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: ->->->->->->->->->->->-> (start in the next line):
The graph shows the times running along the x-axis, and the velocities running along the y-axis.
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• Sketch a straight line segment between these points.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
After plotting the points, and drawing the straight line segment, the line appears to be increasing at an increasing rate.
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• What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The rise of the segment is, (40 cm/s – 10 cm/s) = 30 cm/s.
The run of the segment is, (9s – 4s) = 5s
Therefore the slope is, (30 cm/s) / 5sec = 6 cm/s/s.
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• What is the area of the graph beneath this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The easiest way to find the area under the graph is to treat it like a rectangle and a triangle. The rectangle measures; (5 * 10) = 50. The triangle measures 5 * (30/2) = 75. Added together the total area is 50 + 75 = 125.
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You're better off treating the region as a trapezoid, because the dimensions of the trapezoid have a more direct relationship to the motion than do the dimensions of the rectangle and triangle.
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20 minutes
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No need for a revision but be sure to check out the discussion of the last question at the link given below:
&#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. &#