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A particle of mass m moves with a variable velocity v, which changes with distance covered x along a straight line as $v = k \sqrt{x}$ , where k is a positive constant. The work done by all the forces acting on the particle, during the first t seconds is

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\frac{{mk}^{4} t^{2}}{8}

\frac{{mk}^{4}}{t^{2}}

\frac{{mk}^{4} t^{2}}{16}

\frac{{mk}^{4} t^{2}}{4}

answer is C.

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Detailed Solution

v = k \sqrt{x}

$\frac{d x}{dt} = k \sqrt{x}$

$\frac{dv}{dt} = k \frac{1}{2 \sqrt{x}} v$

$a = \frac{k}{2 \sqrt{x}}$

$k \sqrt{x} = \frac{k^{2}}{2}$

$\int_{0}^{x} \frac{d x}{\sqrt{x}} = k \int_{0}^{t} dt$

$2 \sqrt{x} = kt$

$x = \frac{k^{2} t^{2}}{4}$

$W = m \frac{k^{2}}{2} (\frac{k^{2} t^{2}}{4})$

$W = \frac{{mk}^{4} t^{2}}{8}$

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