Question

A man wants to reach from A to the opposite comer of the square C. The sides of the square are 100 m. A central square of 50 m x 50 m is filled with sand. Outside this square, he can walk at a speed 1 m/s. In the central square, he can walk only at a speed of v m/s (v < 1). What is 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?

Answer 1

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

T_sand = \(\frac{AP+QC}{1}+\frac{PQ}{v}\)

= \(\frac{25\sqrt2+25\sqrt2}{1}+\frac{50\sqrt2}{v}\)

= \(50\sqrt2[\frac{1}{v}+1]\)

The shortest path outside the sand will be ARC.

T_sand = \(\frac{AR+AC}{1}S\)

AR = \(\sqrt{75^2+25^2}=\sqrt{75\times75\times25\times25}\)

= 5 × \(5\sqrt{9+1}=25\sqrt10m\)

RC = AR =\(\sqrt{75^2+25^2}=25\sqrt10m\)

T_outside = 2AR = \(2\times25\sqrt10=50\sqrt10 s\)

For T_sand < T_outside

\(50\sqrt2(\frac{1}{v}+1)<2\times25\sqrt10\)

\(\frac{2\sqrt2}{2}(\frac{1}{v}+1)=\sqrt10\)

⇒\((\frac{1}{v}+1)<\frac{2\sqrt10}{2\sqrt2}=\frac{\sqrt5}{2}\times2=\sqrt5\)

⇒ \(\frac{1}{v}<\frac{\sqrt5}{2}\times2-1\)

⇒\(\frac{1}{v}<\sqrt5-1\)

⇒ v> \(\frac{1}{\sqrt5-1}\) ≈ 0.81 m/s

⇒ v> 0.81 m/s.