Finding graphs from motion model

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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!
 
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chinnie15 said:

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:
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.
 
Oooh, motion at constant acceleration?
 
Xf=xi+(vx)iΔt+(1/2ax)(Δt)2

So.. this one?
 
Ok, got it! ax=1.6m/s2, vxi= -2 m/s, and xi=2.5m
 
V changes at a constant rate of 1.6 m/s2.

And the peak of the parabola is (1.3s, 1.3m)?
 
chinnie15 said:
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.

chinnie15 said:
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