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

Explanation

The distance covered in one round of the field = the perimeter of the field
`= 4 xx 10 m = 40 m`

Time taken by the farmer to complette one round. i.e., 40 m is 40 second.
ln 2 minute 20 second i.e., 140 second the farmer covers a distance of 140m.
One round = 40 m
` therefore m = (140)/(40) =3.4` rounds
`0.5` rounds ` 0.5 xx 40 m = 20 m`
If the farmeer starts from point A, then at the end of 2 min 20 s he will be at the point C.
`therefore` The displacement is AC.
To find the magnitude of AC, use pythagoras. theorem.
In right angled triangle ABC,
`AC ^(2) = AB^(2) + BC ^(2)`
=`(10)^(2) + (10) ^(2)`
`= 100 + 100 = 200`
`therefore AC sqrt (200) = sqrt (100 xx2 ) = 10 sqrt2 m `
The magnitude of displacement of the farmer is `10 sqrt2 m.`
The distance travelled by the farmer is 140 m.