}=\frac{2}{17} \, rads^{- 1}$