A particle moves on a circle of radius r with centripetal acceleration as function of time as a_e = k²rt² where k is a positive constant. Find the resultant acceleration.

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$kr \sqrt{k^{2} t^{4} + 1}$

kt²

$kr \sqrt{k^{2} t^{4} - 1}$

answer is C.

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Detailed Solution

$\begin{array}{l} a_{c} = k^{2} {rt}^{2} = \frac{v^{2}}{r}; v = krt; a_{t} = \frac{dv}{dt} = kr \\ a_{net} = \sqrt{a_{c}^{2} + a_{t}^{2}} = \sqrt{k^{4} r^{2} t^{4} + k^{2} r^{2}} = kr \sqrt{k^{2} t^{4} + 1} \end{array}$

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