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Question
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?
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
The correct option is C 14.14m
Given, side of the square field = 10 m
Therefore, perimeter = 10 m×4=40 m
The farmer moves along the boundary in 40 s.
Displacement after 2 m 20 s(=2×60 s+20s=140s)=?
Since in 40 s the farmer moves 40 m, in 1s distance covered by farmer = 40m40=1m
Therefore, in 140s distance covered = 140 m.
Number of rounds covered =140 m40 m=3.5 rounds
If the farmer starts from point A, after 3.5 rounds he will at point C of the field.
Therefore, displacement AC=√(10 m)2+(10 m)2
=√100m2+100m2
=√200m2
=10√2m
=10×1.414m=14.14m
Given, side of the square field = 10 m
Therefore, perimeter = 10 m×4=40 m
The farmer moves along the boundary in 40 s.
Displacement after 2 m 20 s(=2×60 s+20s=140s)=?
Since in 40 s the farmer moves 40 m, in 1s distance covered by farmer = 40m40=1m
Therefore, in 140s distance covered = 140 m.
Number of rounds covered =140 m40 m=3.5 rounds
If the farmer starts from point A, after 3.5 rounds he will at point C of the field.
Therefore, displacement AC=√(10 m)2+(10 m)2
=√100m2+100m2
=√200m2
=10√2m
=10×1.414m=14.14m
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