Question

A farmer move along the boundary of a square field of side 8 m in 32 s. what will be the magnitude of displacement of farmer at the end of 1 min 20sec​

Answer 1

• Given that: A farmer move along the boundary of a square field of side 8 m in 32 s. What will be the magnitude of displacement of farmer at the end of 1 min 20 seconds

According to statement:

Provided that:

• Side of square field = 8 metres

• Time taken to cover 8 metres = 32 seconds

To calculate:

• The magnitude of displacement of farmer at the end of 1 min 20 seconds

Solution:

• The magnitude of displacement of farmer at the end of 1 min 20 seconds = 8√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 × 8

→ Perimeter of square = 32 cm

~ 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

→ 1 min 20 sec = 60 + 20

→ 80 seconds...henceforth, converted!

~ Now let's see what to do!

Explanation: As it's given that he covers each 8 metres (32 m because as we find the perimeter) in 32 seconds Therefore, he covers

→ 32/32 = 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 = 82/32

→ Rounds = 2.5

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

Explanation: As we are able to see in the attachment that how he moved and we observe that at last movement he take half round that is 2.5 rounds he take means 2 and half rounds.

So here we can apply phythagoras theorm as we are able to see that it is looking like a right angle traingle.

Also we can apply Formula to find diagonal of square here too and see it's also looking like diagonal and here, there is a 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.

→ 8√2

→ Displacement = 8√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 = 8² + 8²

→ H² = 64 + 64

→ H² = 128

→ H = √128

→ H = 8√2 metres

→ Displacement = 8√2 metres