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 ?

Explanation

Given, side of the square field = 10 m
Thererfore, perimeter = 10 m `xx` 4 = 40 m
Farmer moves along the boundary in 40 s
Time = 2 minutes 20 s = 2 `xx` 60 s + 20 s = 140 s
since, in 40 s farmer moves 40 m
Therefore, in 1s distance covered by farmer = 40/ 40 = 1m.
Therefore, in 140s distance covered by farmer = 1 `xx` 140 m = 140 m
Now, number of rotation to cover 140 along the boundry `("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 AC=`sqrt((10m)^(2)+(10m)^(2))`
= `sqrt(100m^(2) + 100m^(2))`
= `sqrt(200m^(2))`
= `10sqrt(2)` m
= `10xx1.414` = 14.14 m
Thus, after 2 min 20 seconds the displacement of farmer will be equal to 14.14 m north east from intial position.