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

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$x=2t$

=>$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.

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