Question

2. A farmer
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

Answer:

800m

Explanation:

because if we plus 40,20 it gives 60.Then 60 multiply by 2 gives 120.Then 120 minus 40 gives 80. Then 80 multiply by 10 gives 800.It means 800

Answer 2

Given,

Side of the given square field = 10m

We know,

$\boxed{\bf Perimeter \:of \:a \:square = 4 \times side }$

So,

⇒ Perimeter = 4 × 10

⇒ Perimeter = 40 m

Given,

Farmer takes 40 s to move along the boundary.

So,

Displacement after 2 minutes 20 s = (2 × 60) + 20

⇒ Displacement = 140 s

Since,

In 40 s farmer moves 40 m .

Therefore,

In 1 s the distance covered by farmer = $\sf \frac{40}{40}$  = 1 m

In 140 s the distance covered by farmer = 1 × 140 m = 140 m.

Now,

Number of rotation to cover 140 along the boundary = $\sf \frac{Total \:\:Distance}{Perimeter}$

⇒ Number of rotations  = $\sf\frac{140}{40}$

⇒ Number of rotations = 3.5 round s

Thus,

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

Therefore,

Displacement AC = $\sf \sqrt{10^2+10^2}$

⇒ Displacement = $\sf \sqrt{100+100}$

⇒ Displacement = $\sf \sqrt{200}$

⇒ Displacement = $10\sqrt2$

⇒ Displacement = $\sf 10\times 1.414$

⇒ Displacement = 14.14 m

Thus,

Displacement of farmer = 14.14 m in north east from initial position.

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