A farmer moves along the boundary of a square filed 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 ?
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Answer:
a farmer moves along the boundary of a square field of side 10 m in 40s 2 minutes 20 seconds means 140 seconds. Distance travelled by farmer = 40 40 4040 × 140 m = 140 m. ∴ Farmer moves for 2 minutes 20 seconds. It means 140 40 14040 = 3.5 rounds From the original point, he moves for 2 minutes 20 seconds. Case i) Original point means any point in the corner of the field. In this case he moves for 2 min 20 seconds from the diagonal of the field. Displacement is equal to diagonal of the field displacement = √ 10 2 + 10 2 102+102 = 14.1 m. Case ii) Original point means any point in the middle of the side of the field. ∴ Displacement is equal to any side of the field = 10 m. It means displacement is between any original point i.e., in between 14.1 m and 10 m.
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
totaldistance
=3.5
At this point, the farmer is at a point say B from the origin O
Thus the displacement s=
10
2
+10
2
from Pythagoras theorem.
s=10
2
=14.14m