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Phy 231
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):
The y coordinates represent the velocities and the x coordinates represent the clock times. The velocity is along the y-axis and the clock time is along the x-axis.
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• Sketch a straight line segment between these points.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Since the velocity increases from initial to final velocity and since the time increases, the straight line on the graph will be increasing and an increasing rate.
A straight line can increase or decrease, but the rate at which is does so is constant.
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• What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The slope of the line on the graph with = rise / run. The rise corresponds to the y coordinates and also the change in velocity. The run corresponds to the x coordinates and also the change in clock time. The slope therefore shows the change in velocity / change in clock time which is also the average acceleration.
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• What is the area of the graph beneath this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The area of the graph beneath the segment is (1/2)*base*height because a triangle can be drawn within the y coordinates and x coordinates and the straight line segment. The base of the triangle is the change in clock time. The height of the triangle is the change in velocity. The resulting area should also equal the change in position, or how far the object traveled.
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15 min
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&#Your work looks very good. Let me know if you have any questions. &#