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This is the second of our many quizzes on Position-time graphs. Click on this to access more such competitive quizzes on Measurement.

Position-time graphs Quiz 2

physics

Find out the incorrect position time graph of a train moving through a rail track from the graphs given below?

Particles cannot exist in several places at the same time.

  After observing the graphs of all the options, it can be easily concluded that in option (a) graph, the particle can be seen at two locations at the same time. This is not practically possible when a train is travelling on the track.

Calculate the change in momentum of a ball having a mass $5\text{ }kg$ incident on a wall using the graph? <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/1885adb0-b53e-47b0-b611-4306d58ca013321550132634879054.png"/> 

The area under the acceleration time graph will be the velocity. 

  $\text{In the x-axis, from 0 to 2,}\\\dfrac{1}{2}bh=\dfrac{1}{2}\times 2\times 10\\\Rightarrow\dfrac{1}{2}bh=10\text{ }m{{s}^{-1}}\\\text{From 2 to 5,}\\bh=\left( 5-2 \right)\times 10\\\Rightarrow bh =30\text{ }m{{s}^{-1}}\\\text{From 5 to 7,}\\\dfrac{1}{2}bh=\dfrac{1}{2}\times \left( 7-5 \right)\times 10 \\A=10\text{ }m{{s}^{-1}}\\\text{Therefore the total velocity will be,}\\v=10+30+10\\\Rightarrow v=50\text{ }m{{s}^{-1}}\\\text{Therefore the change in momentum will be,}\\p=mv\\\Rightarrow p =5\times 50\\\therefore p =250\text{ }kgm{{s}^{-1}}$

 $A$ started to travel from his home to the market with a velocity $3\text{ }m{{s}^{-1}}$. From there he met one of his friends $B$. Both of them went to the house of $B$ with a velocity $5\text{ }m{{s}^{-1}}$ together to visit his parents. The man came back to the market in the same path with a velocity, $1.67\text{ }m{{s}^{-1}}$. Which of the following graphs correctly represents the journey of $A$?

The slope of a position time graph will be the velocity. 

 $\text{For the first journey to the market from home,}\\v =\dfrac{dx}{dt}\\\Rightarrow v =3\text{ }m{{s}^{-1}}\\\text{This is correctly shown in options a and c.}\\\text{From market to the house of B,}\\v =\dfrac{dx}{dt}\\\Rightarrow v =5\text{ }m{{s}^{-1}}\\\text{This has been correctly shown in option c.}\\\text{From house of B to the market,}\\v =\dfrac{dx}{dt}\\\Rightarrow v =\dfrac{5}{3}\text{ }m{{s}^{-1}}\\\therefore v =1.67\text{ }m{{s}^{-1}}$

$A$ and $B$ are participating in a running competition. $A$ is having a velocity of $\dfrac{5}{3}\text{ }m{{s}^{-1}}$ and $B$ is having a velocity of $\dfrac{8}{6}\text{ }m{{s}^{-1}}$.Which of the following graphs can represent the relative velocity between $A$ and $B$ correctly? 

When the slope is higher, it will be towards the y-axis. 

 $\text{As the velocity of A is higher than that of B,}\\{{V}_{A}}=\dfrac{5}{3}\text{ }m{{s}^{-1}} \\\Rightarrow {{V}_{B}}=\dfrac{8}{6}\text{ }m{{s}^{-1}} \\\therefore {{V}_{A}}\rangle {{V}_{B}} \\\text{The slope of a position time graph is giving the velocity.}\\\text{Therefore the graph which is having the higher slope will be towards the y-axis.}\\\text{Hence}\,{{V}_{A}}\,\text{will be towards the y-axis.}$

The displacement of a car on a road has been given as $x={{\left( t-2 \right)}^{2}}$. Plot the velocity time graph of this car. Also mention the acceleration of the car.  

The first derivative of velocity is the acceleration. 

 $x={{\left( t-2 \right)}^{2}} \\\Rightarrow x={{t}^{2}}-4t+4 \\\Rightarrow v=\dfrac{dx}{dt}=2t-4 \\\Rightarrow v=2t-4 \\\Rightarrow y=mx+c \\\therefore m=2, c=-4\\\text{The graph will be linear in nature.}\\\text{The acceleration will be,}\\ a=\dfrac{dv}{dt} \\\therefore a=2\text{ }m{{s}^{-2}}$