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: My graph has a positive slope, has a y intercept of 10cm/sec. The y axis is labeled as velocity in cm/sec and the x axis is labeled as clock time in seconds. The intervals on the y axis are 10 cm/sec and the intervals on the x axis is 1 sec.
• Sketch a straight line segment between these points.
answer/question/discussion: The straight line again indicates a positive slope.
• What are the rise, run and slope of this segment?
answer/question/discussion: rise is the difference in y values or the difference in velocity. In this case, the rise is (40cm/sec- 10cm/sec) which is 30 cm/sec. The run is the difference in x values or clock time. In this case it is 9 seconds-4seconds which equals 5 seconds. The slope of a graph is found by taking rise/run which is equal to (30cm/sec)/5sec giving 6cm/sec/sec. The slope of a velocity vs. time graph represents the acceleration between two points.
• What is the area of the graph beneath this segment?
answer/question/discussion: The area beneath the segment is representative of the positional change and can by found by finding the average velocity and multiplying it by the change in time. This is done by adding 40cm/sec and 10 cm/sec and dividing by 2. This gives an average velocity of 25cm/sec. We can then multiply this by the run of the graph which is 5 seconds. This gives a total positional change of 125cm.
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15 mins
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`gr43
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.
** **
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: My graph has a positive slope, has a y intercept of 10cm/sec. The y axis is labeled as velocity in cm/sec and the x axis is labeled as clock time in seconds. The intervals on the y axis are 10 cm/sec and the intervals on the x axis is 1 sec.
• Sketch a straight line segment between these points.
answer/question/discussion: The straight line again indicates a positive slope.
• What are the rise, run and slope of this segment?
answer/question/discussion: rise is the difference in y values or the difference in velocity. In this case, the rise is (40cm/sec- 10cm/sec) which is 30 cm/sec. The run is the difference in x values or clock time. In this case it is 9 seconds-4seconds which equals 5 seconds. The slope of a graph is found by taking rise/run which is equal to (30cm/sec)/5sec giving 6cm/sec/sec. The slope of a velocity vs. time graph represents the acceleration between two points.
• What is the area of the graph beneath this segment?
answer/question/discussion: The area beneath the segment is representative of the positional change and can by found by finding the average velocity and multiplying it by the change in time. This is done by adding 40cm/sec and 10 cm/sec and dividing by 2. This gives an average velocity of 25cm/sec. We can then multiply this by the run of the graph which is 5 seconds. This gives a total positional change of 125cm.
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15 mins
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&#Very good responses. Let me know if you have questions. &#