Question

A man wants to reach from A to the opposite corner of the square C (Fig) . The sides of the square are 100 m. A central square of 50m 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?

• A. 0.18 $m{s}^{-1}$
• B. 0.87 $m{s}^{-1}$
• C. 0.81 $m{s}^{-1}$
• D. 0.95 $m{s}^{-1}$

Answer 1

The correct option is C 0.81 $m{s}^{-1}$

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

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

$AP=\frac{AC-PQ}{2}=\frac{100\sqrt{2}-50\sqrt{2}}{2}$ = $25\sqrt{2}m$

$RC=AR=25\sqrt{2}m$

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

$T_{sand}$ =$\frac{AP+QC}{1}+\frac{PQ}{v}$= $\frac{25\sqrt{2}+25\sqrt{2}}{1}+\frac{50\sqrt{2}}{v}$

$=50\sqrt{2}[\frac{1}{v}+1]$

The shortest path outside the sand will be ARC. Time is taken to go from $A$ to $C$ via this path

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

$=50\sqrt{10}$

$\because$ $T_{sand}$ < $T_{outside}$

$\therefore$ $50\sqrt{2}[\frac{1}{v}+1]$ < $50\sqrt{10}$

$\Rightarrow$ $\frac{1}{v}+1<\sqrt{5}$

or $\frac{1}{v}<\sqrt{5}-1$

or$v>1\sqrt{5}-1=0.81m{s}^{-1}$