Q2) A farmer moves along the boundary of the square feild of side 10m in 40 seconds. What will be the magnitude of displacement of the famer at the end of 2 minute 20 seconds from its initial position.

 

Let the initial point of the farmer be point P

The distance covered by the farmer in 40 s is 4 × 10 = 40 m. 

The total time given: 2 min and 20 s = (2 × 60) + 20 = 140 s

Since the farmer moves 40 m in 40 s, therefore in 1 s the farmer covers a distance,


Therefore, the distance covered by the farmer in 140 s is:

140 × 1 m = 140 m.

Thus the number of rotations to cover 140 m along the boundary is:

So, the farmer will be at point R after 140 s.

So the displacement = PR =  = 14.1 m