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(D) Given that the displacement $s$ is proportional to the square of the velocity $v$,we can write: $s = kv^2$,where $k$ is a constant of proportional

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$A$ constant

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(D) Given that the displacement $s$ is proportional to the square of the velocity $v$,we can write: $s = kv^2$,where $k$ is a constant of proportionality. Rearranging for $v^2$,we get $v^2 = \frac{1}{k} s$. Let $C = \frac{1}{k}$,so $v^2 = Cs$. Differentiating both sides with respect to time $t$: $\frac{d}{dt}(v^2) = \frac{d}{dt}(Cs)$ $2v \frac{dv}{dt} = C \frac{ds}{dt}$ Since $\frac{dv}{dt} = a$ (acceleration) and $\frac{ds}{dt} = v$ (velocity),we have: $2v a = Cv$ Assuming $v 
eq 0$,we divide by $v$: $2a = C$ $a = \frac{C}{2}$ Since $C$ is a constant,the acceleration $a$ is a constant.

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

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