A farmer moves along the boundary of a square field of side 10 m in 40s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

A farmer moves along the boundary of a square field of side 10 m in 40

1.	A farmer moves along the boundary of a square field of side 10 m in 40s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?
Answer» A farmer moves along the boundary of a square field of side 10 m in 40s 2 minutes 20 seconds means 140 seconds. Distance travelled by farmer = \(\frac{40}{40}\) × 140 m = 140 m. ∴ Farmer moves for 2 minutes 20 seconds. It means \(\frac{140}{40}\) = 3.5 rounds From the original point, he moves for 2 minutes 20 seconds. Case i) Original point means any point in the corner of the field. In this case he moves for 2 min 20 seconds from the diagonal of the field. Displacement is equal to diagonal of the field displacement = \(\sqrt{10^2+10^2}\) = 14.1 m. Case ii) Original point means any point in the middle of the side of the field. ∴ Displacement is equal to any side of the field = 10 m. It means displacement is between any original point i.e., in between 14.1 m and 10 m.