Phy231
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.
** **
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:
13s-5s=8s/2=4/s+3/s=7s
• What is the velocity at the midpoint of this interval?
answer/question/discussion:
slope=(40cm/s - 16cm/s) / (13s-5s)=3
y=3x+b
40=3(13) +b
40-39=b
b=1
y=3(7)+1
y=22cm/s
• How far do you think the object travels during this interval?
answer/question/discussion:
avg velocity=(40cm/s+16cm/s)/2=28cm/s
• By how much does the clock time change during this interval?
answer/question/discussion:
8s
• By how much does velocity change during this interval?
answer/question/discussion:
40cm/s-16cm/s=24cm/s
• What is the average rate of change of velocity with respect to clock time on this interval?
answer/question/discussion:
24cm/s/8s=3cm/s^2
• What is the rise of the graph between these points?
answer/question/discussion:
24cm/s
• What is the run of the graph between these points?
answer/question/discussion:
8s
• What is the slope of the graph between these points?
answer/question/discussion:
3
• What does the slope of the graph tell you about the motion of the object during this interval?
answer/question/discussion:
It moves forward at an increasing rate.
• What is the average rate of change of the object's velocity with respect to clock time during this interval?
answer/question/discussion:
This question is asked and answered already above.
** **
30 minutes
** **
&#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
&#