Question

a farmer moves along the boundary of a square field of side 10 metre in 40 second what will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position

Answer 1

Hi friend,

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 helps......

Answer 2

Side of the square = 10 m
Perimeter of the square = 4×10 = 40 m
He completes 1 round in 40 s.
So, speed = 40/40 = 1 m/s
So, distance covered in 2 min 20 s or 140 s is = 140 × 1 = 140 m
Number of rounds of the square completed in moving through 140 m is = 140/40 = 3.5
In 3 rounds the displacement is zero. In 0.5 round the farmer reaches the diagonally opposite end of the square from his starting point.
Displacement = AC = (AB + BC ) = (100+100) = 10√2 m
.......:)