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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 $(in \;m)$ of the farmer at the end of $2$ $min$ $20\,s$ from his initial position ?
$16.14$
$14.14$
$24.14$
$14.41$
Solution
Given, Side of the square field $= 10\,m$
Therefore, perimeter $= 10 \,m \times 4 = 40 \,m$
Farmer moves along the boundary in $40\,s.$
Displacement after $2\, m$ $20\, s = 2 \times 60\, s + 20\, s = 140\, s =$ ?
Since in $40 \,s$ farmer moves $40 \,m$
Therefore, in $1\,s$ distance covered by farmer $= 40 / 40 \,m = 1\,m$
Therefore, in $140\,s$ distance covered by farmer $= 1 \times 140 \,m = 140 \,m$
Now, number of rotation to cover $140$ along the boundary
$=$ Total Distance $/$ Perimeter
$= 140 \,m / 40\, m = 3.5$ round
Thus, after $3.5$ round farmer will at point $C$ of the field.
Therefore. Displacement $A C=\sqrt{(10 \,m)^{2}+(10 \,m)^{2}}$
$=\sqrt{100 \,m^{2}+100\, m^{2}}$
$=\sqrt{200 \,m^{2}}$
$=10 \sqrt{2} \,m$
$=10 \times 1.414=14.14 \,m$
Thus, after $2$ $min$ $20\,\sec$ the displacement of farmer will be equal to $14.14 \,m$ north east from intial position.
