We have, $S^{2}=a t^{2}+2 b t+c \quad \text{...(i)}$
Differentiating Eq. (i) w.r.t. $t$, we get
$$
\begin{aligned}
&2 S \frac{d S}{d t}=2 a t+2 b \\
&\Rightarrow \frac{d S}{d t}=\left(\frac{a t+b}{S}\right) \quad \text{...(ii)}
\end{aligned}
$$
Differentiating Eq. (ii) w.r.t. $t$, we get
$\begin{aligned} \frac{d^{2} S}{d t^{2}} &=\frac{S-\frac{d}{d t}(a t+b)-(a t+b) \frac{d S}{d t}}{S^{2}} \\ \Rightarrow & \frac{d^{2} S}{d t^{2}}=\frac{a S-(a t+b) \frac{(a t+b)}{S}}{S^{2}} \quad \text { [using Eq. (ii)] } \\ \Rightarrow & \frac{d^{2} S}{d t^{2}}=\frac{a S^{2}-(a t+b)^{2}}{S^{3}} \end{aligned}$
$\Rightarrow \frac{d^{2} S}{d t^{2}}=\frac{a\left(a t^{2}+2 b t+c\right)-\left[(a t)^{2}+b^{2}+2 a t b\right]}{S^{3}}$
$\Rightarrow \frac{d^{2} S}{d t^{2}}=\frac{a^{2} t^{2}+2 a b t+a c-a^{2} t^{2}-b^{2}-2 a t b}{S^{3}}$
$\Rightarrow \frac{d^{2} S}{d t^{2}}=\frac{a c-b^{2}}{S^{3}}$
$\Rightarrow$ Acceleration $=\left(\frac{d^{2} S}{d t^{2}}\right) \propto \frac{1}{S^{3}}$