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  • 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 1 m/s.

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 1 m/s. In the central square, he walk only at a speed of v m/s. 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?

Answers (1)

Since PQ is the diagonal, we can use Pythagoras theorem to find out,

PQ=\sqrt{50\times 50+50\times 50}=50\sqrt{2}\; meters

AC=\sqrt{100\times 100+100\times 100}=100\sqrt{2}\; meters

Now let us calculate the time taken by the man to travel the path A-P-Q-C,

T1=\frac{AC-PQ}{1}+\frac{PQ}{v}=\frac{100\sqrt{2}-50\sqrt{2}}{1}+\frac{50\sqrt{2}}{v}

T1=50\sqrt{2}[1+\frac{1}{v}]

Now let us calculate the time taken by the man to travel the path A-R-C,

T2=2\; AR

AR=\left ( \frac{100\sqrt{2}}{2} \right )^{2}+\left ( \frac{50\sqrt{2}}{2} \right )^{2}

AR=25\sqrt{10}

T2=50\sqrt{10}

The case when ,

T1>T2

We get,

50\sqrt{2}\left [ 1+\frac{1}{v} \right ]<50\sqrt{10}

\frac{1}{v}<\sqrt{5}-1

v<0.82\; m/s

Posted by

infoexpert23

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