Question

A farmer moves along the boundary of a square field of of side 10 metre in 40 second what will be the magnitude of displacement of the former at the end of 2 minutes 20 seconds from his initial point

Answer 1

ANSWER:

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

Time 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 =perimetertotaldistance

=3.5 

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

Thus the displacement s=102+102 from Pythagoras theorem.

s=10

$\sqrt{2}$

=14.14m

Answer 2

Given :-

Side of the square field = 10 m

Time taken to move along the boundary = 40 sec

To Find :-

The magnitude of displacement of the former at the end of 2 minutes 20 seconds from his initial point.

Solution :-

We know that,

• p = Perimeter
• d = Distance
• s = Displacement

Using the formula,

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

Given that,

Side = 10 m

Substituting their values,

⇒ p = 4 × 10

⇒ p = 40 m

According to the question,

Displacement after 2 minutes 20 s = 2 × 60 + 20 = 140 sec

Given, farmer moves 40 m in 40 sec.

Therefore,

The distance covered by farmer = 40/40 = 1 m

Distance covered by farmer in 140 sec = 140 × 1 = 140 m

Now, we have

No. rotation to cover 140 along the boundary = Total Distance / Perimeter

Given that,

Total distance (d) = 140 m

Perimeter (p) = 40 m

Substituting their values,

⇒ 140/40

⇒ 3.5 rounds

Using the formula,

$\underline{\boxed{\sf Displacement \ AC = \sqrt{AB^2+BC^2} }}$

Substituting their values,

⇒ AC = √10²+10²

⇒ AC = √100+100

⇒ AC = √200

⇒ AC = 10√2 m = 14.14 m

Therefore, the displacement of farmer will be equal to 14.14 m north east from initial position.