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(C) Given the distance $s = \sqrt{at^2 + bt + c}$,we have $s^2 = at^2 + bt + c$. Differentiating with respect to $t$: $2s \frac{ds}{dt} = 2at + b$ $\f

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Proportional to $s^{-3}$

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(C) Given the distance $s = \sqrt{at^2 + bt + c}$,we have $s^2 = at^2 + bt + c$. Differentiating with respect to $t$: $2s \frac{ds}{dt} = 2at + b$ $\frac{ds}{dt} = \frac{2at + b}{2s}$ Differentiating again with respect to $t$ to find acceleration $f = \frac{d^2s}{dt^2}$: $2 \left( \frac{ds}{dt} \right)^2 + 2s \frac{d^2s}{dt^2} = 2a$ $s \frac{d^2s}{dt^2} = a - \left( \frac{ds}{dt} \right)^2$ $s \frac{d^2s}{dt^2} = a - \frac{(2at + b)^2}{4s^2} = \frac{4as^2 - (2at + b)^2}{4s^2}$ Substituting $s^2 = at^2 + bt + c$: $s \frac{d^2s}{dt^2} = \frac{4a(at^2 + bt + c) - (4a^2t^2 + 4abt + b^2)}{4s^2}$ $s \frac{d^2s}{dt^2} = \frac{4a^2t^2 + 4abt + 4ac - 4a^2t^2 - 4abt - b^2}{4s^2} = \frac{4ac - b^2}{4s^2}$ $\frac{d^2s}{dt^2} = \frac{4ac - b^2}{4s^3}$ Since $4ac - b^2$ is a constant,the acceleration $f \propto s^{-3}$.

The distance travelled by a particle moving in a straight line in time $t$ is $s = \sqrt{at^2 + bt + c}$. The acceleration of the particle is

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