Phy 201
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: ->->->->->->->->->->->-> :
I have a graph with velocity in cm/sec on the vertical y-axis and time in seconds on the horizontal x-axis. I have one point at (4, 10) and another at (9, 40).
• Sketch a straight line segment between these points.
answer/question/discussion: ->->->->->->->->->->->-> :
With the above graph I just described, I drew a line connecting the two coordinate points.
• What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> :
The rise in the given graph is 30. I get this quantity by determining the difference between the 10cm/s and the 40cm/s on the vertical axis. The run for this particular graph is 5. I get this quantity by determining the difference between the 4 sec and 9 sec quantities on the horizontal axis. The slope of this graph is the rise/run, or 30/5 = 6.
• What is the area of the graph beneath this segment?
answer/question/discussion: ->->->->->->->->->->->-> :
The area of the graph beneath the line segment is 75cm^2. I get this solution by using the formula ½(b * h). For this particular graph, I use ½(5s * 30cm/s) = 75cm^2.
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This assignment took me about 20 minutes.
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