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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11
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A particle moves in xy plane. The position vector of particle at any time t is $\overrightarrow r=\{(2t) i+(2t^2)j\}\; m$. The rate of change of $\theta$ at time $t= 2\;$ seconds is ( $\theta$ is the angle which its velocity vector makes with positive x-axis )

\[(a)\;\frac{2}{17}rad/s\quad (b)\;\frac{1}{14}rad/s \quad (c)\;\frac{4}{7}rad/s \quad (d)\;\frac{6}{5}rad/s\]
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1 Answer

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$x=2t$
=>$v_x=\large\frac{dx}{dt}$$=2$
$y=2t^2$
=>$v_y=\large\frac{dy}{dt}$$=4t$
$\tan \theta=\large\frac{v_y}{v_x}$
$\qquad=\large\frac{4t}{2}$$=2t$
Differentiating with respect to t
$(\sec^2 \theta) \large\frac{d\theta}{dt}$$=2$
$(1+\tan ^2 \theta) \large\frac{d\theta}{dt}$$=2$
$(1+4t^2) \large\frac{d\theta}{dt}$$=2$
$ \large\frac{d\theta}{dt}=\large\frac{2}{1+4t^2}$
at $t=2$
$\large\frac{d\theta}{dt}=\frac{2}{1+4(2)^2}=\frac{2}{17}$$ rad/s$
Hence a is the correct answer.

 

answered Jun 29, 2013 by meena.p
edited Jan 25, 2014 by meena.p
 

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