a farmer moves along the boundary of a square field of side 10 m in 40 seconds 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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1
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)^1/2
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)^1/2
Answered by
2
At the end of 2 min 20 sec farmer had taken rounds= 140/40 3.5 rounds
therfore, we need to find displacement of 0.5 rounds as 3 round displacement is 0
so this displacement= diagnol of square
by applying Pythagorean theorem
Displacement²= 10²+10²
displacement =√200
displacement= √2×10
displacement = 14.2 m
therfore, we need to find displacement of 0.5 rounds as 3 round displacement is 0
so this displacement= diagnol of square
by applying Pythagorean theorem
Displacement²= 10²+10²
displacement =√200
displacement= √2×10
displacement = 14.2 m
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