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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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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: ->->->->->->->->->->->-> scussion (start in the next line):
I made a graph with velocity as the vertical axis. The horizontal axis is time. I have two points, (4,10) and (9,40).
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Sketch a straight line segment between these points.
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
It’s linear and creates a trapezoid.
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What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
The rise is the change in velocity which is 30 cm/s. I got that by subtracting the initial velocity from final velocity. The run is the change in time which is 5 seconds. I subtracted 4 from 9 to get this. Rise/Run = slope, so 30cm/s / 5s = 6cm/s^2. This is the slope.
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What is the area of the graph beneath this segment?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
The area is 250 cm/s. I took the average height, which was the midpoint of the line between the points. That was 25 cm/s. Then the width is 10 s. You multiply these to get 250 cm/s.
@& Good, but the width of the trapezoid is only 5 seconds, not 10 seconds. Your area will change accordingly.*@
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*#&!
&#This looks good. See my notes. Let me know if you have any questions. &#