A person moves along the boundary of a square park of side 10 m in 40 s. What will be the magnitude of
displacement and distance of the person at the end of 100 seconds?
Answers
Explanation:
Given side of square =10m, thus perimeter P=40m
Given side of square =10m, thus perimeter P=40mTime taken to cover the boundary of 40 m =40 s
Given side of square =10m, thus perimeter P=40mTime taken to cover the boundary of 40 m =40 sThus in 1 second, the farmer covers a distance of 1 m
Given side of square =10m, thus perimeter P=40mTime taken to cover the boundary of 40 m =40 sThus in 1 second, the farmer covers a distance of 1 mNow distance covered by the farmer in 2 min 20 seconds = 1×140=140m
Given side of square =10m, thus perimeter P=40mTime taken to cover the boundary of 40 m =40 sThus in 1 second, the farmer covers a distance of 1 mNow distance covered by the farmer in 2 min 20 seconds = 1×140=140mNow the total number of rotation the farmer makes to cover a distance of 140 meters =
Given side of square =10m, thus perimeter P=40mTime taken to cover the boundary of 40 m =40 sThus in 1 second, the farmer covers a distance of 1 mNow distance covered by the farmer in 2 min 20 seconds = 1×140=140mNow the total number of rotation the farmer makes to cover a distance of 140 meters = perimeter
Given side of square =10m, thus perimeter P=40mTime taken to cover the boundary of 40 m =40 sThus in 1 second, the farmer covers a distance of 1 mNow distance covered by the farmer in 2 min 20 seconds = 1×140=140mNow the total number of rotation the farmer makes to cover a distance of 140 meters = perimetertotaldistance
Given side of square =10m, thus perimeter P=40mTime taken to cover the boundary of 40 m =40 sThus in 1 second, the farmer covers a distance of 1 mNow distance covered by the farmer in 2 min 20 seconds = 1×140=140mNow the total number of rotation the farmer makes to cover a distance of 140 meters = perimetertotaldistance =3.5
Given side of square =10m, thus perimeter P=40mTime taken to cover the boundary of 40 m =40 sThus in 1 second, the farmer covers a distance of 1 mNow distance covered by the farmer in 2 min 20 seconds = 1×140=140mNow the total number of rotation the farmer makes to cover a distance of 140 meters = perimetertotaldistance =3.5 At this point, the farmer is at a point say B from the origin O
Given side of square =10m, thus perimeter P=40mTime taken to cover the boundary of 40 m =40 sThus in 1 second, the farmer covers a distance of 1 mNow distance covered by the farmer in 2 min 20 seconds = 1×140=140mNow the total number of rotation the farmer makes to cover a distance of 140 meters = perimetertotaldistance =3.5 At this point, the farmer is at a point say B from the origin OThus the displacement s=
Given side of square =10m, thus perimeter P=40mTime taken to cover the boundary of 40 m =40 sThus in 1 second, the farmer covers a distance of 1 mNow distance covered by the farmer in 2 min 20 seconds = 1×140=140mNow the total number of rotation the farmer makes to cover a distance of 140 meters = perimetertotaldistance =3.5 At this point, the farmer is at a point say B from the origin OThus the displacement s= 10 2 +10
Given side of square =10m, thus perimeter P=40mTime taken to cover the boundary of 40 m =40 sThus in 1 second, the farmer covers a distance of 1 mNow distance covered by the farmer in 2 min 20 seconds = 1×140=140mNow the total number of rotation the farmer makes to cover a distance of 140 meters = perimetertotaldistance =3.5 At this point, the farmer is at a point say B from the origin OThus the displacement s= 10 2 +10 2 from Pythagoras theorem.
Given side of square =10m, thus perimeter P=40mTime taken to cover the boundary of 40 m =40 sThus in 1 second, the farmer covers a distance of 1 mNow distance covered by the farmer in 2 min 20 seconds = 1×140=140mNow the total number of rotation the farmer makes to cover a distance of 140 meters = perimetertotaldistance =3.5 At this point, the farmer is at a point say B from the origin OThus the displacement s= 10 2 +10 2 from Pythagoras theorem.s=10
Given side of square =10m, thus perimeter P=40mTime taken to cover the boundary of 40 m =40 sThus in 1 second, the farmer covers a distance of 1 mNow distance covered by the farmer in 2 min 20 seconds = 1×140=140mNow the total number of rotation the farmer makes to cover a distance of 140 meters = perimetertotaldistance =3.5 At this point, the farmer is at a point say B from the origin OThus the displacement s= 10 2 +10 2 from Pythagoras theorem.s=10 2
Given side of square =10m, thus perimeter P=40mTime taken to cover the boundary of 40 m =40 sThus in 1 second, the farmer covers a distance of 1 mNow distance covered by the farmer in 2 min 20 seconds = 1×140=140mNow the total number of rotation the farmer makes to cover a distance of 140 meters = perimetertotaldistance =3.5 At this point, the farmer is at a point say B from the origin OThus the displacement s= 10 2 +10 2 from Pythagoras theorem.s=10 2 =14.14m