Finding graphs from motion model

  • Thread starter chinnie15
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  • #1
chinnie15
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Homework Statement



I have a problem where I am given the model:

x(t) = 2.5 - 2t + 0.8t2

From this I have to determine the x vs t, v vs t, and a vs t graphs. I have to do this without plotting points. Specifically it says, "Draw the graphs by identifying and interpreting the parameters, not by plotting points."


Homework Equations



None, I don't think, except y=Ax2+Bx+C


The Attempt at a Solution


I wish I could say I have an attempt, but I honestly don't know what to do. What am I looking for? This is algebra based physics, so I can't simply use calculus. Just a little guidance would be awesome.

Thank you so much for any help!
 

Answers and Replies

  • #2
ehild
Homework Helper
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Homework Statement



I have a problem where I am given the model:

x(t) = 2.5 - 2t + 0.8t2

From this I have to determine the x vs t, v vs t, and a vs t graphs. I have to do this without plotting points. Specifically it says, "Draw the graphs by identifying and interpreting the parameters, not by plotting points."


Homework Equations



None, I don't think, except y=Ax2+Bx+C


The Attempt at a Solution


I wish I could say I have an attempt, but I honestly don't know what to do. What am I looking for? This is algebra based physics, so I can't simply use calculus. Just a little guidance would be awesome.

Thank you so much for any help!

The independent variable is t, the dependent one is x. You are right, x(t) is the same kind of curve as y=Ax2+Bx+C. What is the name for that type of curve?

What kind of motion is when the displacement is quadratic in time?

ehild
 
Last edited:
  • #3
chinnie15
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Thanks for your response! :)

It would be a parabola.

'What kind of motion is when the displacement is quadratic in time? '

Projectile? Which confuses me because the paper says the model is describing the position of a particular boat as a function of time.
 
  • #4
ehild
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Something like projectile, but more general - think of the acceleration...

ehild
 
  • #5
chinnie15
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Oooh, motion at constant acceleration?
 
  • #6
ehild
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Oooh, motion at constant acceleration?

Yes! You certainly know the formula of the displacement in terms of time...


ehild
 
  • #7
chinnie15
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Xf=xi+(vx)iΔt+(1/2ax)(Δt)2

So.. this one?
 
  • #8
ehild
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Xf=xi+(vx)iΔt+(1/2ax)(Δt)2

So.. this one?

Yes. Now you can find the value of ax, vxi and xi, don't you?


ehild
 
  • #9
chinnie15
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Ok, got it! ax=1.6m/s2, vxi= -2 m/s, and xi=2.5m
 
  • #10
ehild
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Right! a is a constant, and how does v change in time?
It is easy to plot a(t) and v(t), and you know that x(t) is a parabola. Can you find its peak?

ehild
 
  • #11
chinnie15
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V changes at a constant rate of 1.6 m/s2.

And the peak of the parabola is (1.3s, 1.3m)?
 
  • #12
ehild
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V changes at a constant rate of 1.6 m/s2.

Yes, but what is the v(t) function? Remember, there is some initial velocity.

And the peak of the parabola is (1.3s, 1.3m)?

Close, but not quite so. The peak of the parabola is at time when the velocity is zero. When does that happen?

You can also write up x(t) in the form x= a(t-k)2+b, then the peak is at t=k, x=b.


ehild
 

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