Finding graphs from motion model

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Homework Help Overview

The discussion revolves around a motion model described by the equation x(t) = 2.5 - 2t + 0.8t². Participants are tasked with determining the graphs of position (x vs t), velocity (v vs t), and acceleration (a vs t) without plotting points, focusing instead on interpreting the parameters of the equation.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants explore the nature of the curve represented by the equation, identifying it as a parabola and questioning the type of motion it describes. There is discussion about the implications of quadratic displacement in time and its relation to constant acceleration.

Discussion Status

The conversation has progressed with participants identifying key parameters such as acceleration and initial velocity. Some guidance has been provided regarding the relationship between displacement, velocity, and acceleration, but there remains uncertainty about the exact formulation of the velocity function and the timing of the peak of the parabola.

Contextual Notes

Participants are working within the constraints of algebra-based physics, avoiding calculus in their reasoning. There is an emphasis on interpreting the motion model without direct calculations or plotting points.

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.
 
Something like projectile, but more general - think of the acceleration...

ehild
 
Oooh, motion at constant acceleration?
 
chinnie15 said:
Oooh, motion at constant acceleration?

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


ehild
 
Xf=xi+(vx)iΔt+(1/2ax)(Δt)2

So.. this one?
 
chinnie15 said:
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
 
Ok, got it! ax=1.6m/s2, vxi= -2 m/s, and xi=2.5m
 
  • #10
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
V changes at a constant rate of 1.6 m/s2.

And the peak of the parabola is (1.3s, 1.3m)?
 
  • #12
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
 

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