A farmer movers along the boundary of a square field of side 10 m in 40 seconds of shown in the figure below what will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds for his initial position?
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Q. A farmer movers along the boundary of a square field of side 10 m in 40 seconds of shown in the figure below what will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds for his initial position?
Answer :
Given, side of the square field = 10 m
Therefore, perimeter = 10×4 = 40 m
Farmer moves along the boundary in 40 s.
Total time of journey = 2×60+20 = 140 s
Since, in 40 s farmer moves 40 m.
Therefore, in 1 s distance covered by farmer = 40/40 m = 1 m
Therefore, in 140 s distance covered by farmer = 10×4 = 40 m
Now,
Number of rotation to cover 140 along the boundary
= (total distance)/(Perimeter)
= 140/40 = 3.5 m
Thus, after 3.5 round farmer will at point C of the field.
Therefore displacement
(AC)² = (10)² + (10)²
(AC)² = 100+100
(AC)² = 200
AC = √200
AC = 14.143 m
Thus, after 2 minutes 20 seconds the displacement of farmer will be equal to 14.143 m north east.
Answer :
Given, side of the square field = 10 m
Therefore, perimeter = 10×4 = 40 m
Farmer moves along the boundary in 40 s.
Total time of journey = 2×60+20 = 140 s
Since, in 40 s farmer moves 40 m.
Therefore, in 1 s distance covered by farmer = 40/40 m = 1 m
Therefore, in 140 s distance covered by farmer = 10×4 = 40 m
Now,
Number of rotation to cover 140 along the boundary
= (total distance)/(Perimeter)
= 140/40 = 3.5 m
Thus, after 3.5 round farmer will at point C of the field.
Therefore displacement
(AC)² = (10)² + (10)²
(AC)² = 100+100
(AC)² = 200
AC = √200
AC = 14.143 m
Thus, after 2 minutes 20 seconds the displacement of farmer will be equal to 14.143 m north east.
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