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A farmer moves along the boundary of a square field of side 10 met in 40 seconds then what will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds?
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A farmer moves along the boundary of a square field of side 10 met in ...
Answer: Total distance 40m in 40 sec. Total time taken by the farmer = 2 min 20 sec. =140 seconds. Total rounds completed= 140/40=3.5 rounds That means if the farmer starts from point A of the square field, he reaches point C. Therefore, displacement is AC. Assume ABC is a right angled triangle. Therefore, AC2= AB2 +BC2 AC2= (10)2 +(10)2 AC2= 100+100 AC2=200 AC = (200)1/2 AC= 10× (2) ½
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A farmer moves along the boundary of a square field of side 10 met in ...
Problem:
A farmer moves along the boundary of a square field of side 10 meters in 40 seconds. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds?
Approach:
To solve the problem, we need to find the distance traveled by the farmer in 2 minutes 20 seconds and the direction of his final displacement. The farmer moves along the boundary of a square field of side 10 meters, which means that he covers a distance of 40 meters in each round. Therefore, the total distance covered by the farmer in 2 minutes 20 seconds is:
Calculations:
Total time = 2 minutes 20 seconds = 140 seconds
Total distance covered = (40 meters/round) x (140 seconds/40 seconds/round) = 140 meters
Now, to find the magnitude of displacement, we need to find the straight-line distance between the starting point and the ending point of the farmer. As the farmer moves along the boundary of a square field, he covers the same distance in all four directions. Therefore, the displacement of the farmer is equal to the length of one of the diagonals of the square field. The length of the diagonal can be found using the Pythagorean theorem:
Diagonal of the square field:
Length of one side of the square = 10 meters
Length of diagonal = √(10² + 10²) = √200 ≈ 14.14 meters
Answer:
Therefore, the magnitude of the displacement of the farmer at the end of 2 minutes 20 seconds is approximately 14.14 meters.
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A farmer moves along the boundary of a square field of side 10 met in 40 seconds then what will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds?
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A farmer moves along the boundary of a square field of side 10 met in 40 seconds then what will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about A farmer moves along the boundary of a square field of side 10 met in 40 seconds then what will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A farmer moves along the boundary of a square field of side 10 met in 40 seconds then 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 met in 40 seconds then what will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about A farmer moves along the boundary of a square field of side 10 met in 40 seconds then what will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A farmer moves along the boundary of a square field of side 10 met in 40 seconds then what will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds?.
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