The acceleration a of a particle moving with an initial velocity u varies with distance x as $a = k \sqrt{x}$ where k is a constant. The distance covered by the particle when its velocity becomes 3u is given by

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${(\frac{u^{2}}{3 k})}^{\frac{2}{3}}$

${(\frac{6 u^{2}}{k})}^{\frac{2}{3}}$

${(\frac{3 u^{2}}{k})}^{\frac{2}{3}}$

${(3 k u^{2})}^{\frac{4}{3}}$

answer is C.

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Detailed Solution

Given $a = k \sqrt{x}$

$\Rightarrow \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = k \sqrt{x}$

$\Rightarrow vdv = k \sqrt{x} dx$

Integrating on both sides

$\Rightarrow \int_{u}^{3 u} vdv = \int_{0}^{x} k \sqrt{x} dx$

$\Rightarrow {|\frac{v^{2}}{2}|}_{u}^{3 u} = \frac{2 k}{3} {|x^{3 / 2}|}_{0}^{x}$

$\Rightarrow x^{3 / 2} = \frac{3}{4} \frac{8 u^{2}}{k}$

$x = {(\frac{6 u^{2}}{k})}^{\frac{2}{3}}$

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