As the given diagram :
$$
\begin{aligned}
&P Q=\sqrt{50^2+50^2}=50 \sqrt{2} \\
&A C=\sqrt{100^2+100^2}=100 \sqrt{2}
\end{aligned}
$$
So, time taken from $A$ to $C$ via straight line path $A P Q C$ through the $S$ and


$T_{\text {sand }}=\frac{A P+Q C}{1}+\frac{P Q}{v}$
$=\frac{25 \sqrt{2}+25 \sqrt{2}}{1}+\frac{50 \sqrt{2}}{v}$
$=50 \sqrt{2}+\frac{50 \sqrt{2}}{v}=50 \sqrt{2}\left(\frac{1}{v}+1\right)$
So, its shown that in figure the shortest path outside the sand will be $A R C$.
Time taken to go from $A$ to $C$ via this path
$$
T_{\text {outside }}=\frac{A R+R C}{1} \mathrm{~s}
$$
Clearly, $A R^2=O A^2+O R^2(\operatorname{In} \triangle \mathrm{AOR})$
$$
\begin{aligned}
A R &=\sqrt{75^2+25^2}=\sqrt{75 \times 75+25 \times 25} \\
&=5 \times 5 \sqrt{9+1}=25 \sqrt{10} \mathrm{~m} \\
R C &=A R=\sqrt{75^2+25^2}=25 \sqrt{10} \mathrm{~m} \\
\Rightarrow T_{\text {cutside }} &=2 A R=2 \times 25 \sqrt{10} \mathrm{~s}=50 \sqrt{10} \mathrm{~s}
\end{aligned}
$$
For $T_{\text {sand }} < T_{\text {outside }}$
So, $50 \sqrt{2}\left(\frac{1}{v}+1\right) < 2 \times 25 \sqrt{10}$
$$
\begin{aligned}
&\Rightarrow \frac{2 \sqrt{2}}{2}\left(\frac{1}{v}+1\right) < \sqrt{10} \\
&\frac{1}{v}+1 < \frac{2 \sqrt{10}}{2 \sqrt{2}}=\frac{\sqrt{5}}{2} \times 2=\sqrt{5} \\
&\frac{1}{v} < \frac{\sqrt{5}}{2} \times 2-1 \Rightarrow \frac{1}{v} < \sqrt{5}-1 \\
&v < \frac{1}{(\sqrt{5}-1)} \times \frac{(\sqrt{5}+1)}{(\sqrt{5}+1)}=\frac{(2 \cdot 3+1)}{(5-1)}=\frac{(3 \cdot 3)}{4} \\
&v < 0.82 \mathrm{~m} / \mathrm{sec}
\end{aligned}
$$