Question

PLz solve question 9​

Answer 1

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Figure (3-E4) shows the graph of the x-coordinate of a particle going along the X-axis as a function of time. Find (a) the average velocity during 0 to 10 s, (b) instantaneous velocity at 2, 5, 8 and 12s.

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This is a question from HC Verma - Concepts of Physics Part 1 - Rest and Motion : Kinematics - Exercises - Question 9

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$\sf{\red{\underline{\large{Answer:-}}}}$

We don't need calculus to solve this!

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We know,

$\boxed{\sf{\small{Average \:Velocity\: =\: \dfrac{Total\: displacement\:in\:given\:time}{Time\:taken}}}}$

$\texttt{\underline{First answer:}}$

We know that slope of displacement- time graph gives velocity

So by above formula,

$\sf{\small{Average \:Velocity\:during\:0\:to\:10s\: =\: \dfrac{Total\: displacement\:in\:0\:to\:10s\:}{Time\:taken}}}$

From graph, we can observe that displacement from 0 to 10s is 10m

And time taken is = (10 - 0)s = 10s

$\implies\sf{\small{Average \:Velocity\:during\:0\:to\:10s\: =\: \dfrac{10m}{10s}}}$

$\implies\sf{\small{Average\:Velocity\:during\:0\:to\:10s\: = 10m/s}}$

$\texttt{\underline{Second answer:}}$

$\texttt{\underline{Instantaneous velocity at 2s:}}$

We know slope of displacement time graph gives velocity.

Looking at the graph, as the part of the graph from 0 to 2.5s is a straight line, it shows that the velocity is constant.

$\sf{Slope\:=\:tan\theta\:=\:\dfrac{{y}_{2}\:-\:{y}_{1}}{{x}_{2}\:-\:{x}_{1}}}$

$\sf{Slope\:=\:\dfrac{50\:-\:0}{2.5\:-\:0}}$

$\sf{Slope\:=\:20m/s}$

This is actually average velocity, but since the velocity is constant throughout the journey from 0 to 10s, it will also be the instantaneous velocity at 2s

$\bf{Instantaneous\: velocity\:at\:2s\:=\:20m/s}$

$\texttt{\underline{Instantaneous velocity at 5s:}}$

We know slope of displacement time graph gives velocity.

Looking at the graph, as the part of the graph from 2.5 to 7.5s is a straight line, it shows that the velocity is constant.

$\sf{Slope\:=\:tan\theta\:=\:\dfrac{{y}_{2}\:-\:{y}_{1}}{{x}_{2}\:-\:{x}_{1}}}$

$\sf{Slope\:=\:\dfrac{50\:-\:50}{7.5\:-\:2.5}}$

$\sf{Slope\:=\:0m/s}$

This is actually average velocity, but since the velocity is constant throughout the journey from 2.5 to 7.5s, it will also be the instantaneous velocity at 5s

$\bf{Instantaneous\: velocity\:at\:5s\:=\:0m/s}$

$\texttt{\underline{Instantaneous velocity at 8s:}}$

We know slope of displacement time graph gives velocity.

Looking at the graph, as the part of the graph from 7.5 to 10s is a straight line, it shows that the velocity is constant.

$\sf{Slope\:=\:tan\theta\:=\:\dfrac{{y}_{2}\:-\:{y}_{1}}{{x}_{2}\:-\:{x}_{1}}}$

$\sf{Slope\:=\:\dfrac{100\:-\:50}{10\:-\:7.5}}$

$\sf{Slope\:=\:20m/s}$

This is actually average velocity, but since the velocity is constant throughout the journey from 7.5 to 10s, it will also be the instantaneous velocity at 8s

$\bf{Instantaneous\: velocity\:at\:8s\:=\:20m/s}$

$\texttt{\underline{Instantaneous velocity at 12s:}}$

We know slope of displacement time graph gives velocity.

Looking at the graph, as the part of the graph from 10 to 15s is a straight line, it shows that the velocity is constant.

$\sf{Slope\:=\:tan\theta\:=\:\dfrac{{y}_{2}\:-\:{y}_{1}}{{x}_{2}\:-\:{x}_{1}}}$

$\sf{Slope\:=\:\dfrac{0\:-\:100}{15\:-\:10}}$

$\sf{Slope\:=\:-20m/s}$

This is actually average velocity, but since the velocity is constant throughout the journey from 10 to 15s, it will also be the instantaneous velocity at 12s

$\bf{Instantaneous\: velocity\:at\:12s\:=\:-20m/s}$