a farmer moves along the boundary of square field of side 10m in 40 sec .what will be the magnitude displacement of the farmer at the end of 2minutes and 20sec what is the total distance covered by him.
balaji7:
help me as fast
Answers
Answered by
2
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) ½
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) ½
Not 40m in 40s? Thats 10.
it is physics problem
plz help me i need it urgent
Answered by
1
Speed=10m/40s=0.25m/s.
Time=140 s.
Distance=140*0.25=35m(3 sides and half of one side).
Displacement=-5m(He covered half of the last side, 5 m behind starting point in definite direction )
Time=140 s.
Distance=140*0.25=35m(3 sides and half of one side).
Displacement=-5m(He covered half of the last side, 5 m behind starting point in definite direction )
Similar questions