Question

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

Answer 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 =

$\frac{40}{40}$

Distance covered in 140 s =

$\frac{40}{40} \times 140$

=140 m

Therefore,

Total round completed=

$\frac{40}{140}$

=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 ☺️