# The central difference method! Help!

1. Sep 25, 2007

### Winner

Hi guys. Ok I'm working on this problem, it's more theoretical than anything.:yuck:

I'll try to explain as best I can. Lets say you have a runner going past the following points up till and past 30 m:

Distance (x) of x1=0, x2=4, x3=9, x4=20, and x5=30m
the times to these intervals are 0, 1.0s, 2.0s, 3.0s and 5.0s.

Now if all I did to find velocity was V=d/t, this would only give me an average velocity over that time. Now if I wanted to find the velocity right at that time point, I was told to use the central difference method:

V at time 3.0s = (Distance at x4- distance at x2)/(time from distance x2 to distance x4)

This should give me the velocity at that point?
Now the BIG question is? how am I suppose to do it for those values at the end and at the beginning, ie. when time is 5.0 secs?

Thanks guys.

2. Sep 25, 2007

### Staff: Mentor

Well, x = 0 at t = 0, and I believe, v(0) = 0. x (0) =0 and v(0) = 0 are specified initial conditions.

Often one is solving the differential equation via central difference, e.g.

www.mate.tue.nl/~piet/inf/disco/pdf/cendif.pdf [Broken]

http://mceer-nt4.mceer.buffalo.edu/...esearch/ANCER/Activities/2004/wu_b_edpmrc.pdf

But in this case, one is starting with data.

One possibility would be to estimate the acceleration at t5 and extrapolate a velocity at t6. In the case of a runner, one would probably find that v reaches some asymptote.

Last edited by a moderator: May 3, 2017
3. Nov 13, 2007

### cyberjaya

ya good.

ok, here is my questions.

the CDM (Central Difference Method) is used to try to find the velocity of the point at specific time. in numerical analysis, above case is actually first order derivatives generated by CDM of first order, am i correct?

so my question is

1) how to generate 3rd order derivatives using CDM of first order.... or is it possible?
2) if first order derivatives means velocity in physics, then please give the meaning of 2nd order and 3rd order.

thanks