# 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?

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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