Question

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

Physics class 9​

Answer 1

Given that: A farmer move along the boundary of a square field of side 10 metres in 40 s. What will be the magnitude of displacement of farmer at the end of 2 min 20 seconds

According to statement:

Provided that:

• Side of square field = 10 metres

• Time taken to cover 10 metres = 40 seconds

To calculate:

• The magnitude of displacement of farmer at the end of 2 min 20 seconds

Solution:

• The magnitude of displacement of farmer at the end of 2 min 20 seconds = 10√2 m

Full solution:

~ Firstly let us find out perimeter of the given square field!

${\small{\underline{\boxed{\sf{Perimeter \: of \: square \: = 4 \times side}}}}}$

Therefore, according to formula

→ Perimeter of square = 4 × side

→ Perimeter of square = 4 × 10

→ Perimeter of square = 40 m

~ Now let's convert minutes into seconds by applying suitable formula!

${\small{\underline{\boxed{\sf{1 \: minute \: = 60 \: seconds}}}}}$

Therefore, according to formula

→ 1 minute = 60 seconds

→ 2 min 20 sec = 60 × 2 + 20

→ 2 min 20 sec = 120 + 20

→ 140 seconds...henceforth, converted!

~ Now let's see what to do!

Explanation: As it's given that he covers each 10 metres (40 m because as we find the perimeter) in 40 seconds Therefore, he covers

→ 40/40 = 1 m in 1 second

~ Now let's find how many rounds he take in his field that is in square shape!

→ Rounds = Time/Perimeter

→ Rounds = 140/40

→ Rounds = 3.5

~ Now let's find out the magnitude of displacement!

Explanation: As we get it that he moved and we observe that at last movement he take half round that is 3.5 rounds he take means 3 and half rounds.

So here we can apply phythagoras theorm as then his position must be looked like a right angled triangle

Also we can apply formula to find diagonal of square here as it's a square field and looking like diagonal is cutting square field.

Don't forget! Displacement is said to be the shortest distance!

(Choice may yours!)

By applying diagonal of sq. formula we get the following results!

${\small{\underline{\boxed{\sf{Diagonal \: of \: square \: = a \sqrt{2}}}}}}$

Here, a denotes side.

→ 10√2

→ Displacement = 10√2 metres

By applying phythagoras theorm we get the following results!

${\small{\underline{\boxed{\sf{(H)^{2} \: = (P)^{2} \: + (B)^{2}}}}}}$

Here, H is hypotenuse, P is perpendicular, B is the base.

→ H² = P² + B²

→ H² = (10)² + (10)²

→ H² = 100 + 100

→ H² = 200

→ H = √200

→ H = 10√2 metres

→ Displacement = 10√2 metres

Answer 2

Answer:

10√2 m or 14.142 m

Explanation:

As per the provided information in the given question, we have :

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

We are asked to calculate the magnitude of displacement of farmer at the end of 2 minutes 20 seconds from his initial point.

In order to calculate the displacement, firstly we need to find its final position. In order to do that so, we have to find the rounds taken by him, so we can find his final position easily.

→ Time taken = 2 minutes 20 seconds

→ Time taken = (2 × 60) seconds + 20 seconds

→ Time taken = 120 seconds + 20 seconds

→ Time taken = 140 s

According to the question, he covers 1 round of the field in 40 seconds.

→ 1 round = Length of the boundary of square

→ 1 round = 4 × 10 m

→ 1 round = 40 m

Now,

→ 1 round = 40 seconds

→ $\sf \dfrac{1}{40}$ rounds = 1 second

Now, if he covers $\sf \dfrac{1}{40}$ rounds, so rounds covered by him in 140 seconds will be given by,

→ No. of rounds in 140 s = $\sf \dfrac{1}{40}$ × 140

→ No. of rounds in 140 s = 3.5

Henceforth, it covers 3.5 rounds in 140 seconds.

Now, suppose he starts from point A after covering 3 rounds he'll come back again to point A then he'll cover 0.5 or half round, and after 3.5 rounds it'll come to point C, which is his final position.

Displacement is the shortest distance from initial to final position. Here, initial position is A and final position is C. The shortest distance from two points is always a straight line. So, AC or the diagonal of the square is the shortest distance from A to C.

By using Pythagoras property,

→ (AC)² = (AB)² + (BC)²

→ (AC)² = (10 m)² + (10 m)²

→ (AC)² = 100 m² + 100 m²

→ (AC)² = 200 m²

→ AC = √(200 m²)

→ AC = 10√2 m

∴ The magnitude of displacement of farmer at the end of 2 minutes 20 seconds from his initial point is 10√2 m.