Question

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 $50\,m \times 50\,m$ is filled with sand.

Outside this square, he can walk at a speed $1\,m\,s^{-1}$. In the central square, he can walk only at a speed of $v\,ms^{-1}\, (v < 1)$. 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. $0.18\,ms^{-1}$
• B. $0.81\,ms^{-1}$
• C. $0.5\,ms^{-1}$
• D. $0.95\,ms^{-1}$

Answer 1

The Correct Option is B

$AC=\sqrt{\left(100\right)^{2}+\left(100\right)^{2}}$ $=100\sqrt{2}\,m$ $PQ=\sqrt{\left(50\right)^{2}+\left(50\right)^{2}}$ $=50\sqrt{2}\,m$ $AP=\frac{AC-PQ}{2}$ $=\frac{100\sqrt{2}-50\sqrt{2}}{2}$ $=25\sqrt{2}\,m$

$QC=AP=25\sqrt{2}\,m$, $AR=\sqrt{\left(75\right)^{2}+\left(25\right)^{2}}$ $=25\sqrt{10}\,m$ $RC=AR=25\sqrt{10}\,m$ Consider the straight line path $APQC$ through the sand. Time taken to go from $A$ to $C$ via this path $T_{sand}=\frac{AP+QC}{1}+\frac{PQ}{v}$ $=\frac{25\sqrt{2}+25\sqrt{2}}{1}+\frac{50\sqrt{2}}{v}$ $=50\sqrt{2}\left[\frac{1}{v}+1\right]$ The shortest path outside the sand will be $ARC$. Time taken to go from $A$ to $C$ via this path $T_{outside}=\frac{AR+RC}{1}$ $=\frac{25\sqrt{10}+25\sqrt{10}}{1}$ $=50\sqrt{10}$ $\because T_{sand} < T_{outside}\, $ $\therefore 50\sqrt{2}\left[\frac{1}{v}+1\right] < 50\sqrt{10}$ $\Rightarrow \frac{1}{v}+1 < \sqrt{5}$ or $\frac{1}{v} < \sqrt{5}-1$ or $v > \frac{1}{\sqrt{5} -1} $ $\approx0.81\,ms^{-1}$