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phy 121
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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seed 41
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course Phy 121
9/9 around 5:00
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):
the y axis is the and the x axis is the distance.
You have that reversed.
&&&& y axis it distance and x axis is time
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Sketch a straight line segment between these points.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
the line increasing at a steep rate.
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What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
rise is 10, run is 5 and the slope is (9 s - 4 s) / (40 cm/s - 10 cm/s) = 5s / 30cm/s
=1/6 cm
you got your axes reversed, but otherwise your calculation is detailed and correct; however s / (cm/s) would be s^2 / cm, not cm.
&&&rise is 5, run is 10 and slope is (40cm/s – 10cm/s) / (9s – 4s) = 30cm/s / 5s = 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):
????? i don't know how to do this.
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&&&& region beneath this segment forms a trapezoid, with 'altitudes' being the initial and final velocities 10 cm/s and 40 cm/s. The average 'altitude' of the trapezoid is (10 cm/s + 40 cm/s) / 2 = 25 cm/s. Since the graph is a straight line, the average altitude therefore represents the average velocity on the interval. The width of the trapezoid is equal to the 'run' between the two points, which as seen earlier is 5 s. This represents duration of the time interval. The area of the trapezoid is equal to average altitude * width = 25 cm/s * 5 s = 125 cm.
Hopefully I've got the link right. See my notes, then check the link below, and in this case do resubmit. Let me know if the link isn't right.
&#See any notes I might have inserted into your document. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your 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. &#
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revision
&#This looks good. Let me know if you have any questions. &#
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