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

Explanation

Given side of the square
Field =10 m

Therefore, perimeter = 10x 4 = 40m
Farmer moves along the boundry in 40s
Displacement after 2m20s = `2x 60 + 20 = 140s`= ?
Since, in 40s farmer moves 40m
Therefore in 1s idstance covered by farmer
= 40/40= 1m
Therefore, in 140s distance covered by farmer = `1 xx 140 = 140m`
Now, number of rotation to cover 140 along the boundry= Total distance /perimeter
= 140/40 = 3.5 round
Thus, after 3.5 round farmer will be at point C of the field
Therefore Displacement
`AC = sqrt((10)^(2) + (10)^(2))`
`=sqrt(100 + 100) = sqrt200`
`=10 sqrt2m`
`=10 xx 1.141`
= 14.14m
Thus, after 2min 20 seconds the displacement of farmer will be equal to 14.14 m north East from initial position.