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?

Given,

Time period of travel = 2 minutes 20 seconds

Convert min into sec

1 min = 60 sec

So 2 min 20 sec = (2 x 60) sec + 20 sec = 120 + 20 = 140 sec

Side of the given square field = 10 m

Hence, the perimeter of a square = 4 x side = 4 x 10 = 40 m

Time taken by the farmer to cover the boundary of 40 m = 40 s

So, in 1 s, the farmer covers a distance of = $\dfrac{40}{40}$ x 1 = 1 m

Hence, distance covered by the farmer in 140 sec = 140 m

The total number of rotations taken by the farmer to cover a distance of 140 m

= $\dfrac{\text{total distance}}{\text{perimeter}}$

= $\dfrac{140}{40}$

= 3.5

If the farmer starts from point A, after 3.5 rounds he will be at point C of the field.

∴ Displacement (from Pythagoras theorem)

$\text{AC} = \sqrt{(\text{AB})^2 + (\text{BC})^2} \\[1em] = \sqrt{(10)^2 + (10)^2} \\[1em] = \sqrt{100 + 100} \\[1em] = \sqrt{200} \\[1em] = 10\sqrt{2} \\[1em] = 10 \times 1.414 \\[1em] = 14.14 \text{ m}$

Hence, the magnitude of displacement = 14.14 m