and a2=.20 which is plugged into the formula , repeating steps in part I with a more
difficult graph to follow on the computer. Part III of the lab was done by one of us
dropping a strip of picket fence by a photogate and then both of us measuring the
acceleration of the picket fence in free fall. The end result of percent discrepancy between
the slope of the our velocity versus time plot (9.73) and the accepted value of the
acceleration of a free falling object (g=9.8) was 0.70% while the percent discrepancy
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between the mean of the acceleration from the table (9.7) and the accepted value of the
acceleration of a free falling object (g=9.8) was 1.00%. Possible uncertainty in our results
could be due to in part I and II the motion sensor could’ve been calibrated wrong and we
could’ve messed up trying to follow the lines in the original graph by moving too fast or
too slow. In part III uncertainty comes from Distance from the ground at which the picket
fence was being dropped (elevation) and possibility of air resistance being different each
trial.
Questions:
Part I:
1.
In the graph, what is the slope of the line of best fit for the middle section of the
plot? M=.179.0031
2. Describe your motion. Start at time=0 and end when the data recorder stopped
(Example: “Constant speed for 2 seconds followed by no motion for 3 seconds, etc.”)
We stood still from 02 seconds at .40m. Then we moved back with constant velocity for
four seconds. We then stayed there for four seconds at 1.10 meters to finish off with a
straight line on the graph.
Part II:
1. For your best attempt, how well did your
plot of motion fit the plot that was already on
the graph? Calculate the percent difference of the linear slope of your motion and the
slope of the given graph.
Expected acceleration a
.222
Actual acceleration a2
.20
Percent error=
Percent error=
Part III:
1. How does the slope of your velocity versus time plot compare to the accepted value of
the acceleration of a free falling object (g=9.8)?
•
Reminder: percentdiscrepancy=*100%
•
0.70%
2. How does the mean of the acceleration from the table compare to the accepted value of
the acceleration of a free falling object (g=9.8)?
•
Reminder: percentdiscrepancy=*100%
•
1.00%
3. What factors do you think may cause the experimental value to be different from the
accepted value? Factors that may have caused the experimental value to be different from
the accepted value are: Distance from the ground at which the picket fence was being
dropped (elevation) and possibility of air resistance being different each trial.
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 Fall '11
 BrunoBauer
 Physics, Derivative, Acceleration, Velocity, Picket fence

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