# When going from velocity to position..

• Apr 30th 2010, 03:08 PM
KanedaSyndrome
When going from velocity to position..
Hi, I finally sat down and got to understand what the workings are when you derive position -> velocity -> acceleration -> jerk etc.

I made some graphs on a paper, since I work better when I get a visual representation, and the thing is, I can't get it to make sense in the link between position and velocity.

I have a movement that goes through: 1 m/s, 2 m/s, 3 m/s and 4 m/s plottet on the position vs time graph. So it's an expontentially growing curve.

When plotting on the velocity vs time graph, it's a growing curve with 'x' (inclination of 1), since for each second passing the object is moving 1 meter extra per second compared to the previous second (acceleration).

So looking at it visually, it makes sense to me at this point. But when I go to derive the 'x' function it becomes '1/2x^2', naturally, and that function doesn't fulfill my position vs. time graph.

The values at the position vs. time plot would be 1/2, 2, 4.5, 8, when I'd want it to be 1, 2, 3, 4 to fit my graphical representation in the position vs. time plot.

So that's why I'm a bit confused and would love a pointer to what I'm not seeing. So if anyone could help me understand.

I am wondering if the position vs. time graph is wrong compared to the velocity vs. time one, that's the only explanation I can think of, and it might be that my position function can't be described by a simple function?

I'm going to keep thinking about this, and I might have found the answer by the time I get an answer here, but well, if not then I have this to look forward to :)
• Apr 30th 2010, 03:26 PM
KanedaSyndrome
Ok, well found a function matching my position vs. time plot, that of:

1/2*x^2+1/2*x

The thing is now:

(1/2*x^2+1/2*x)' = x+1/2

So the inclination in my velocity vs. time graph is fine now at 'x', but the +'1/2' is curious I think.

Is there a certain thing to understand about this '1/2m/s' added to the velocities? Since I can see from the position graph that it is 1m/s, 2m/s, 3m/s, 4m/s etc. So I wonder what this '1/2m/s' means. Do I just disregard it since the growth rate matches?
• Apr 30th 2010, 03:44 PM
Hi Kenada, the function you are describing is one with a constant acceleration of 1m/s². The curve is actually a parabola, the x² term is an indicator that it is a parabola.

By calculus we can derive the equations of constant acceleration just by some integration and a knowledge of differential calculus. We know that acceleration a is rate of change of velocity.

$\displaystyle a = 1 \frac{m}{s^2}$

$\displaystyle a = \frac{dv}{dt}$

$\displaystyle \frac{dv}{dt} \ = 1$

$\displaystyle dv = 1 dt = a dt$

$\displaystyle \int dv = \int a dt$

$\displaystyle v = at + c$

$\displaystyle v = v_0 + at$

Velocity, in turn, is just the rate of change of position with respect to time.

$\displaystyle \frac{dx}{dt} = v_0 + at$

$\displaystyle dx = (v_0 + at)dt$

$\displaystyle \int dx = \int (v_0 + at)dt$

$\displaystyle x = v_0t + \frac{1}{2} at^2 + c$

$\displaystyle x = x_0 + v_0t + \frac{1}{2} at^2$

If your function had no acceleration term it would just be nice and clean, if you draw a graph of the velocity you'll notice a pretty clean looking graph.

YouTube - 1. Course Introduction and Newtonian Mechanics

This is a great lecture on what is going on here, I'd really advise you to check it out as the man is great at explaining things.

If this is a bit confusing just let us know!
• Apr 30th 2010, 04:00 PM
zzzoak
$\displaystyle v(t)=v_0+at$

$\displaystyle 1/2\: = \: v_0 - the\:\: initial\:\: speed\:\: at\:\: t=0.$
• May 1st 2010, 02:58 AM
HallsofIvy
Quote:

Originally Posted by KanedaSyndrome
Hi, I finally sat down and got to understand what the workings are when you derive position -> velocity -> acceleration -> jerk etc.

I made some graphs on a paper, since I work better when I get a visual representation, and the thing is, I can't get it to make sense in the link between position and velocity.

I have a movement that goes through: 1 m/s, 2 m/s, 3 m/s and 4 m/s plottet on the position vs time graph. So it's an expontentially growing curve.

This is meaningless unless you tell us what time, t, those correspond to. Are we to assume they are at t= 1, 2, 3, 4 seconds also?

Quote:

When plotting on the velocity vs time graph, it's a growing curve with 'x' (inclination of 1), since for each second passing the object is moving 1 meter extra per second compared to the previous second (acceleration).
Okay, so they were at one second intervals. It would have helped if you had told us that!

Quote:

So looking at it visually, it makes sense to me at this point. But when I go to derive the 'x' function it becomes '1/2x^2', naturally, and that function doesn't fulfill my position vs. time graph.
The integral or "anti-derivative" of x is (1/2)x^2+ C, where C is an arbitrary constant.

Quote:

The values at the position vs. time plot would be 1/2, 2, 4.5, 8, when I'd want it to be 1, 2, 3, 4 to fit my graphical representation in the position vs. time plot.
Again, you haven't told us what t those correspond to. If it is again at one second intervals, you can't fit it to that- that corresponds to moving at a constant speed, 1 m/s, and that does not have the changing v you have above.

Quote:

So that's why I'm a bit confused and would love a pointer to what I'm not seeing. So if anyone could help me understand.

I am wondering if the position vs. time graph is wrong compared to the velocity vs. time one, that's the only explanation I can think of, and it might be that my position function can't be described by a simple function?

I'm going to keep thinking about this, and I might have found the answer by the time I get an answer here, but well, if not then I have this to look forward to :)