#$&*
phy 201
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Tests
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Ive asked before on emails i guess you havent recieved them but where do we take test 2 . In the testing area? It doesnt explain on the program when ive looked at it. And when do we take the exam for this class and is it comprehensive?
@&
You stopped by the lab today and I think we got that straight.
Do reread the instructions under 'Testing'.
*@
#$&*
phy 201
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
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#$&*
For the 2-domino ramp, the ramp descends through a distance equal to the height of two dominoes while rolling a distance of about 30 cm. In the space below, give in the first line the total descent of the ball for the entire 30 cm of roll. In the second line give the amount of descent per cm of roll. Starting in the third line explain how you obtained your two results.
Note that the word 'descent' is clearly defined in the above paragraph being equal to the height of two dominoes; 'descent' refers to the vertical distance through which the ball descends. If the ball rolls 30 cm, it does not descend 30 cm. 30 cm would be the distance of roll for the trial in which the domino rolled the entire length of the ramp. Its descent for the entire 30-cm roll will in this case be about equal to the height of the two dominoes (perhaps a little more, depending on exactly where the dominoes were positioned).
------>>>>> dist of descent, descent per cm, 2 dom
Your answer (start in the next line):
1.65 total descent
.055 cm down per cm
#$&*
For the 2-domino ramp, use the preceding results to determine the amount of descent that would correspond to rolls of 10 cm, 15 cm, 20 cm and 30 cm. Give these results as 4 numbers in the first line, delimited by commas. In the second line explain how you obtained your results.
To check whether your results make sense, note that the more centimeters the ball rolls, the further it will descend. You just got done calculating the amount of descent per cm of roll, which provides a basis for the present calculation.
------->>>>>>> descent for 10, 15, 20 and 30 cm on 2 dom ramp
Your answer (start in the next line):
.165 cm down per cm
.11 cm down per cm
.0825 cm down per cm
.055 cm down per cm
#$&*
Using a similar strategy find the descent corresponding to rolls of 10 cm, 15 cm, 20 cm and 30 cm on a 4-domino ramp and report in similar syntax in the space below:
----->>>>>> descent for 10, 15, 20 and 30 cm four-dom ramp
Your answer (start in the next line):
.33 cm down per cm
.22 cm down per cm
.165 cm down per cm
.11 cm down per cm
#$&*
For your remaining calculations, assume that the mass of the ball is 100 grams. This isn't completely accurate, but the mass of the ball isn't critical and if necessary results could later be adjusted very easily for the accurate mass.
Determine how much work it would take to raise the mass of the ball, against the downward gravitational pull, through each of the distances of descent you have calculated. In the space below indicate in the first line the first distance of descent, then the work. In subsequent lines give the remaining distances of descent and corresponding amounts of work. In the first line following your data lines, specify the units of the quantities you have given, and explain how you obtained your results. Include a set of sample calculations for one of your lines.
----->>>> work to raise 100 g thru each dist of descent, each line dist of descent and work to raise
Your answer (start in the next line):
** **
From the conservation of energy on an incline experiment
I just dont understand what we are supposed to use (On the last question ive included) for the distance it seems like its referring to the questions before (which i included) but those are rates not distances and so work cant be figured for them
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What are we to use we cant use the rates so i dont know?
This lab exercise is based on the observations you previously made of a ball rolling down ramps of various
slopes. We further investigate the relationship between ramp slope and acceleration.
The mean time reported to complete this exercise is 2 hours. The most frequently reported times range from 1
hour to 3 hours, with some reports of shorter or longer times.
Note that there are a number of repetitive calculations in this exercise. You are encouraged to use a spreadsheet
as appropriate to save you time, but be sure your results check out with a handwritten analysis of at least a few
representative trials.
Document your data
For ramps supported by 1, 2 and 3 dominoes, in a previous exercise you reported time intervals for 5 trials of the
ball rolling from right to left down a single ramp, and 5 trials for the ball rolling from left to right.
If in that experiment you were not instructed to take data for all three setups in both directions, report only the
data you were instructed to obtain.
(Note: If you did the experiment using the short ramp and coins, specify which type of coin you used. In the
instructions below you would substitute the word 'coins' for 'dominoes').
