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A particle whose velocity is given as $\vec{v} = \hat{i} + 6 t \hat{j} m / s$ is moving in $x - y$ plane. At $t = 0,$ particle is at origin. Find the radius of curvature of path at point $(\frac{\sqrt{2}}{3} m, \frac{2}{3} m)$

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$3.0 m$

$1.5 m$

$4.5 m$

$6.0 m$

answer is C.

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Detailed Solution

$v_{x} = 1 \Rightarrow x = t a n d v_{y} = 6 t \Rightarrow y = 3 t^{2} \Rightarrow y = 3 x^{2} \Rightarrow \frac{d y}{d x} = 6 x, \frac{d^{2} y}{d x^{2}} = 6 \Rightarrow {\frac{d y}{d x} |}_{x = \frac{\sqrt{2}}{3}} = 2 \sqrt{2}$

As we know that $R = \frac{{[1 + {(\frac{d y}{d x})}^{2}]}^{3 / 2}}{\frac{d^{2} y}{d x^{2}}} = \frac{{(1 + 8)}^{3 / 2}}{6} = 4.5 m$

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