Ncert Exercises & Solutions

4.37. A man wants to reach from A to the opposite comer of the square C. The sides of the square are 100 m. A central square of 50 m × 50 m is filled with sand. Outside this square, he can walk at a speed of 1 m/s. In the central square, he walks only at a speed of v m/s. What is the 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?

Hint: Calculate time separately for both of the paths and compare.

Step 1: Find the time taken to go through the sand

Consider adjacent diagram,

Time is taken to go from A to C via straight-line path APOC through the sand

$\begin{array}{l} T_{sand} = \frac{AP + QC}{1} + \frac{PQ}{v} = \frac{25 \sqrt{2} + 25 \sqrt{2}}{1} + \frac{50 \sqrt{2}}{v} \\ = 50 \sqrt{2} + \frac{50 \sqrt{2}}{v} = 50 \sqrt{2} (\frac{1}{v} + 1) \end{array}$

Step 2: Find the time taken to go through the path outside of the sand.

Clearly from the figure, the shortest path outside the sand will be ARC.
Time is taken to go from A to C via this path

$\begin{array}{l} \cdot T_{outside} = \frac{AR + RC}{1} s \\ Clearly, AR = RC = \sqrt{75^{2} + 25^{2}} = \sqrt{75 \times 75 + 25 \times 25} \\ = 5 \times 5 \sqrt{9 + 1} = 25 \sqrt{10} m \\ RC = AR = \sqrt{75^{2} + 25^{2}} = 25 \sqrt{10} m \\ \Rightarrow T_{outside} = 2 AR = 2 \times 25 \sqrt{10} s = 50 \sqrt{10} s \end{array}$

Step 3: Find the minimum value of v

$\begin{array}{lc} For & T_{sand} < T_{outside} \\ \Rightarrow & 50 \sqrt{2} (\frac{1}{v} + 1) < 2 \times 25 \sqrt{10} \\ \Rightarrow & \frac{2 \sqrt{2}}{2} (\frac{1}{v} + 1) < \sqrt{10} \\ \Rightarrow & \frac{1}{v} + 1 < \frac{2 \sqrt{10}}{2 \sqrt{2}} = \frac{\sqrt{5}}{2} \times 2 = \sqrt{5} \\ \Rightarrow & \frac{1}{v} < \frac{\sqrt{5}}{2} \times 2 - 1 \Rightarrow \frac{1}{v} < \sqrt{5} - 1 \\ \Rightarrow & v > \frac{1}{\sqrt{5} - 1} \approx 0 .81 m / s \\ \Rightarrow & v > 0 .81 m / s \end{array}$

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