Question

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
2minutes 20 seconds from his initial position?

Answer 1

A farmer moves along the boundary of a square field of side 10 m in 40 s.

Side of square = 10 m and time = 40 sec

Perimeter of square = area × area

= 4 × 10 = 40 m

We have to find the displacement of the farmer at the end of 2 minutes 20 seconds.

Time = 2 min 20 sec

1 min = 60 sec

2 min = 2(60) = 120 sec

= 120 sec + 20 sec = 140 sec

Now,

In 1 sec distance covered by farmer = 40/40 = 1 m

So, in 140 sec distance covered by farmer = 1 × 140 = 140 m

Number of rotations to cover 140 m along the boundary = Distance/Perimeter

= 140/40 = 3.5 rounds

Therefore, the farmer takes 3.5 revolutions.

Let us assume that farmer is at the point A from the origin of the square field.

Now,

Displacement = diagonal of square

And from above we have a side of square = 10 m

So, displacement = 10√2 m

Answer 2

Given:-

Side of the square field = 10 m

Time taken by the farmer to cover 10 m = 40 s

To find:-

The magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position.

Solution:-

2 minutes 20 seconds = (2*60) + 20     (∴2 minutes = 2*60 secs)

                                      = 120 + 40

                                      = 140 s

Total rounds completed by the farmer = 140/40 = 14/4 = 3.5 rounds

Perimeter of the square = 4a = 4*10 = 40 m

Now, the farmer moves from A(his initial point) and after 3.5 rounds, stops at C. (please refer the attachment)

So, the distance covered by the farmer becomes =

3*perimeter of the square + 1/2 * perimeter of the square

= (3*40) + (1/2 * 40)

= 120 + 20

= 140 m

Now, for displacement, the shortest length covered by the farmer becomes AC.

In triangle ABC,

$AC^2=AB^2+BC^2$    (∠B = 90° as it is a square field)

$AC^2=(10)^2+(10)^2\\AC^2=100+100\\AC=\sqrt{200}$

$AC=\sqrt{100*2} \\AC=10\sqrt{2} m$

Thus, the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position is 10√2 m.