# A farmer moves along the boundary of a square field of side 10 m in 40 s.

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

by (128k points) Given, side of the square field = 10 m

Therefore, perimeter = 10 m × 4 = 40 m

Farmer moves along the boundary in 40 s

Time = 2 minutes 20 s = 2 × 60 s + 20 s = 140 s

since, in 40 s farmer moves 40 m

Therefore, in 1s distance covered by farmer = 40 ÷ 40 = 1m.

Therefore, in 140s distance covered by farmer = 1 × 140 m = 140 m

Now, number of rotation to cover 140 along the boundary = Total distance / perimeter

= 140 m ÷ 40 m = 3.5 round

Thus after 3.5 round farmer will at point C (diagonally opposite to his initial position) of the field.

Therefore, Displacement AC = √102 + 102 = √200 = 10√2 m

Thus, after 2 minute 20 second the displacement of farmer will be equal to 10√2 m north east from initial position.