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# Question 2 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?

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Solution

## Given, side of the square field,s = 10 m Therefore, perimeter = 4×s=4×10=40 m The farmer moves along the boundary in 40 s. Seconds in 2 m 20 s=2×60 s+20 s=140 s Since in 40 s, the farmer moves 40 m, in 1 s distance covered by farmer = 4040=1 m Therefore, in 140 s, distance covered = 140 m. Number of rounds covered =140 m40 m=3.5 rounds1 round = 40 m, 0.5 round = 20 m. In 20 m two sides are covered. If the farmer starts from point A, after 3.5 rounds he will be at point C of the field. Therefore, displacement AC=√(AB)2+(BC)2 =√(10)2+(10)2 =√100+100 =√200 =10√2 =10×1.414=14.14 m

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