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A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Q 2.  A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

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Side of the square field = 10\ m
\therefore Perimeter of the square = 4\times10 = 40\ m
According to question,
He completes 1 round in 40\ s.
\therefore Speed of the farmer = \frac{40}{40}= 1\ m/s
\therefore Distance covered in 2\ min\ 20\ s\ (=140\ s) = 140 \times 1 = 140\ m
Now,
Number of round trips completed travelling 140\ m = \frac{140}{40} = 3.5
We know, in 3 round trips the displacement will be zero.
In 0.5 round, the farmer will reach diametrically opposite to his initial position.
\therefore Displacement = AC = \sqrt{(AB^2 + BC^2 )} = \sqrt{(100^2+100^2)} = 10\sqrt{2}\ m

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