Question

 A player moves along the boundary of a square ground of side 50 m in 200 sec. The magnitude of displacement of the farmer at the end of 11 minutes 40 seconds from his initial position is


a). 50 m

b). 150 m

c). 200 m

d). 50√2 m
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Answer 1

$\color{red}{\large\underline{\underline\mathtt{Question:}}}$

$\textit{A player moves along the boundary}$ $\textit{of a square ground of side 50 m in 200 sec}$. $\textit{The magnitude of displacement of the farmer at the end of 11}$ $\textit{minutes 40 seconds from his initial position is.}$

• $\mathtt{50 m}$

• $\mathtt{150 m}$

• $\mathtt{200 m}$

• $\mathtt{50√2 m}$

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$\color{purple}{\large\underline{\underline\mathtt{To\:Find:}}}$

$\textsf{The magnitude of displacement}$

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$\color{blue}{\large\underline{\underline\mathtt{Concept:}}}$

$\textsf{Here the velocity will be uniform in both the cases}$

$\text{The distance travelled = 4 × 50 = 200m}$

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$\color{magenta}{\large\underline{\underline\mathtt{Solution:}}}$

$we\:know$

$\textit{If a body moving with uniform velocity}$$\mathtt{\overrightarrow{v}}$$\textit{has displacement}$$\mathtt{\overrightarrow{s}}$$\textit{in a time interval t then by definition }$

$\mathtt{\overrightarrow{v} \rightarrow \dfrac{\overrightarrow{s}}{t}}$

so , we get,

$\Rightarrow v = \left(\dfrac{200}{200}\right) ms^{-1}$

$\Rightarrow v = 1 ms^{-1}$

$\boxed{\mathtt{velocity = 1 ms^{-1}}}$

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$Time\:Taken = 11 min 40 sec = 700 sec$

$speed = 1 ms^{-1}$

From the relation,

$\mathtt{\overrightarrow{v} \rightarrow \dfrac{\overrightarrow{s}}{t}}$

we get ,

$\mathtt{\overrightarrow{s} \rightarrow \overrightarrow{v} \times t}$

$\Rightarrow s = 1 \times 700$

$\Rightarrow s = 700 m$

We know that , the displacement can be found by the Pythagoras theorem.

i.e,

$BD^{2} = BC^{2} + DC^{2}$

$\Rightarrow BD^{2} = 50^{2} + 50^{2}$

$\Rightarrow BD = \sqrt{50^{2} + 50^{2}}$

$\Rightarrow BD = \sqrt{2500 + 2500}$

$\Rightarrow BD = \sqrt{5000}$

$\Rightarrow BD = 50\sqrt{2}m$

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Answer 2

Answer:

Explaination:time taken to travel 200m( perimeter of square is 4*side)=200s. i.e time taken to travel 1m is 1sec.i.e speed of player is 1m/s so The distance travelled is (1*(11*60)+40)=700m i.e 3.5 rounds of the field so displacement is AC(see in photo)=diagonal of square According to pythagorus theorem. (AD)^2+(DC)^2=(AC)^2. So AC=(2500+2500)^1/2. AC=50root2ANSWER:option D

Explanation: