- Home
- Standard 11
- Physics
A man wants to reach from $A$ to the opposite corner of the square $C$. The sides of the square are $100\, m$. A central square of $50\, m\,\times \,50\, m$ is filled with sand. Outside this square, he can walk at a speed $1\,ms^{-1}$. 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 ?
Solution
Consider adjacent diagram,
Time taken to go from A to $C$ via straight line path APQC through the $S$ and
$\mathrm{T}_{\text {sand }} =\frac{\mathrm{AP}+\mathrm{QC}}{1}+\frac{\mathrm{PQ}}{\mathrm{V}}$
$=\frac{25 \sqrt{2}+25 \sqrt{2}}{1}+\frac{50 \sqrt{2}}{\mathrm{~V}}$
$=50 \sqrt{2}+\frac{50 \sqrt{2}} {\mathrm{~V}}=50 \sqrt{2}\left(\frac{1}{\mathrm{~V}}+1\right)$
Clearly from figure the shortest path outside the sand will be ARC. Time taken to go from $\mathrm{A}$ to $\mathrm{C}$ via this path,
$\mathrm{T}_{\text {outside }}=\frac{\mathrm{AR}+\mathrm{RC}}{1} \mathrm{~s}$
Clearly, $\quad$ AR $=\sqrt{75^{2}+25^{2}}=\sqrt{75 \times 75+25 \times 25}$ $\quad=5 \times 5 \sqrt{9+1}=25 \sqrt{10} \mathrm{~m}$ $\quad \mathrm{RC}=\mathrm{AR}=\sqrt{75^{2}+25^{2}}=25 \sqrt{10} \mathrm{~m}$
