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 sketched this graph velocity is on the y axis and time is on the x axis the point I plotted were (4s,10cm/s) and (9s, 40cm/s)
• Sketch a straight line segment between these points.
answer/question/discussion: I drew a straight line from my two points.
• What are the rise, run and slope of this segment?
answer/question/discussion: The rise is 40 cm/s – 10cm/s = 30 cm/s. the rise represents the change in the position coordinate. The run represents the change in the clock time so the run is 9s – 4s = 5s. So the slope of the segment is the rise/run = 30cm/s/5s=6cm
Right procedure but your units in the final answer are not quite right:
(cm/s) / s = (cm/s) * (1/s) = cm/s^2, not cm. The correct answer would be 6 cm/s^2.
• What is the area of the graph beneath this segment?
answer/question/discussion: The change in velocity between two clock times is represented as an area beneath a graph of acceleration vs. clock time so the area beneath the graph is 40cm/s – 10cm/s= 30cm/s
25 min
Your statement
'change in velocity between two clock times is represented as an area beneath a graph of acceleration vs. clock time'
is correct, and you are doing a good job of working from the definitions. The area beneath an acceration vs. clock time graph for this interval would be exactly as you say. Again this is very good.
However you don't have a graph of acceleration vs. clock time. The given graph is of velocity vs. clock time, and the calculation of the area of the given graph is very different.
The document at the link below should help you clarify the calculation and interpretation of the area, and should also reinforce your correct thinking on the slope.
&#At least part of your solution does not agree with the solution and comments given at the link below. You should view the solution at that link and self-critique as indicated there.
Solution
This link also expands on these topics and alerts you to many of the common errors made by students in the first part of this course. &#