Go to your original data or to the 'readable' version that should have been posted to your access page, and copy
your data as indicated in the boxes below:
Copy the 10 trials for the 1-domino setups, which you should have entered into your original lab submission in
the format specified by the instruction
'In the box below, give the time interval for each trial, rounded to the nearest .001 second. Give 1 trial on each
line, and give the 5 trials for the first system, then the 5 trials for the second system. You will therefore give 10
numbers on 10 lines.'
In the 'readable' posted version this data will follow the boldfaced heading
'5 trials each way 1 domino'
Enter your 10 numbers on 10 lines below, and on the first subsequent line briefly indicate the meaning of the data:
------>>>>>> ten trials for 1-domino setups
Your answer (start in the next line):
1.621
1.766
1.605
1.621
1.808
1.515
1.512
1.609
1.578
1.594
#$&*
Enter your data for the 2-domino setups in the same format, being sure to include your brief explanation:
On the 'readable' posted version this data will follow the boldfaced heading
'5 trials each way 2 dominoes'
------>>>>>> 2 domino results
Your answer (start in the next line):
1.219
1.273
1.28
1.29
1.234
1.303
1.277
1.16
1.2
1.168
#$&*
Enter your data for the 3-domino setups in the same format, including brief explanation.
On the 'readable' posted version this data will follow the boldfaced heading
'5 trials each way 3 dominoes'
------>>>>>> 3 domino results
Your answer (start in the next line):
1.059
1.043
1.078
1.121
1.063
1
.964
1.059
1.137
1.184
#$&*
Calculate mean time down ramp for each setup
In the previous hypothesis testing exercise, you calculated and reported the mean and standard deviation of times
down each of the two 1-domino setups, one running right-left and the other left-right.
You may use any results obtained from that analysis (provided you are confident that your results follow correctly
from your data), or you may simply recalculate this information, which can be done very quickly and easily using
the Data Analysis Program at
http://www.vhcc.edu/dsmith/genInfo/labrynth_created_fall_05/levl1_15\levl2_51/dataProgram.exe\
In any case, calculate as needed and enter the following information, in the order requested, giving one mean and
standard deviation per line in comma-delimited format:
Mean and standard deviation of times down ramp for 1 domino, right-to-left.
Mean and standard deviation of times down ramp for 1 domino, left-to-right.
Mean and standard deviation of times down ramp for 2 dominoes, right-to-left.
Mean and standard deviation of times down ramp for 2 dominoes, left-to-right.
Mean and standard deviation of times down ramp for 3 dominoes, right-to-left.
Mean and standard deviation fof times down ramp or 3 dominoes, left-to-right.
On the first subsequent line briefly indicate the meaning of your results and how they were obtained:
------>>>>>> mean, std dev each setup each direction
Your answer (start in the next line):
1.684 , .09523
1.562 , .04526
1.259 , .03091
1.222 , .06486
1.073 , .02969
1.069 , .09184
#$&*
Calculate average ball velocity for each setup
Assuming that the ball traveled 28 cm from release until the time it struck the bracket, determine each of the
following, using the mean time required for the ball to travel down the ramp:
Average ball velocity for 1 domino, right-to-left.
Average ball velocity for 1 domino, left-to-right.
Average ball velocity for 2 dominoes, right-to-left.
Average ball velocity for 2 dominoes, left-to-right.
Average ball velocity for 3 dominoes, right-to-left.
Average ball velocity for 3 dominoes, left-to-right.
Report your six results in the box below, one result per line, in the order requested above.
Starting in the seventh line explain how you obtained your results, giving the details of how you obtained at least
one of your results. These details should include the definition of the average velocity, and should explain how
you used the mean time and the distance down the ramp to arrive at your result, and should show the numbers
used and the numbers obtained in each step.
------>>>>>> ave velocities each of six setups
Your answer (start in the next line):
16.627 m/s
17.926 m/s
22.239 m/s
22.913 m/s
26.095 m/s
26.193 m/s
#$&*
Calculate average ball acceleration for each setup
Assuming that the velocity of the ball changed at a constant rate in each trial, use the mean time interval and the 28
cm distance to determine the average rate of change of velocity with respect to clock time. You will determine
your results in the following order:
Average rate of change of ball velocity with respect to clock time for 1 domino, right-to-left.
