A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?
Answer: 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
Answers & Comments
Hey mate,.
Total distance covered 10m in 40sec
Total time taken= 2min 20sec= 140sec
Total round completed = 140/40=3.5 rounds
So If an object moves from A to D, the shortest path will be from A to C. So displacement is AC, that we will find by phythogoras theorem.
AC2= AB2 +BC2
AC2= (10)2 +(10)2
AC2= 100+100
AC2=200
AC = (200)1/2
AC= 10× (2) ½
Hope it will help you
Answer: 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