Question

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

Answer 1

Total distance 40m in 40 sec.

Total time taken by the farmer = 2 min 20 sec. =140 seconds.

Total rounds completed= 140/40=3.5 rounds

That means if the farmer starts from point A of the square field, he reaches point C.

Therefore, displacement is AC.

Assume ABC is a right angled triangle.

Therefore,

AC2= AB2 +BC2  

AC2= (10)2 +(10)2

AC2= 100+100

AC2=200

AC = (200)1/2

 

AC= 10× (2) ½  

Answer 2

$let \: points \: be \: abcd \: and \: farmers \: start \: from \: point \: a \: \: 2min \: 20sec = 140sec \\ p erimeter = 40 \\ round \: taken = \frac{140}{40} = > 3.5 \\ therefore \: \: displcement = .5 \times 40m = 20m \\ \\ therefore \: famer \: is \: on \: point \: c$