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InterviewSolution
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A farmer moves along the boundry of a square field of side 10M in 40s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position? |
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Answer» Solution :The distance covered in one round of the field = the perimeter of the field `= 4 xx 10 m = 40 m`  Time taken by the farmer to complette one round. i.e., 40 m is 40 second. ln 2 minute 20 second i.e., 140 second the farmer covers a distance of 140m. One round = 40 m ` therefore m = (140)/(40) =3.4` rounds `0.5` rounds ` 0.5 xx 40 m = 20 m` If the farmeer starts from POINT A, then at the end of 2 min 20 s he will be at the point C. `therefore` The displacement is AC. To find the magnitude of AC, use pythagoras. theorem. In right ANGLED triangle ABC, `AC ^(2) = AB^(2) + BC ^(2)` =`(10)^(2) + (10) ^(2)` `= 100 + 100 = 200` `therefore AC SQRT (200) = sqrt (100 xx2 ) = 10 sqrt2 m ` The magnitude of displacement of the farmer is `10 sqrt2 m.` The distance travelled by the farmer is 140 m. |
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