Question

A farmer moves along the boundary of square field of side 10m in 40 second what will the magnitude of displacement of the farmer at the end of 2minutes 20 second

Answer 1

Answer:

Answer: 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

here's the answer ⤵️⤵️

Side of square=10m

Perimeter=4×side

=4×10

=40m.

He moves 10m in 40 sec=40/40=1m.

Distance covered in 140m=140×1=140m.

Displacement of farmer at the end of 2min 20 sec=2×60+20=140sec.

For 140m=140/4=3.5m

Displacement of his initial position=√10m²+√10m²

=√200m²

=10√2m²

Note:√ is the sign of under root.

hope it will help you..................✌️