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Re: Kinematics



Ludwik,

It may be superfluous to the pedagogical intent of your exercise, but
I would suggest that the best way to answer the question, "Was the
acceleration (reasonably) constant?" (admittedly, a question you
don't claim to be asking) is "simply" to fit the raw data directly to
a constant acceleration curve and decide whether the fit is "good."

For instance, I took your hypothetical data for t(x) and fit it to
the constant acceleration function

t(x) = t0 + sqrt[2*(x-x0)/a]

to obtain

t0 = -1.17 s
x0 = -0.452 m
a = 0.669 m/s^2

This fit gives the following results:

x(m) t_exp (s) t_fit (s) res = t_exp-t_fit (s) [res/(0.05 s)]^2
0 0 -0.00 0.00 0.00
2 1.5 1.54 -0.04 0.64
4 2.5 2.48 0.02 0.16
6 3.3 3.22 0.08 2.56
8 3.8 3.86 -0.06 1.44
----
chi^2 = 4.80

To my eye this is a good fit: The residuals show no obvious trend
and they hover around or below the minimum implied experimental error
in the time measurements of 0.05 s.

On the other hand it is a three parameter fit to five data points
leaving only two degrees of freedom so perhaps we shouldn't be too
smug.

Indeed, under the assumption that the basic time uncertainties *is*
0.05 s I get a chi^2 per degree of freedom of 2.4. Not particularly
great, but given all the assumptions and the small amount of data ...

--
John Mallinckrodt mailto:ajm@csupomona.edu
Cal Poly Pomona http://www.csupomona.edu/~ajm