Question

A farmer moves along the boundary of a square field of side 10 m in 40s .what will be the magnitude of displacement of the farmer at the end of minutes 20 second from his initial position.​

Answer 1

$\sf\star\bold\red{{QUESTION:-}}$

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

$\sf\star\bold\purple{{ANSWER:-}}$

$\implies\boxed{\boxed{{\sf\:=14.1m}}}$

$\sf\star\bold\green{{EXPLANATION:-}}$

length of square filed = 10 m

Time taken to move along square filed = 40 sec

Total time the Farmer walked on square field =2min 20 sec

$\sf=2×60+20$⠀⠀⠀ [ 1 min = 60 sec]

$\sf= 120+20$

$\sf= 140 sec$

In 40 sec Farmer covers 40 m

Thus,

Distance covered in 40s = 40m

Distance covered in 1 sec $=\sf\dfrac{40}{40}m$

Distance covered in 140s $=\sf\dfrac{40}{40}×140m$

$\sf=140m$

Thus,

Total round $\boxed{\boxed{{=\sf\dfrac{Total\: distance\: covered}{Distance\: covered\:in\:1\:round}}}}$

$=\dfrac{140}{40}$

$\sf=3.5 round$

∴ The Farmer covered 3.5 rounds

If he started from A he will be at C after 20min 20 sec

Displacement = shortest distance

$\sf=AC$

By Pythagoras theorem,

$\sf\implies{Hypotenuse}^{2}={Height}^{2}+{Base}^{2}$

$\sf\implies{AC}^{2}={AB}^{2}+{BC}^{2}$

$\sf\implies{AC}^{2}={10}^{2}+{10}^{2}$

$\sf\implies{AC}^{2}=100+100$

$\sf\implies{AC}^{2}=200$

$\sf\implies\:AC={ \sqrt{200}}$

$\sf\implies\:AC={ \sqrt{2×100}}$

$\sf\implies\:AC=10{ \sqrt{2}}m$

$\sf\implies\:AC=10×1.41$

$\implies\boxed{\boxed{{\sf\:AC=14.1m}}}$

Answer 2

Correct Question :

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

Solution :

First of all convert the given time 2 minutes 20 seconds into seconds.

Total Time = 2 minutes 20 seconds

→ 2 × 60 seconds + 20 seconds

→ 120 seconds + 20 seconds

→ 140 seconds

Now,

In 40 seconds,Round made = 1

So,

In 140 seconds,Round made = $\sf\dfrac{1}{40}\: \times 140$

In 140 seconds = 3.5 rounds

Thus,the farmer will make three and half rounds of the square field. If the farmer starts from position A,then after completion of 3 rounds, he'll be at starting position A. But in the next half round,Farmer will move from A to B,and B to C,so that his final position will be at C. Thus,the net displacement of farmer will be AC. Now ABC is a right angled triangle in which AC is the hypotenuse.

[Refer to the attachment]

So,

$\sf \longrightarrow \: ( {AC})^{2} = ({AB})^{2} + ({BC})^{2} \:$

$\sf \longrightarrow \: ( {AC})^{2} = ( {10)}^{2} + ({10})^{2}$

$\sf \longrightarrow \: ( {AC})^{2} = 100 + 100$

$\sf \longrightarrow \: ( {AC})^{2} = 200$

$\sf \longrightarrow \: {AC} = \sqrt{200}$

$\sf \longrightarrow \red{ {AC} =14.143 \: m}$

$\therefore$ The magnitude of the displacement of the farmer at the end of 2 minutes 20 seconds will be 14.143 metres.