# Finding graphs from motion model

1. Sep 26, 2012

### chinnie15

1. The problem statement, all variables and given/known data

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."

2. Relevant equations

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

3. 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!

2. Sep 26, 2012

### ehild

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: Sep 26, 2012
3. Sep 26, 2012

### chinnie15

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. Sep 26, 2012

### ehild

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

ehild

5. Sep 26, 2012

### chinnie15

Oooh, motion at constant acceleration?

6. Sep 26, 2012

### ehild

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

ehild

7. Sep 26, 2012

### chinnie15

Xf=xi+(vx)iΔt+(1/2ax)(Δt)2

So.. this one?

8. Sep 26, 2012

### ehild

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

ehild

9. Sep 26, 2012

### chinnie15

Ok, got it! ax=1.6m/s2, vxi= -2 m/s, and xi=2.5m

10. Sep 26, 2012

### ehild

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. Sep 26, 2012

### chinnie15

V changes at a constant rate of 1.6 m/s2.

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

12. Sep 26, 2012

### ehild

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

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