5) A farmer moves along the boundary of a square field
of side 10 m in 40s. What will be the magnitude of the
displacement of the farmer at the end of 2 minutes 20
s from his initial position?
Answers
Answer:
Here, Side of the given square field = 10m
so, perimeter of a square = 4*side = 10 m * 4 = 40 m
Farmer takes 40 s to move along the boundary.
Displacement after 2 minutes 20 s = 2 * 60 s + 20 s = 140 seconds
since in 40 s farmer moves 40 m
Therefore, in 1s the distance covered by farmer = 40 / 40 m = 1m
Therefore, in 140s distance covered by farmer = 1 � 140 m = 140 m. Now, number of rotation to cover 140 along the boundary= Total Distance / Perimeter
= 140 m / 40 m = 3.5 round
Thus, after 3.5 round farmer will at point C of the field.
Thus, after 2 min 20 seconds the displacement of farmer will be equal to 14.14 m north east from initial position.
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Given - Farmer covers 10 m in 40s
Given - Farmer covers 10 m in 40s And he is following a sqare path with side length 10 m
Given - Farmer covers 10 m in 40s And he is following a sqare path with side length 10 m So after 2 min 20s
Given - Farmer covers 10 m in 40s And he is following a sqare path with side length 10 m So after 2 min 20s Total time will be (60+60+20) s
Given - Farmer covers 10 m in 40s And he is following a sqare path with side length 10 m So after 2 min 20s Total time will be (60+60+20) sHe will take 120s to come at its initial point
Given - Farmer covers 10 m in 40s And he is following a sqare path with side length 10 m So after 2 min 20s Total time will be (60+60+20) sHe will take 120s to come at its initial point Then we have total displacement for only 20s,
Given - Farmer covers 10 m in 40s And he is following a sqare path with side length 10 m So after 2 min 20s Total time will be (60+60+20) sHe will take 120s to come at its initial point Then we have total displacement for only 20s,So in 20s he will cover only half of its one side
Given - Farmer covers 10 m in 40s And he is following a sqare path with side length 10 m So after 2 min 20s Total time will be (60+60+20) sHe will take 120s to come at its initial point Then we have total displacement for only 20s,So in 20s he will cover only half of its one side That is 5 m.
Given - Farmer covers 10 m in 40s And he is following a sqare path with side length 10 m So after 2 min 20s Total time will be (60+60+20) sHe will take 120s to come at its initial point Then we have total displacement for only 20s,So in 20s he will cover only half of its one side That is 5 m. So total diaplacemnt will be 5 m.
Given - Farmer covers 10 m in 40s And he is following a sqare path with side length 10 m So after 2 min 20s Total time will be (60+60+20) sHe will take 120s to come at its initial point Then we have total displacement for only 20s,So in 20s he will cover only half of its one side That is 5 m. So total diaplacemnt will be 5 m. Hope it will help you.