Question

a farmer moves along the boundary of a square field of side 1 m in 40 seconds what will be the magnitude of the farmer at the end of 2 minutes 20 seconds​

Answer 1

Answer:

14.14 metres

Explanation:

Given:

• Side of the square field = 10 metres

• Time taken to move around the square field = 40 seconds

To find:

• Displacement at the end of 1 minute 40 seconds has to be found

Perimeter of the square field = 4×Side of the square field  = 4×10 = 80 metres

2 minute 20 seconds = 140 seconds

At the end of 140 seconds the farmer will cover = 140/40 = 3.5 rounds

Since it is a square field displacement = will be hypotenuse

Hypotenuse²=10²+10²

Hypotenuse²=100+100

Hypotenuse = √200 = 14.14 metres

Answer 2

Answer:

Given side of square =10m

thus, perimeter P=40m

Time taken to cover the boundary of 40 m =40 s

Thus in 1 second, the farmer covers a distance of 1 m

Now distance covered by the farmer in 2 min 20 seconds = 1×140=140m

Now the total number of rotation the farmer makes to cover a distance of 140 meters =   perimeter

/total distance  =3.5  

At this point, the farmer is at a point say B from the origin O

Thus the displacement s=$\sqrt$$10^{2}$+

from Pythagoras theorem.

s=102

​ =14.14m

may it helps you