A farmer moves along the
boundary of a square field of side
10 min 40 s. What will be the
magnitude of displacement

1 thought on “A farmer moves along the<br />boundary of a square field of side<br />10 min 40 s. What will be the<br />magnitude of displacement”

1. Answer:

   Given,

   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.

   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 =

   [tex] \frac{40}{40} [/tex]

   Distance covered in 140 s =

   [tex] \frac{40}{40} \times 140[/tex]

   =140 m

   Therefore,

   Total round completed=

   [tex] \frac{40}{140} [/tex]

   =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

   Hypotenuse^2= Perpendicular^2 + Base^2

   AC²=AB² +BC²

   AC² = 10²+ 10²

   AC²=100+100

   AC= √200

   AC² = √200

   AC =√ 2 x 100

   AC=10√2 m

   AC 10 x 1.41

   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

   Hope will be helpful ☺️

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