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(C) Given the velocity $v = \alpha \sqrt{x}$.We know that acceleration $a = v \frac{dv}{dx}$.First,find $\frac{dv}{dx}$:$\frac{dv}{dx} = \alpha \cdot

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(C) Given the velocity $v = \alpha \sqrt{x}$.We know that acceleration $a = v \frac{dv}{dx}$.First,find $\frac{dv}{dx}$:$\frac{dv}{dx} = \alpha \cdot \frac{1}{2\sqrt{x}} = \frac{\alpha}{2\sqrt{x}}$.Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula:$a = (\alpha \sqrt{x}) \cdot \left( \frac{\alpha}{2\sqrt{x}} \right) = \frac{\alpha^2}{2}$.Since $\alpha$ is a constant,the acceleration $a = \frac{\alpha^2}{2}$ is also a constant.Therefore,the graph of acceleration $a$ versus time $t$ is a horizontal straight line parallel to the time axis,which corresponds to option $C$.

The velocity $v$ of a particle moving along the $x$-axis varies with its position $x$ as $v = \alpha \sqrt{x}$,where $\alpha$ is a constant. Which of the following graphs represents the variation of its acceleration $a$ with time $t$?

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