A man wants to reach from A to the opposite c | vedclass

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(C) The total distance from $A$ to $C$ is along the diagonal of the $100\, m 	imes 100\, m$ square. The diagonal length is $100\sqrt{2}\, m$. The cen

Vedclass · Accepted Answer

$1/\sqrt{2}$

Vedclass · Answer

(C) The total distance from $A$ to $C$ is along the diagonal of the $100\, m 	imes 100\, m$ square. The diagonal length is $100\sqrt{2}\, m$. The central sand square has side $50\, m$,so its diagonal is $50\sqrt{2}\, m$. The distance outside the sand along the diagonal is $100\sqrt{2} - 50\sqrt{2} = 50\sqrt{2}\, m$. Time taken through the sand path $T_{	ext{sand}} = \frac{50\sqrt{2}}{1} + \frac{50\sqrt{2}}{v} = 50\sqrt{2}(1 + \frac{1}{v})$. The shortest path outside the sand is along the boundary of the sand square. The distance from $A$ to the corner of the sand square is $\sqrt{25^2 + 75^2} = \sqrt{625 + 5625} = \sqrt{6250} = 25\sqrt{10}\, m$. Total distance outside is $2 	imes 25\sqrt{10} = 50\sqrt{10}\, m$. Time taken $T_{	ext{outside}} = \frac{50\sqrt{10}}{1} = 50\sqrt{10}\, s$. For the sand path to be faster,$T_{	ext{sand}} $50\sqrt{2}(1 + \frac{1}{v}) $v > \frac{1}{\sqrt{5} - 1} = \frac{\sqrt{5} + 1}{4} \approx 0.809$. Given the options,the question implies a comparison with the path along the boundary. Re-evaluating the path $A 	o P 	o R 	o C$ where $P, R$ are corners of the sand square: $AP = \sqrt{25^2 + 25^2} = 25\sqrt{2}$,$PR = 50$,$RC = 25\sqrt{2}$. $T = \frac{25\sqrt{2} + 25\sqrt{2}}{1} + \frac{50}{v} = 50\sqrt{2} + \frac{50}{v}$. Setting $50\sqrt{2} + \frac{50}{v} $v > 0.57$. The closest value is $1/\sqrt{2} \approx 0.707$.

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