Question

Happy Friendship Day But solve this question please ....

A man walk 10 m south 20 m north and then 6 m east . In it's full walking he take 3 mins ...
Then find the following:

(i) Distance
(ii) Displacement
(iii) Velocity per min .

$\textsf{No Spams}$​

Answer 1

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

$\textit{A man walk 10 m south 20 m north and then 6 m}$$\textit{ .In it's full walking he take 3 mins}$

$\textit{Then find the following:}$

• $\mathrm{Distance}$
• $\mathrm{Displacement}$
• $\mathrm{Velocity (In\:m\:min^{-1})}$

______________________________________

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

$\textsf{We have to find the:}$

• $\mathrm{Distance}$
• $\mathrm{Displacement}$
• $\mathrm{Velocity (In\:m\:min^{-1})}$

______________________________________

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

$\textit{In this case , the body is traveling 10 m south and 20 north}$ $\textit{so whole distance will be 10 m towards north.}$

______________________________________

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

$\underline\mathrm{Distance}$

We Know:

$\textit{Distance = Sum of initials positions}$

Here , the initial positions are.

$10m\:,\:20 m\:and \:6m$

$\textit{Distance = (10 + 20 + 6)m}$

$\textit{Distance = 36m}$

${\boxed{\therefore Distance = 36m}}$

______________________________________

$\underline{\mathrm{Displacement}}$

We know at the final positions , the figure formed is a right - angled triangle.

By Pythagoras theorem :

${\boxed{h^{2} = p^{2} + b^{2}}}$

Where,

• $\mathtt{h = hypotenuse \rightarrow 10m}$
• $\mathtt{p = height \rightarrow x\:m}$
• $\mathtt{b = base \rightarrow 6 m}$

We get ,

$\Rightarrow 10^{2} = x^{2} + 6^{2}$

$\Rightarrow 10^{2} - 6^{2} = x^{2}$

$\Rightarrow \sqrt{10^{2} - 6^{2}} = x$

$\Rightarrow \sqrt{100 - 36} = x$

$\Rightarrow \sqrt{64} = x$

$\Rightarrow 8 = x$

${\boxed{\therefore Displacement = 8 m}}$

______________________________________

$\underline\mathrm{Velocity}$

We know ,

${\boxed{Velocity = \dfrac{Displacement}{time}}}$

So , by using this we get:

$\Rightarrow Velocity = \dfrac{8}{3}$

$\Rightarrow Velocity = 2\dfrac{2}{3}m\:min^{-1}$

${\boxed{\therefore Velocity = 2\dfrac{2}{3}m\:min^{-1}}}$

______________________________________