Average rate of change of ball velocity with respect to clock time for 1 domino, left-to-right.
Average rate of change of ball velocity with respect to clock time for 2 dominoes, right-to-left.
Average rate of change of ball velocity with respect to clock time for 2 dominoes, left-to-right.
Average rate of change of ball velocity with respect to clock time for 3 dominoes, right-to-left.
Average rate of change of ball velocity with respect to clock time for 3 dominoes, left-to-right.
Report your six results in the box below, one result per line, in the order requested above.
Starting in the seventh line explain how you obtained your results, giving the details of how you obtained at least
one of your results. These details should include the definition of the average rate of change of velocity with
respect to clock time and should explain, step by step, how you used the mean time and the distance down the
ramp to arrive at your result, and should show the numbers used and the numbers obtained in each step.
------>>>>>> ave roc of vel each of six setups
Your answer (start in the next line):
9.874 m/s^2
11.476 m/s^2
17.664 m/s^2
18.75 m/s^2
24.32 m/s^2
24.502 m/s^2
#$&*
Average left-right and right-left velocities for each slope
For the 1-domino system you have obtained two values for the average rate of change of velocity with respect to
clock time, one for the right-left setup and one for the left-right. Average those two values and note your result.
For the 2-domino system you have also obtained two values for the average rate of change of velocity with
respect to clock time. Average those two values and note your result.
For the 3-domino system you have also obtained two values for the average rate of change of velocity with
respect to clock time. Average those two values and note your result.
Report your results in the box below, giving one average rate of change of velocity with respect to clock time per
line, in the order requested. Starting the the first subsequent line, briefly indicate how you obtained your results
and what you think they mean.
------>>>>>> ave of right-left, left-right each slope
Your answer (start in the next line):
10.675 m/s^2
18.207 m/s^2
24.411 m/s^2
#$&*
Find acceleration for each slope based on average of left-right and right-left times
Average the mean time required for the right-to-left run with the mean time for the left-to-right run.
Using this average mean time, recalculate your average rate of velocity change with respect to clock time for the
1-domino trials
Do the same for the 2-domino results, and for the 3-domino results.
Report your results in the box below, giving one average rate of change of velocity with respect to clock time per
line, in the order requested. In the subsequent line explain how you obtained your results and what you think they
mean.
------>>>>>> left-right, right-left each setup, ave mean times and give ave accel
Your answer (start in the next line):
10.644 m/s^2
18.199 m/s^2
24.411 m/s^2
#$&*
Compare acceleration results for the two different methods
You obtained data for three basic setups, each with a different slope. Each basic setup was done with a right-left
and a left-right version.
You previously calculated a single average rate of change of velocity with respect to clock time for each slope, by
averaging the right-left rate with the left-right rate.
You have now calculated a single average rate of change of velocity with respect to clock time for each slope, but
this time by using the average of the mean times for the right-left and left-right versions.
Answer the following questions in the box below:
Since both methods give a single average rate of change of velocity with respect to clock time, would you
therefore expect these two results to be the same for each slope?
Are the results you reported here, based on the average of the two mean times, the same as those you obtained
previously by average the two rates? Are they nearly the same?
Why would you expect that they would be the same or nearly the same?
If they are not exactly the same, can you explain why?
------>>>>>> ave of mean vel, ave based on mean of `dt same, different, why
Your answer (start in the next line):
They are very near the same
Yes i would expect some amount of accuracy
Since we are dealing with averages and with some rounding there will be some error
#$&*
Associate acceleration with ramp slope
Your results will clearly indicate that, as expected, acceleration increases when ramp slope increases. We want to
look further at just how the acceleration changes with ramp slope.
If you set up the ramps according to instructions, then the ramp slopes for 1-, 2- and 3-domino systems should
have been approximately equal to .03, .06 and .09 (if you used coins and the 15 cm ramp instead of dominoes
and the 30-cm ramp, your ramp slopes will be different; each dime will correspond to a ramp slope of about
.007, each penny to a slope of about .010, each quarter to a slope of about .013).
For each slope you have obtained two values for the average rate of change of velocity with respect to clock time
on that slope. You may use below the values obtained in the preceding box, or the values you obtained in the box
preceding that one. Use the one in which you have more faith.
