Sarthaks Test
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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?

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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. 

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