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If the acceleration is constant, then the averagecut
velocity during a time interval (t1 < t < t2)
equals the instantaneous velocity at the midpoint
of the time interval [t =3D (t1 + t2)/2]. If the
midpoint of the time interval is used with the
data given by Ludwik, then the calculated
acceleration is inconsistent with the assumption
that it is constant.
I obtain the following values:
t (s) 0.75 2.0 2.9 3.55
Vav (m/s) 1.33 2.0 2.5 4.0
Aav (m/s^2) 0.536 0.556 2.31
The first two calculated values of acceleration
are consistent with the assumption that the
acceleration is constant, but the third value
is clearly inconsistent with that assumption.
If the acceleration is increasing (decreasing)
monotonically, then the average velocity equals
the instantaneous velocity at a time later
(earlier) than the midpoint of the interval.
If the final time reading was 4.0 s, then the
final average velocity would be 2.86 m/s and
would correspond to the instantaneous velocity
at t =3D 3.65 s, assuming that the acceleration
was constant. In that case, the acceleration
for the final interval would be 0.514 m/s^2,
which would be more consistent with the
assumption of constant acceleration.
Having students run provides a kinesthetic
experience that can aid learning, but it is
difficult for a runner to maintain a constant
acceleration for an extended time.
We perform a somewhat similar experiment
using a cart accelerated across a table by a
hanging mass. We use a spark timer to record
the displacement of cart every 0.1 s. This
experiment provides an acceleration that is
essentially constant, but does not provide
a kinesthetic experience.
Daniel Crowe
Oklahoma School of Science and Mathematics
Ardmore Regional Center
dcrowe@sotc.org
-----Original Message-----
=46rom: Ludwik Kowalski [mailto:kowalskil@MAIL.MONTCLAIR.EDU]