Question

A farmer moves along the boundarof a square field of side 10 m in 40 s what will be the magnitude of displacement of the farmer at the end of 2 minute and 20 second

Answer 1

Given that, side of square = 10 m and time = 40 sec

Now, 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:-

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

To find:-

The magnitude of displacement of the farmer at the end of 2 minute and 20 second.

Solution:-

Here, side of the square field = 10 m

So,

It's boundary length = its perimeter  

perimeter of the square = 4 × side

= 4 × 10 = 40 m

Now,

perimeter of the square = 40 m

According to the question,

Farmer moves 40 meter in 40 seconds

Given that,

Farmer continued moving on the boundary of the square field for 2 minutes 20 seconds.

= 120 + 20 secs

= 140 seconds

Now,

Number of rounds of square field in 140 s:

140 = 40 + 40 + 40 + 20

The farmer moved 3 rounds and another half. Half means, If he starts from A of the field ABCD, he stops at C.

So,

Displacement = AC = diagonal of the square field

Diagonal of square = √2 a

Where,

• a is the side of the square

Length of diagonal = $\boxed{10\sqrt{2}\ m}$

∴Displacement of the farmer in 2 minutes and 20 seconds is 10√2 m.

Check the attachment for more reference.