Question

${\huge{\boxed{\red{\sf{Question :}}}}}$
A farmer moves along the boundary of a square field of side 10m in 40s. What will be the magnitude of displacement of the farmer at the end of 2 min 20s. Find its initial position.


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Answer 1

${\huge{\sf{\underline{\red{Given :}}}}}$

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

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${\huge{\sf{\underline{\blue{To \: Find :}}}}}$

What will be the magnitude of displacement of the farmer at the end of 2 min 20s. Find its initial position.

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${\sf{\huge{\green{\underline{Formula \: used :}}}}}$

${\sf{\pink{Average \: speed = \frac{total \: distance \: covered}{total \: time \: taken} }}}$

${\sf{\pink{Using \: Pythagoras \: theorem }}}$

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${\sf{\huge{\underline{\blue{Solution :}}}}}$

First,

Side of square = 10m

Perimeter of square = 4 × side = 4 × 10

= 40m

Time taken = 40s

${\sf{average \: speed = \frac{40m}{40s} = 1}}$

2 min 20 s in seconds = 140s

${\sf{Rounds covered in 140s = \frac{140}{40} = 3.5 \: rounds }}$

Than,

Shortest path is A to C .So,we will find it using Pythagoras theorem :

${\sf{ {ac}^{2} = {ab}^{2} + {bc}^{2} }}$

${\sf{ {ac}^{2} = {10}^{2} + {10}^{2} }}$

${\sf{ {ac}^{2} = 100 + 100 }}$

${\sf{ {ac}^{2} = 200 }}$

${\sf{AC = 10 \sqrt{2} }}$

${\mathfrak{\red{\boxed{AC = 10\sqrt{2} }}}}$

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Answer 2

Calculate the net distance travelled by the farmer using the given values and then use it to find the point on the square which is the final position of the farmer. The length of the line joining directly the initial and final position of the farmer is the net displacement.

Complete step-by-step answer:

Given the problem, a farmer moves along the boundary of a square field of side 10m in 40 seconds.

We need to find the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position.

First, we calculate the perimeter of the square.

The side of square ABCD is x=10m.

Perimeter of the square =4×side=4x=4×10=40m.

It is given that the farmer takes 40 seconds to complete one round of this square field.

Hence the distance travelled by the farmer in 40 seconds is 40m.

We know that speed=distancetime.

Using the above values in this formula we get,

Speed of farmer=40m40s=1 m/s.

Now we need to calculate the distance travelled by the farmer in 2 minutes 20 seconds or 140 seconds.

Using the same speed formula, we get

speed=distancetime⇒1=distance140⇒distance = 140m

Hence the distance travelled by the farmer in 2 minutes 20 seconds is 140m.

Now we need to calculate the position of the farmer on the boundary of square ABCD after 2 minutes 20 seconds.

We know that,

Perimeter of the square = One round of the square

⇒40m=1 round

⇒1m=140 round⇒140m=140×140=3.5 rounds

Hence the farmer completes 3.5 rounds of the square field ABCD in 2 minutes 20 seconds.

When the farmer completes one round of square field ABCD starting from vertex A

clockwise, after completion the net displacement from the starting position is zero.

Hence the net displacement after the completion of 3 rounds is also zero.

After 0.5 rounds, the farmer reaches the diagonally opposite point of the starting position,

that is if the farmer starts from vertex A of square field ABCD, it reaches to point C.

The net displacement from the starting point is equal to the diagonal AC of the square

field.

Hence the net displacement if the farmer after 3.5 rounds is equal to the diagonal AC of the square field.

Applying Pythagoras theorem in right angle triangle ADC, we get

Hypotenuse2=Base2+Height2⇒AC2=AD2+DC2⇒AC2=102+102=200⇒AC=200−−−√=102–√m

Hence the net displacement if the farmer after 3.5 rounds is equal to 102–√m.

Therefore, the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position is 102–√m.