A particle moves in xy plane. The position vector at any time / is r→=(2t)i^+2t2j^m. The rate of change of θ at time t = 2 second (where θ is the angle which its velocity vector makes with positive x-axis) is

x=2t⇒vx=dxdt=2Y=2t2⇒vy=dydt=4ttan⁡θ=vyvx=4t2=2t∴Differentiating with respect to time we get,sec2⁡θdθ2=2or 1+tan2⁡θdθ2=2; or 1+4t2dθdt=2or dθdt=21+4t2;dθdt at t=2 is dθdt=21+4(2)2=217rad/s

A particle moves in xy plane. The position vector at any time / is r→=(2t)i^+2t2j^m. The rate of change of θ at time t = 2 second (where θ is the angle which its velocity vector makes with positive x-axis) is

x=2t⇒vx=dxdt=2Y=2t2⇒vy=dydt=4ttan⁡θ=vyvx=4t2=2t∴Differentiating with respect to time we get,sec2⁡θdθ2=2or 1+tan2⁡θdθ2=2; or 1+4t2dθdt=2or dθdt=21+4t2;dθdt at t=2 is dθdt=21+4(2)2=217rad/s

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A particle moves in xy plane. The position vector at any time / is $\vec{r} = \{(2 t) \hat{i} + (2 t^{2}) \hat{j}\} m$ . The rate of change of $θ$ at time t = 2 second (where $θ$ is the angle which its velocity vector makes with positive x-axis) is

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$\frac{2}{17} rad / s$

$\frac{1}{14} rad / s$

$\frac{4}{7} rad / s$

$\frac{6}{5} rad / s$

answer is A.

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Detailed Solution

$\begin{array}{l} x = 2 t \Rightarrow v_{x} = \frac{d x}{d t} = 2 \\ Y = 2 t^{2} \Rightarrow v_{y} = \frac{d y}{d t} = 4 t \\ \tan θ = \frac{v_{y}}{v_{x}} = \frac{4 t}{2} = 2 t \end{array}$

$∴$ Differentiating with respect to time we get,

$\begin{array}{l} (\sec^{2} θ) \frac{d θ}{2} = 2 \\ or (1 + \tan^{2} θ) \frac{d θ}{2} = 2; or (1 + 4 t^{2}) \frac{d θ}{d t} = 2 \\ or \frac{d θ}{d t} = \frac{2}{1 + 4 t^{2}}; \frac{d θ}{d t} at t = 2 is \\ \frac{d θ}{d t} = \frac{2}{1 + 4 (2)^{2}} = \frac{2}{17} rad / s \end{array}$