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PHY 241
Your 'cq_1_04.1' 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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The problem:
A ball is moving at 10 cm/s when clock time is 4 seconds, and at 40 cm/s when clock time is 9 seconds.
Sketch a v vs. t graph and represent these two events by the points (4 sec, 10 cm/s) and (9 s, 40 cm/s).
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
I sketched a graph with the x axis from 0 to 9 with 1 s increments and the y axis from 0 to 40 with 10 cm/s increments. I plotted to points on this graph (4,10) and (9,40)
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Sketch a straight line segment between these points.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
I sketched a straight line segment from (4,10) to (9,40)
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What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
rise = 30 cm/s
run = 5 s
slope = rise/run = 6 cm/s^2
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What is the area of the graph beneath this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
y - y_1 = m(x - x_1)
v(t) = 6t - 14
integral from 4 to 9 of 6t -14 = 125 cm/s^2
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25 minutes
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9pm 10/21/2011
self-critique #$&*
#$&* self-critique
self-critique rating
rating #$&*:
&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#
@& Good.
The graph in this point is a straight line, so it wasn't actually necessary to find the integral. A straight-line v vs. t graph occurs if and only if acceleration is uniform, in which case it's possible to solve using the equations of uniformly accelerated motion (or, equivalently, analysis by trapezoids).
If the graph isn't a straight line, then the integral is typically required.
No revision necessary, but take a look at the discussion at the link below.
&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#
*@