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PHY 201
Your 'cq_1_09.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Seed Question 9.1
You should enter your answers using the text editor or word processor. You will then copy-and-paste it into the box below, and submit.
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A ball accelerates uniformly as it rolls 20 cm down a ramp, starting from rest, in 2 seconds.
• What are its average velocity, final velocity and acceleration?
answer/question/discussion: ->->->->->->->->->->->-> :
Given: ds = 20 cm, v0 = 0, dt = 2 s
vAve = ds / dt,
vAve = 20 cm / 2 s
vAve = 10 cm/s
vAve = (v0 + vf) / 2
vf = 2 * (vAve) - v0
vf = (2 * 10 cm/s) - 0
vf = 20 cm/s
vf = v0 + a * dt
a = (vf - v0) / dt
a = (20 cm/s - 0 cm/s) / 2 s
a = 10 cm/s^2
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• If the time interval is in error so that it is 3% longer than the actual time interval, then what are the actual values of the final velocity and acceleration?
answer/question/discussion: ->->->->->->->->->->->-> :
0.6 sec = 1%, so 1.8 sec = 3%, 2 sec + 1.8 sec = 3.8 sec
vAve = ds / dt,
vAve = 20 cm / 3.8 s
vAve = 5.26 cm/s
vf = v0 + a * dt
a = (vf - v0) / dt
a = (20 cm/s - 0 cm/s) / 3.8 s
a = 5.26 cm/s^2
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@& 1% of a 2 second interval is .02 second, not .6 sec.
3% of a 2 second interval would be .06 sec.
Your calculations follow correctly from your percent calculations, but the percent calculations are incorrect.*@
• What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> :
% error = (estimate - actual) / actual * 100
I am confused with calculating the percent error for the velocity and acceleration.
#$&*
@& The percent error is the error, as a percent of the actual value.
The error is the difference between the observed and actual values.
You got an actual value of 5.25 cm/s^2, when the observed value was 10 cm/s^2.
What therefore is the error, and what is the percent error?*@
@& You should then correct your previous calculations and recalculate the percent error.*@
• If the percent error is the same for both velocity and acceleration, explain why this must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
I believe the percent errors will be the same because the velocity and the acceleration are the same value with different units.
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*#&!
@& You know how to calculate acceleration. However you have an error in calculating your percents, and as you indicate you need a little more information on calculating percent error.
&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions 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.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you. &#
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*#&!