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A farmer moves along the boundary of a square field of the side 10 m in 40 second.what will be the magnitude of displacement of the farmer at the end of ,2 minute 20 second from his initial position?
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A farmer moves along the boundary of a square field of the side 10 m i...
Solution:
Given: Side of the square field = 10 m
Time taken by the farmer to move along the boundary of the field = 40 seconds
Time taken by the farmer to move along the boundary of the field for 2 minutes 20 seconds = 140 seconds
To find: Magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position
Explanation:
Distance covered by the farmer in moving along the boundary of the field for 1 round = Perimeter of the square field
= 4 × 10 m = 40 m
Time taken by the farmer to move along the boundary of the field for 1 round = 40 seconds
Therefore, the speed of the farmer = Distance covered / Time taken = 40 m / 40 s = 1 m/s
Now, the time taken by the farmer to move along the boundary of the field for 2 minutes 20 seconds = 140 seconds
Therefore, the number of rounds made by the farmer = Total time taken / Time taken for 1 round
= 140 s / 40 s = 3.5 rounds
Hence, the farmer will be at the initial position at the end of 3 rounds, and in the 4th round, he will be at a distance of 10 m from the initial position in the direction perpendicular to his initial position.
So, the magnitude of the displacement of the farmer at the end of 2 minutes 20 seconds from his initial position = Distance between initial position and final position
= √[(10 m)2 + (10 m)2] = √200 m ≈ 14.14 m
Therefore, the magnitude of the displacement of the farmer at the end of 2 minutes 20 seconds from his initial position is approximately 14.14 m.
Summary:
Initial position of the farmer = at one corner of the square field
Distance covered by the farmer in moving along the boundary of the field for 1 round = Perimeter of the square field = 40 m
Speed of the farmer = Distance covered / Time taken = 1 m/s
Time taken by the farmer to move along the boundary of the field for 2 minutes 20 seconds = 140 seconds
Number of rounds made by the farmer = Total time taken / Time taken for 1 round = 3.5 rounds
Magnitude of the displacement of the farmer at the end of 2 minutes 20 seconds from his initial position = Distance between initial position and final position = √[(10 m)2 + (10 m)2] = √200 m ≈ 14.14 m
Given: Side of the square field = 10 m
Time taken by the farmer to move along the boundary of the field = 40 seconds
Time taken by the farmer to move along the boundary of the field for 2 minutes 20 seconds = 140 seconds
To find: Magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position
Explanation:
Distance covered by the farmer in moving along the boundary of the field for 1 round = Perimeter of the square field
= 4 × 10 m = 40 m
Time taken by the farmer to move along the boundary of the field for 1 round = 40 seconds
Therefore, the speed of the farmer = Distance covered / Time taken = 40 m / 40 s = 1 m/s
Now, the time taken by the farmer to move along the boundary of the field for 2 minutes 20 seconds = 140 seconds
Therefore, the number of rounds made by the farmer = Total time taken / Time taken for 1 round
= 140 s / 40 s = 3.5 rounds
Hence, the farmer will be at the initial position at the end of 3 rounds, and in the 4th round, he will be at a distance of 10 m from the initial position in the direction perpendicular to his initial position.
So, the magnitude of the displacement of the farmer at the end of 2 minutes 20 seconds from his initial position = Distance between initial position and final position
= √[(10 m)2 + (10 m)2] = √200 m ≈ 14.14 m
Therefore, the magnitude of the displacement of the farmer at the end of 2 minutes 20 seconds from his initial position is approximately 14.14 m.
Summary:
Initial position of the farmer = at one corner of the square field
Distance covered by the farmer in moving along the boundary of the field for 1 round = Perimeter of the square field = 40 m
Speed of the farmer = Distance covered / Time taken = 1 m/s
Time taken by the farmer to move along the boundary of the field for 2 minutes 20 seconds = 140 seconds
Number of rounds made by the farmer = Total time taken / Time taken for 1 round = 3.5 rounds
Magnitude of the displacement of the farmer at the end of 2 minutes 20 seconds from his initial position = Distance between initial position and final position = √[(10 m)2 + (10 m)2] = √200 m ≈ 14.14 m
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A farmer moves along the boundary of a square field of the side 10 m in 40 second.what will be the magnitude of displacement of the farmer at the end of ,2 minute 20 second from his initial position?
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A farmer moves along the boundary of a square field of the side 10 m in 40 second.what will be the magnitude of displacement of the farmer at the end of ,2 minute 20 second from his initial position? 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 the side 10 m in 40 second.what will be the magnitude of displacement of the farmer at the end of ,2 minute 20 second from his initial position? 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 the side 10 m in 40 second.what will be the magnitude of displacement of the farmer at the end of ,2 minute 20 second from his initial position?.
A farmer moves along the boundary of a square field of the side 10 m in 40 second.what will be the magnitude of displacement of the farmer at the end of ,2 minute 20 second from his initial position? 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 the side 10 m in 40 second.what will be the magnitude of displacement of the farmer at the end of ,2 minute 20 second from his initial position? 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 the side 10 m in 40 second.what will be the magnitude of displacement of the farmer at the end of ,2 minute 20 second from his initial position?.
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