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):
This graph has velocity on the y-axis and seconds on the x-axis. The slope of the graph would be `dv/`dt which is the acceleration.
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• Sketch a straight line segment between these points.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
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• What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
This graph has velocity on the y-axis and seconds on the x-axis. The slope of the graph would be `dv/`dt which is the acceleration.
the rise is the velocity measure along the y-axis it is the change in velocity which is represented by `dv. The run is the change in time across the x-axis. This is represented by `dt.
The rise is 40cm/s-10cm/s= 30cm/s and the run is 9s-4s= 5s. The slope is 30cm/s/5s= 6cm/s^2 which is 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 is the distance that the abject it traveling represented by ‘ds
The points create a trapezoid. The area of the trapezoid is is 1/2the rise * run so (10+40)/2=25cm/s
25cm/s* run (5s)= 125 cm. This is the distance.
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15 min
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&#Your work looks very good. Let me know if you have any questions. &#