Question

a farmer moves along the boundary of a square field in 40 seconds . what will be the magnitude of displacement of the farmer at the end of 2 min and 20 seconds

Answer 1

Correct Question

A farmer moves along the boundary of a square field of side 10m 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?

Given:

Length of side of square field, a= 10m

Time taken by farmer to complete 1 round of field, t= 40 s

To Find:

Magnitude of the displacement at the end of 2 min 20 s

Solution:

We know that,

• Length of diagonal d of square is given by

$\pink{\boxed{\bf{d=\sqrt{2}\ a}}}$

where a is the length of side of square

• Displacement is the shortest distance between starting and final position.

$\rule{190}{1}$

Given time t is

$\longrightarrow\rm{t=2\:min\:+20\:s}$

$\longrightarrow\rm{t=2\times60\:s\:+20\:s}$

$\longrightarrow\rm{t=120\:s\:+20\:s}$

$\longrightarrow\rm{t=140\:s}$

$\rule{190}{1}$

Now,

Rounds completed by farmer in 40s= 1

Rounds completed by farmer in 1s=$\dfrac{1}{40}$

Rounds completed by farmer in 140s=$\rm{\dfrac{1}{40}\times140}$

or

Rounds completed by farmer in 140s=$3.5$

That means farmer covers 3 and a half round in 140s

$\rule{190}{1}$

Let the diagonal of given square be d

So,

$\longrightarrow\rm{d=\sqrt{2}\ a}$

$\longrightarrow\rm{d=\sqrt{2}\times10 }$

$\longrightarrow\rm{d=10\sqrt{2}\ m}$

$\rule{190}{1}$

Let the displacement of farmer be 'D'

Considering that farmer has started taking round from one of the corner of square field

Since, after covering 3 and half rounds of field final position of farmer is opposite corner to starting position

So,

$\longrightarrow\rm{D=d}$

$\longrightarrow\rm\green{D=10\sqrt{2}\:m}$

Hence, the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds is 10√2 m.

Answer 2

Answer:

Displacement = 14.142 m/s

Explanation:

Given-

Side of square = 10 m

Time taken to take one round around the field= 40 secs

Total time taken = 2mins 20s

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To Find-

Second Initial position.

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Solution -

Converting mins to s =

2 mins 20s = 2(60)+20= 120+20 = 140 s

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Rounds taken in 140 s = 140÷40 = 3.5 rounds

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[make a square ABCD and mark every side as 10m for better understanding]

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By observing the question, we get to know that the intial point(displacement) is the diagonal of the field.

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So now finding the diagonal through Pythagoras theorem -

Hypotenuse² = height² +base²

${x}^{2} = {10}^{2} + {10}^{2} \\ \\ {x}^{2} = 100 + 100 \\ \\ {x}^{2} = 200 \\ \\ {x}^{2} = 10 \sqrt{2} \\ \\ x = 14.142 \: m {s}^{ - 1}$

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Displacement = 14.142 m/s