A man wants to reach from $A$ to the opposite corner of the square $C$(Figure). The sides of the square are $100 m$ each. A central square of  $50m\times 50m$ filled with sand. Outside this square, he can walk at a speed $1m{s}^{-1}.$

In the central square, he can walk only at a speed of $v m{s}^{-1}(v<1)$. What is the smallest value of $v$ for which he can reach faster via a straight path through the sand than any path in the square outside the sand?

$AC=\sqrt{(100{)}^2+(100{)}^2}=100\sqrt{2} \mathrm{m}$

$PQ=\sqrt{(50{)}^2+(50{)}^2}=50\sqrt{2} \mathrm{m}$

$AP=\frac{AC-PQ}{2}=\frac{100\sqrt{2}-50\sqrt{2}}{2}$

$=25\sqrt{2} \mathrm{m}$

$QC=AP=25\sqrt{2} \mathrm{m},AR=\sqrt{(75{)}^2+(25{)}^2}=25\sqrt{10} \mathrm{m}$

$RC=AR=25\sqrt{10} \mathrm{m}$

Consider the straight line path $APQC$ through the sand. Time taken to go from $A$ to $C$ via this path

$T_{\text{sand }}=\frac{AP+QC}{1}+\frac{PQ}{v}=\frac{25\sqrt{2}+25\sqrt{2}}{1}+\frac{50\sqrt{2}}{v}=50\sqrt{2}\left[\frac{1}{v}+1\right]$

The shortest path outside the sand will be $ARC.$

Time taken to go from $A$ to $C$ via this path

$T_{\text{outside }}=\frac{AR+RC}{1}=\frac{25\sqrt{10}+25\sqrt{10}}{1}=50\sqrt{10}$

$\because T_{\text{sand }}<T_{\text{outside }} \therefore 50\sqrt{2}\left[\frac{1}{v}+1\right]<50\sqrt{10}$

$\Rightarrow \frac{1}{v}+1<\sqrt{5}\text{ or }\frac{1}{v}<\sqrt{5}-1\text{ or }v>\frac{1}{\sqrt{5}-1}\approx 0.81 \mathrm{m}{ \mathrm{s}}^{-1}$