A particle starts moving with initial velocity of 3 m/s along x-axis from origin. Its acceleration is varying with position $(x)$ in parabolic nature with vertex at the origin as shown in figure. A tangent to the graph is drawn at $x = \sqrt{3} m$ . The tangent makes an angle of $60^{0}$ with the x-axis. At this position the velocity of particle is $\sqrt{\sqrt{3} + k} m / s$ the value of $k$ is

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answer is 9.

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Detailed Solution

$a = k x^{2}$
$\frac{d a}{d x} = 2 k x$
At $x = \sqrt{3} \frac{d a}{d x} = \sqrt{3} \Rightarrow k = \frac{1}{2}$
$v \frac{d v}{d x} = \frac{1}{2} x^{2} \Rightarrow \int v d v = \int \frac{x^{2} d x}{2}$
$\frac{v^{2} - u^{2}}{2} = \frac{1}{2} {(\frac{x^{3}}{3})}_{0}^{\sqrt{3}}$
$v^{2} = 9 + \sqrt{3} \Rightarrow v = \sqrt{9 + \sqrt{3}}$

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