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 to 20 seconds from his initial position? ​

Answer 1

$\huge \underline\textbf{Answer}$

first of all we will convert the total time of 2 minutes 20 seconds into seconds .

$\blue{\bold{ total \:time \:=\: 2 \:minutes \:20 \:seconds}}$

$\blue{\bold{=\:2\:×\:60\:second\:+\:20\: seconds}}$

$\blue{\bold{=\:120\:seconds\:+\:20\:seconds}}$

$\blue{\bold{=\:140\:seconds}}$

Now, In 40 s, number of rounds made = 1

So, In 140 s, number of rounds made $\blue{\bold{ \frac{1}{40}\: ×\: 140}}$

$\large\bf{\underline\green{=\:3.5}}$

Thus, the farmers will make three and half round (3.5 rounds) of the square field. if the farmers starts from position A .

(see above attachment)

then after three complete rounds, he will reach at starting position A. but in the next half round, the farmer will move from A to B, and B to C, so that his final position will be C. Thus, the net displacement of the farmer will be AC. Now, ABC is a right angled triangle in which AC is the hypotenuse.

So,

$\blue{\bold{( {AC})^{2}\:=\:( {AB})^{2}\:+\:( {BC})^{2}}}$

$\blue{\bold{( {AC})^{2}\:=\:( {10})^{2}\:+\:( {10})^{2}}}$

$\blue{\bold{( {AC})^{2}\:=\:100\:+\:100}}$

$\blue{\bold{( {AC})^{2}\:=\:200}}$

$\blue{\bold{AC\:=\: \sqrt{200}}}$

$\blue{\bold{AC \:=\: 14.143\:m}}$

Thus, the magnitude of displacement of the farmer at the end of 2 minutes and 20 seconds will be 14.143 meters.

$\huge\bf{\underline\green{14.143\:meters}}$

Answer 2

Answer:

$\red{\bold{Displacement\:=\:14.143 m}}$

Step-by-step explanation:

$\red{\bold{Correct\:Question-}}$

A farmer moves along the boundary of a square field of side 10 metre in 40 seconds. What will be the magnitudeof displacement of the farmer at the end of 2 minutes 20 seconds from his initial position ?

$\green{\bold{Solution-}}$

$\orange{\bold{Round\:covers \:in\:40\:second\:=\:1\:(one\:round)}}$

$\orange{\bold{Round\:covers\:in\:140\:second\:=\:⟱}}$

$\orange{\bold{⟹140×1/40=3.5\:rounds}}$

$\green{\bold{i. e.}}$ 3 full round and 1 half round

$\orange{\bold{In\:the\: 3 \:full\: rounds\:displacement\:=\:0}}$ because initial and final position are same.

$\green{\bold{But\:in\:the\:next\half\:round\:displacement\:will\:be}}$

$\red{\bold{by\:Pythagoras\:Theorem \:,}}$

⟹ ${(ac)}^{2} = {(ab)}^{2} + {(bc)}^{2}$

⟹ ${(ac)}^{2} = {(10)}^{2} + {(10)}^{2}$

⟹ ${(ac)}^{2} = 100 + 100$

⟹ ${(ac)}^{2} = 200$

⟹ $ac = \sqrt{200}$

⟹ $ac = 14.143m$

$\red{\bold{Hence , }}$ Displacement will be 14.143m