A man wants to reach from $A$ to the opposite corner of the square $C$ (Figure). The sides of the sq

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 $50 \mathrm{~m} 	imes 50 \mathrm{~m}$ is filled with sand. Outside this square, he can walk at a speed $1 \mathrm{~m} \mathrm{~s}^{-1}$. In the central square, he can walk only at a speed of $v \mathrm{~m} \mathrm{~s}^{-1}(v<1)$. 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?

Professor PC Thomas Classes · Answer

(b) : $A C=\sqrt{(100)^2+(100)^2}=100 \sqrt{2} \mathrm{~m}$ \ $\begin{aligned} P Q & =\sqrt{(50)^2+(50)^2}=50 \sqrt{2} \mathrm{~m} \ A P & =\frac{A C-P Q}{2}=\frac{100 \sqrt{2}-50 \sqrt{2}}{2} \ & =25 \sqrt{2} \mathrm{~m}\end{aligned}$ \ $Q C=A P=25 \sqrt{2} \mathrm{~m}$ \ $A R=\sqrt{(75)^2+(25)^2}=25 \sqrt{10} \mathrm{~m}$ \ $$ R C=A R=25 \sqrt{10} \mathrm{~m} $$ Consider the straight line path $A P Q C$ through the sand. Time taken to go from $A$ to $C$ via this path $T_{\mathrm{sand}}=\frac{A P+Q C}{1}+\frac{P Q}{v}=\frac{25 \sqrt{2}+25 \sqrt{2}}{1}+\frac{50 \sqrt{2}}{v}$ \ $=50 \sqrt{2}\left[\frac{1}{v}+1ight]$ $\because \quad T_{	ext {sand }}<T_{	ext {outside }} \quad 	herefore \quad 50 \sqrt{2}\left[\frac{1}{v}+1ight]<50 \sqrt{10}$ $\Rightarrow \frac{1}{v}+1<\sqrt{5}$ or $\frac{1}{v}<\sqrt{5}-1$ or $v>\frac{1}{\sqrt{5}-1} \approx 0.81 \mathrm{~m} \mathrm{~s}^{-1}$