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 2 minutes 20 seconds from his initial position?

Answer 1

Here:

Side of the given square field = 10 m

So:

Perimeter of a square:

$\implies$ 4 × side

$\implies$ 10 m × 4

$\implies$ 40 m

Note: Farmer takes 40 s to move along the boundary.

Displacement after 2 minutes 20 s:

$\implies$ 2 × 60 s + 20 s

$\implies$ 120 s + 20 s

$\implies$ 140 seconds

Since in 40 s farmer moves 40 m:

Therefore:

In 1s the distance covered by farmer:

$\implies$ $\frac{40}{40}$

$\implies$ 1 m

Therefore:

In 140 s distance covered by farmer:

$\implies$ 1 × 140 m

$\implies$ 140 m

Now:

Number of rotation to cover 140 along the boundary:

$\implies$ $\sf{ \frac{Total \: distance}{Perimeter}}$

$\implies$ $\frac{140}{40}$

$\implies$ 3.5

Thus:

After 3.5 round farmer will at point C of the field.

Therefore:

Displacement AC:

$\implies$ $\sf{ \sqrt{ {(10 \: m ^{2} ) + (10 {m} \: ^{2} )} }}$

$\implies$ $\sf{\sqrt{100 + 100}}$

$\implies$ $\sf{ \sqrt{200}}$

$\implies$ $\sf{ 10 \sqrt{2}}$

$\implies$ $\sf{ 10 \times 1.414}$

$\implies$ $\sf{14.14 \: m}$

Thus:

After 2 min 20 seconds, the displacement of farmer will be equal to 14.14 m north-east from initial position.