Phy 201
Your 'cq_1_02.2' 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 graph is constructed representing velocity vs. clock time for the interval between clock times t = 5 seconds and t = 13 seconds. The graph consists of a straight line from the point (5 sec, 16 cm/s) to the point (13 sec, 40 cm/s).
• What is the clock time at the midpoint of this interval?
answer/question/discussion: the clock time would be 9 because you subtract the 13 and the 5 which gives you 8 then divide that by 2 which gives you 4 then move 4 spaces to the right of 5 or 4 spaces to the left of 13 which puts you at 9
Good. Alternatively add the two clock times and divide by 2; same result and it's good to understand it both ways.
What is the velocity at the midpoint of this interval?
answer/question/discussion: do the same type of thing subtract the 40 and the 16 to get 24 then divide that by 2 to get 12 then move 12 spaces below 40 or 12 spaces above 16 to get 28
• How far do you think the object travels during this interval?
answer/question/discussion: Im not sure if this is right but I think that the distance traveled is the average speed multiplied by the time. So it would be 28cm/s * 9 sec = 252 cm
That would be vAve * midpoint clock time, not vAve * `dt.
• By how much does the clock time change during this interval?
answer/question/discussion: If my previous answer is correct then to get the time interval you would divide the distance traveled/by the average speed so it would be 252cm/28cm/s = 9 sec
• By how much does velocity change during this interval?
answer/question/discussion: velocity is the rate of change of position divided by the change in clock time so it would be 252/9=28cm/s
• What is the average rate of change of velocity with respect to clock time on this interval?
answer/question/discussion: well the average rate of change is a change in something divided by a change in something else so 28/9 =3.11cm/s
28 cm/s is average velocity, not the change in velocity
• What is the rise of the graph between these points?
answer/question/discussion: well you first take velocities and subtract them to get the rise so 40-16 = 24 so the rise is 24
• What is the run of the graph between these points?
answer/question/discussion: to get the run you take the times and subtract them so 13-5=8 so the run is 8s
• What is the slope of the graph between these points?
answer/question/discussion: the slope = rise/run so 24/8=3 so the slope is 3cm
• What does the slope of the graph tell you about the motion of the object during this interval?
answer/question/discussion: the object is rising during this interval at 3 cm per sec
• What is the average rate of change of the object's velocity with respect to clock time during this interval?
answer/question/discussion: the rate of change is the change in something divided by a change in something else. So 8/3 = 2.67
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40 min
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Most of your answers are good, and you'll understand the disussion in the link below.
&#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. &#