Question

A farmer moves along the
boundary of Sauare
field of side 10 minute in 40 seconds. what will be the magnitude of the
displacement of farmer at
the end of 2 minutes and 20 seconds?​

Answer 1

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

Side of square = 10 m and time = 40 sec

Perimeter of square = 4 × side

= 4 × 10 = 40 m

We have to find the displacement of the farmer at the end of 2 min 20 sec.

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 of the farmer at the end of 2 min and 20 seconds is 10√2 m

Answer 2

Correct Question -

A farmer moves along the boundary of a square field of side 10 metres in 40 seconds.

What will be the magnitude of the displacement of farmer at

the end of 2 minutes and 20 seconds?

Solution -

In the above question , the following information is given -

A farmer moves along the boundary of a square field of side 10 metres in 40 seconds.

We can use this information to find the speed of the farmer .

Here ,

The boundary of the given square is 10m .

Therefore the perimeter of the square -

=> 4 × Boundary

=> 4 × 10

=> 40m .

Now, the farmer moves this distance in 10 sec .

So, Velocity Of Farmer -

=> (40 / 40) m/s

=> 1 m/s.

40 Now that we have the velocity of the farmer, let us calculate the distance travelled by the farmer in 2 minutes 20 seconds .

Given time = 2 minute 20 secondd

=> 140 seconds.

The farmer moves 140 metres in 140 seconds around the square.

So,

After travelling 40m, he reaches the innitial point.

Travelling another 40m, he again reaches the innitial point .

Similarly travelling another 40m, he again reaches the innitial point .

So, he travels 20 m { 140 - 3 × 20 = 20 m }

Now,

10n is the length of each side of the square .

He travels 1 side , 10m and reaches the end of the consecutive side ( travelling another 10m ) .

So,

The farmer is at the diagonal of the square .

Displacement = ✓ [ 100 + 100 ] = 10✓ 2 m.

Hence , displacement = 10✓2 m.