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(D) Given that $s \propto v^2$, we can write $s = kv^2$, where $k$ is a constant.Rearranging for velocity, $v^2 = s/k$, which implies $v = \frac{1}{\s

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a constant

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(D) Given that $s \propto v^2$, we can write $s = kv^2$, where $k$ is a constant.Rearranging for velocity, $v^2 = s/k$, which implies $v = \frac{1}{\sqrt{k}} s^{1/2}$.Acceleration $a$ is given by $a = v \frac{dv}{ds}$.Differentiating $v = \frac{1}{\sqrt{k}} s^{1/2}$ with respect to $s$, we get $\frac{dv}{ds} = \frac{1}{\sqrt{k}} \cdot \frac{1}{2} s^{-1/2}$.Substituting these into the acceleration formula: $a = (\frac{1}{\sqrt{k}} s^{1/2}) \cdot (\frac{1}{2\sqrt{k}} s^{-1/2})$.Simplifying, $a = \frac{1}{2k} \cdot s^{1/2 - 1/2} = \frac{1}{2k}$.Since $k$ is a constant, $a$ is a constant.

If a particle moves such that the displacement $s$ is proportional to the square of the velocity $v$, then its acceleration $a$ is

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