In the box below, report in the first line the ramp slope and the average rate of change of velocity with respect to
clock time for the 1-domino system. Use comma-delimited format.
Using the same format report your results for the 2-domino system in the second line, and for the 3-domino
system in the third.
In your fourth line specify the units of these quantities. Ramp slope is a unitless quantity; be sure you report this.
Also briefly explain how you got your results and what they tell you about this system:
------>>>>>> ramp slope ave roc of vel each system
Your answer (start in the next line):
.03 , 10.675 m/s^2
.06 , 18.207 m/s^2
.09 , 24.411 m/s^2
#$&*
Graph acceleration vs. ramp slope
A graph of acceleration vs. ramp slope will contain three data points. The graph will visually represent the way
acceleration changes with ramp slope. A straight line through your three data points will have a slope and a y-
intercept, each of which has a very significant meaning.
Your results constitute a table with three rows and two columns, representing rate of velocity change vs. ramp
slope.
Sketch in your lab notebook a graph of the table you have just entered. The graph will be of rate of change of
velocity with respect to clock time vs. ramp slope. Be sure to follow the y vs. x convention to put the right
quantities on the horizontal and vertical axes (if it's y vs. x, then y is on the vertical, x on the horizontal axis).
Your graph might look something like the following. Note, however, that this graph is a little too long for its
height. On a good graph the region occupied by the data points should be about as high as it is wide. To save
space on the page, graphs depicted here are often not high enough for their width
Sketch the best possible straight line through your 3 data points. Unless the points lie perfectly along a straight
line, which due to experimental uncertainty is very unlikely, the best possible line will not actually pass through
any of these points. The best-fit line can be constructed reasonably well by sketching the line which passes as
close as possible, on the average, to the 3 points.
For reference, other examples of 3-point graphs and best-fit lines are shown below.
Describe your best-fit line by giving the following:
On the first line, the horizontal intercept of your best-fit line. The horizontal intercept will be specified here by a
single number, which will be the coordinate at which the line passes through the horizontal axis of your graph.
On the second line, the vertical intercept of your best-fit line. The horizontal intercept will be specified here by a
single number, which will be the coordinate at which the line passes through the vertical axis of your graph.
On the third line, give the units of your horizontal intercept and the meaning of that intercept.
On the fourth line, give the units of your vertical intercept and the meaning of that intercept.
Starting in the fifth line, give a brief written description of your graph and an explanation of what you think it
might tell you about the system:
------>>>>>> horiz int, vert int, units and meaning of horiz, then vert int
Your answer (start in the next line):
-.05
1
that would be in slope since the x axis is the slope the meaning is the slope of the graph to be at that acceleration
the acceleration at the point of the slope being zero
The graph is fairly linear with one of the points running across the line which shows the accel is fairly constant
#$&*
Mark the point on your best-fit line which would correspond to a ramp slope of .10. Determine as accurately as
you can the rate of velocity change that goes with this point, so that you have both the horizontal and vertical
coordinates of the point.
Report the horizontal and vertical coordinates of that point on the first line below, in the specified order, in
comma-delimited format. Starting at the second line, explain how you made your estimate and how accurate you
think it might have been. Explain, briefly, what your numbers mean and how you got them.
------>>>>>> mark and report best fit line coord for ramp slope .10
Your answer (start in the next line):
.10 , 29
The measurement is directly from the graph by extending the line of best fit , I would say withing 20%
They are what the graph would be at if the slope was at .10
#$&*
Determine the slope of the best-fit line
We defined rise, run and slope between graph points:
The 'run' from one graph point to another is the change in the horizontal coordinate, from the first point to the
second.
The 'rise' from one graph point to another is the change in the vertical coordinate, from the first point to the
second.
The slope between the two graph points is the rise-to-run ratio, calculated as slope = rise / run.
As our first point we will use the horizontal intercept of your best-fit line, the point where that line goes through
the horizontal axis.
As our second point we will use the point on that line corresponding to ramp slope .10.
In the box below give on the first line the run from the first point to the second.
On the second line give the rise from the first point to the second.
On the third line give the slope of your best-fit straight line.
Starting in the fourth line, give a brief explanation and an indication of what you think the slope might tell you
about the system.
