# Question #50b52

May 19, 2017

Farmer's Speed $= \frac{10}{40} = 0.25 m {s}^{-} 1$

Let the farmer be at the origin initially.
Total time of travel $= 60 \times 2 + 20 = 120 + 20 = 140 \sec$
Total distance traveled$= 0.25 \times 140 = 35 m$

Field being square of side $= 10 m$
$\therefore$Total number of sides traveled in the given time$= 3.5$

We know that after traveling $4$ sides farmer would be back at the origin.
As such magnitude of displacement after given time $= 4 - 3.5 = \frac{1}{2} s i \mathrm{de}$
$= \frac{1}{2} \times 10 = 5 m$