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

Length of the side of a square field = 10 m

Time taken to move along square field = 40 s

Total time taken by the farmer to walk around the square field = 2 min 20 sec. = (60 × 2 + 20) = 140 seconds. $\left(\because 1m=60sec\right)$

In 40 sec farmer covers 40 m (perimeter of the square = 4 × 10 m = 40 m)

Thus,

Distance covered in 40 s = 40 m

Distance covered in 1 s = $\frac {40}{40}m$

Distance covered in 140 s = $\frac {40}{40}\times 140\ m$ = 140 m

Therefore, 

$Total\ round\ completed=\frac{Total\ distance\ covered}{Distance\ covered\ in\ 1\ round}$

$Total\ round\ completed=\frac{140}{40}$

$Total\ round\ completed= 3.5$

Hence the farmer covered 3.5 rounds.

Therefore, if the farmer starts moving from point A of the square field, then he reaches point C on covering 3.5 rounds.

Now, we know that,

Displacement = Shortest distance

                           = AC.

Here, ABC is a right-angled triangle.

Therefore, by Pythagoras theorem

Hypotenuse² = Perpendicular² + Base²

$A{C}^{2}=A{B}^{2}+B{C}^{2}$

$A{C}^{2}={10}^{2}+{10}^{2}$

$A{C}^{2}=100+100$

$A{C}^{2}=200$

$\sqrt{A{C}^{2}}=\sqrt{200}$            (applying root both side)

$AC=\sqrt{200}$

$AC=\sqrt{2\times 100}$

$AC=10\sqrt{2}\ m$

$AC=10\times 1.41$               $\left(\because \sqrt{2}=1.41\right)$

$AC=14.1\ m$

Thus, the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position is 14.1m.

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Updated on: 10-Oct-2022

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