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 min 20 sec from his initial position ?​

Answer 1

Answer:

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The magnitude of displacement of the farmer at the end of 2 min 20 sec from his initial position is 14.14m.

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Explanation:

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Given side of square = 10m, thus perimeter P = 40m

Time is taken to cover the boundary of 40 m = 40 s

Thus in 1 second, the farmer covers a distance of 1 m

Now distance covered by the farmer in 2 min 20 seconds = 1 × 140 = 140m

Now the total number of rotation the farmer makes to cover a distance of 140 meters =  

=   $\frac{total distance}{perimeter}$

= 3.5  

At this point, the farmer is at a point say B from the origin O

Thus the displacement s=  $\sqrt{10^{2} +10^{2} }$ from Pythagoras theorem.

s=10$\sqrt{2}$  

=14.14m

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Answer 2

Given :-

Side of the square = 10 m

Time taken by the farmer for 1 round = 40 sec

Total time taken for the displacement = 2 min 20 sec

To Find :-

The magnitude of displacement of the farmer at the end of 2 min 20 sec from his initial position.

Solution :-

We know that,

• s = Side
• t = Time
• s = Displacement
• p = Perimeter
• b = Base
• h = Height

Using the formula,

$\underline{\boxed{\sf Perimeter \ of \ the \ square=4 \times Side}}$

Given that,

Side (s) = 10 m

Substituting their values,

⇒ p = 4s

⇒ p = 4 × 10

⇒ p = 40 m

Given that,

Farmer takes 40 seconds to move along the boundary.

Displacement after 2 min 20 sec,

2 min 20 sec = 140 sec

Distance covered in 1 sec,

40/40 = 1 m

140 sec = 1 × 140 = 140 m

Number of rotation,

$\underline{\boxed{\sf Rotation=\dfrac{Total \ distance}{Perimeter} }}$

Given that,

Distance (s) = 140 m

Perimeter (p) = 40 m

Substituting their values,

⇒ 140/40

⇒ 3.5 rounds

Using the Pythagoras theorem,

$\underline{\boxed{\sf (Hypotenuse)^2 =(Base^2)+(Altitude^2)}}$

Given that,

Base (b) = 10 m

Altitude = 10 m

Substituting their values,

⇒ h² = 10² + 10²

⇒ h² = 100 + 100

⇒ h² = 200 m

⇒ h = √200

⇒ h = 10√2 m = 14.14 m

Therefore, the magnitude of displacement of the farmer at the end of 2 min 20 sec from his initial position is 14.14 m.