------>>>>>> slope of graph based on horiz int, ramp slope .10 point
Your answer (start in the next line):
.15
29
193.333
This slope gives us an estimation of the rate at which acceleration is changing
#$&*
Assess the uncertainties in your result
The rest of this exercise is optional for Phy 121 and Phy 201 students whose goal is a C grade
Calculate average of mean times and average of standard deviations for 1-domino ramp
Since there is uncertainty in the timing data on which the velocities and rates of velocity change calculated in this
experiment have been based, there is uncertainty in the velocities and rates of velocity change.
We first estimate this uncertainty for the 1-domino case.
In the box below, report in the first line the right-to-left mean time, the left-to-right mean time and the average of
these two mean times on the 1-domino ramp. This third number, which you also calculated previously, will be
called 'the average of the mean times'.
In the second line report the standard deviation of right-to-left times, the standard deviation of left-to-right times
and the average of these standard deviations for the 1-domino ramp. This third number will be called 'the average
of the standard deviations'.
Starting in the next line give a brief explanation and speculate on the significance of these results.
------>>>>>> 1 dom ramp mean rt-left and left-rt, then std def of both
Your answer (start in the next line):
1.684 , 1.562 , 1.623
.09523 , .04526 , .07023
They are the average for the movements on the ramp and give us a pretty good estimate
#$&*
Use average time and standard deviation to estimate minimum and maximum possible velocity and acceleration
for first ramp
We will use the average of the mean times and the average of the standard deviations to estimate our error in the
average velocity and in the acceleration on the 1-domino ramp.
We will assume that the actual time down the ramp is within in the interval defined by mean +- std dev, where
'mean' is in this case the average of the mean times, and 'std dev' is the average of the standard deviations.
Using these values for mean and std dev:
Sketch a number line and sketch the interval from mean - std dev to mean + std dev. The interval will be centered
at the average of the mean times as you reported it in the previous box, and will extend a distance equal to the
average of the standard deviations (as also reported in the previous box) on either side.
So for example if the average of the mean times was 1.93 seconds and the standard deviation .11 second, the
interval would extend from 1.93 sec - .11 sec = 1.82 sec to 1.93 sec + .11 sec to 2.04 sec.. This interval would be
bounded on the left by 1.82 sec and on the right by 2.04 sec..
Report in the first line of the box below the left and right boundaries of your interval. Starting in the second line
explain briefly, in your own words, what these numbers represent.
------>>>>>> boundaries of intervals rt-left, left-rt
Your answer (start in the next line):
1.55277 , 1.69323
These are the possible max and min times it takes for the ball to go down the ramp
#$&*
Instead of 'rate of velocity change with respect to clock time' we will now begin to use the word 'acceleration'.
So 'average acceleration' means exactly the same thing as 'average rate of velocity change with respect to clock
time', and vice versa.
Since we are assuming here that acceleration is constant on a straight ramp, in this context we can simply say
'acceleration' rather than 'average acceleration'.
Using this terminology:
If the time down the ramp is equal to that of the left-hand boundary of the interval you just sketched, then what
would be the average velocity and the acceleration of the ball? Report in comma-delimited format on the first line
below.
Find the same quantities for the right-hand boundary of your interval, and report in similar format on the second
line.
In the third line report the resulting minimum and maximum possible values of acceleration on this interval, using
comma-delimited format. Your results will just be a repeat of the results you just obtained.
Starting on the fourth line, explain what your numbers represent and why it is likely that the actual acceleration of
the ball on a 1-domino ramp, if set up carefully so that right-left symmetry is assured, would be between the two
results you have given.
------>>>>>> 1-dom vel and accel left boundary of interval, rt boundary, min and max possible accel
Your answer (start in the next line):
18.032 m/s , 11.613 m/s^2
16.536 m/s , 9.766 m/s^2
#$&*
Repeat for 2- and 3-domino ramps
Do the same for the 2-domino data, and report in identical format, including explanations:
------>>>>>> 2-dom vel and accel left boundary of interval, rt boundary, min and max possible accel
Your answer (start in the next line):
1.192615 left time , 1.288385 right time , 23.478 left vel , 21.732 right vel , 19.686 left accel , 16.868 right accel
#$&*
Do the same for the 3-domino data, and report in identical format, including explanations:
------>>>>>> 3-dom vel and accel left boundary of interval, rt boundary, min and max possible accel
Your answer (start in the next line):
1.010235 left time , 1.131765 right time , 27.716 left vel , 24.74 right vel , 27.436 left accel , 21.86 right accel
#$&*
Now make a table of your results, as follows.
