2. A farmer moves along the boundary of a squarefield of side 10 m in

1.	2. A farmer moves along the boundary of a squarefield of side 10 m in 40 s. What will be themagnitude of displacement of the farmer atthe end of 2 minute 20 seconds from his initialposition?
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 = total distance/perimetre=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 therom s=10 √ 2 =14.14m

2. A farmer moves along the boundary of a squarefield of side 10 m in 40 s. What will be themagnitude of displacement of the farmer atthe end of 2 minute 20 seconds from his initialposition?

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 = total distance/perimetre=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 therom

s=10 √ 2 =14.14m

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2. A farmer moves along the boundary of a squarefield of side 10 m in 40 s. What will be themagnitude of displacement of the farmer atthe end of 2 minute 20 seconds from his initialposition?​

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2. A farmer moves along the boundary of a squarefield of side 10 m in 40 s. What will be themagnitude of displacement of the farmer atthe end of 2 minute 20 seconds from his initialposition?