You will recall that slopes of .03, .06 and .09 correspond to the 1-, 2- and 3-domino ramps.
In the first line report the slope and the lower limit on acceleration for the 1-domino ramp.
In the second line report the slope and the lower limit on acceleration for the 2-domino ramp.
In the third line report the slope and the lower limit on acceleration for the 3-domino ramp.
In the fourth line report the slope and the upper limit on acceleration for the 1-domino ramp.
In the fifth line report the slope and the upper limit on acceleration for the 2-domino ramp.
In the sixth line report the slope and the upper limit on acceleration for the 3-domino ramp.
Starting in the seventh line give a brief explanation, in your own words, of what these numbers mean and what
they tell you about the system:
------>>>>>> slope and lower limit 1, 2, 3 dom; slope and upper limit 1, 2, 3 dom
Your answer (start in the next line):
.03 , 9.766
.06 , 16.868
.09 , 21.86
.03 , 11.613
.06 , 19.686
.09 , 27.436
#$&*
Plot acceleration vs. ramp slope using vertical segments to represent velocity ranges
On your graph of acceleration vs. ramp slope, plot the points specified by this table.
When you are done you will have three points lying directly above the .03 label of your horizontal axis. Connect
these three points with a line segment running vertically from the lowest to the highest.
You will also have three points above the .06 label, which you will similarly connect with a segment, and three
points above the .09 label, which you will also connect.
Your graph will now contain the best-fit straight line you made earlier, and the three short vertical line segments
you have just drawn. Your graph will look something like the one below, though your short vertical line segments
will probably be a little thinner than the ones shown here, and unlike yours the graph shown here does not contain
the best-fit line. And of course your points won't be the same as those used in constructing this graph:
Does your graph fit this description?
Does your best-fit straight line pass through the three short vertical segments?
Give your answer and be sure to include a couple of sentences of explanation.
------>>>>>>
best-fit line thru error bars?
Your answer (start in the next line):
Yes it does maybe a little out of it
It fits a fairly well maybe an error in judment could skew it though
#$&*
Determine max and min possible slopes of acceleration vs. ramp slope graph
It should be possible to draw a number of straight lines which pass through all three vertical segments. Some of
these lines will have greater slopes than others. For example note that the figures below show two lines which
pass through all three vertical segments, with the line in the second graph being steeper than the line in the first.
Draw the steepest possible straight line which passes through all three vertical segments on your graph.
Using the x-intercept of this line and the point on this line corresponding to ramp slope .10, determine the slope
of the line.
In the first line below report the rise, run and slope of your new line. Use comma-delimited format.
Starting in the second line give a brief statement of what your numbers mean, including an explanation of how you
obtained your slope. Be sure to include the coordinates of the two points you used and the resulting rise and run.
------>>>>>>
max possible graph slope
Your answer (start in the next line):
274.77
#$&*
Now draw the least-steep possible straight line which passes through all three vertical segments.
Follow the same instructions as before, and report your results for this line in the same way, including a brief
explanation:
------>>>>>> min possible graph slope
Your answer (start in the next line):
from the way my graph looks it seems the original slope gotten was the least steep
#$&*
Your instructor is trying to gauge the typical time spent by students on these experiments. Please answer the
following question as accurately as you can, understanding that your answer will be used only for the stated
purpose and has no bearing on your grades:
Approximately how long did it take you to complete this experiment?
------>>>>>>
Your answer (start in the next line):
3 hours
#$&*
@&
It isn't clear how you calculated your values for the accelerations.
On a 30-cm ramp a ball that accelerates the length of the ramp from rest in 1.5 seconds would average 20 cm/second, reach a final velocity of 40 cm/second, and have an average acceleration of about 26 cm/s^2.
On two of your ramps the ball accelerated the length of the ramp in less than 1.5 seconds, but none of your accelerations is as great as 26 cm/s^2.
Times on the first ramp weren't that much greater than 1.5 seconds, so your 10 cm/s^2 acceleration doesn't appear justified for those trials either.
*@
@&
Everything else looks pretty good.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
*@