Phy 121
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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cq_1_022
Phy 121
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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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: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
The clock time at the midpoint of this interval would be 9 sec.
### The clock time at the midpoint is determined by finding the total of the two clock times and dividing by two. 5s+ 13s = 18 s. 18 s/2= 9s. ####
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• What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
I would guess that the velocity of the object a this interval would be 3cm/sec.
### I think what threw me off on this question was that I determined the average velocity by taking the change in position and dividing it by the change in time. This is how I arrived at 3 cm/s. Good; be sure you are aware that you weren't given any information about position. Instead the question is simply asking for the velocity at the midpoint. This is determined very similarly to how we determined the clock time at midpoint. We take 16 cm/s + 40 cm/s = 56 cm/s. 56cm/s /2= 28 cm/s ####
very good
Given velocities are 16 cm/s and 40 cm/s. Average velocity would be more than 3 cm/s.
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• How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
If we are talking about how far the object travels between the start point and midpoint, then I would say that the object travels 12 cm.
if the ave velocity was 3 cm/s (it isn't but this was the result you gave earlier) and time interval is 4 sec (which is the interval between start and midpoint), then 12 cm would be the correct conclusion.
If we are talking about how far it travels from our point of 16 cm to 40 cm then I would say that the object travels 24 cm.
16 cm and 40 cm are not quantities involved in this problem; 16 cm/s and 40 cm/s are, but their difference is 24 cm/s, not 24 cm
#### Initially when solving this problem, I was not sure if it was asking how far the object traveled from start to the given midpoint or from start to end. After reading the explanation in the solution, I understand that it was asking how far the object traveled from start to finish. This is determined by figuring out the change from 16cm/s to 40 cm/s. This is a change of 24 cm/s.24 cm/s is the change in velocity, and isn't directly related to how far the object travels. To figure out how far this object travels you would use the average velocity, which is 28 cm/s.
I also switched up the units for the given numbers when first giving an answer. It is important to pay close attention to the units assigned in problems or else it changes the work completely and is incorrect. ####
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• By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
The clock time changes 4 seconds during the time interval from start to midpoint.
The clock time changes a total of 9 seconds from the start point to finish point.
#### The clock time changes a total of 8 seconds for the duration of the time interval. 13 seconds-5 seconds = 8 seconds. ####
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• By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
Velocity changes at 3cm/ second from start to midpoint.
The velocity changes 2.6 cm/sec from start to finish.
####Velocity changes a total of 24cm/s during this time interval. 40 cm/s – 16 cm/s = 24 cm/s.####
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• What is the average rate of change of velocity with respect to clock time on this interval?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
the average rate of change can be found by change in a/change in b.
So the average rate of change of velocity from start to midpoint would be 3cm/sec
The average rate of change of velocity from start to finish would be 2.6 cm/sec.
### The average rate of change can be found by change in a/change in b. So the average rate of change of velocity would be vAve= ‘ds/ ‘dt.
vAve = 24 cm/s / 8 s = 3 cm/s
24 cm/s / (8 s) = 3 cm/s / s, not 3 cm/s.
3 cm/s / s is 3 cm/s^2, as you correctly state below.
3cm/s / s = 3 cm/s * 1s = 3 cm/s^2
####
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• What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
The rise on the graph was determined by completing the following: 13-6=9
So rise = 9
###The rise of the graph is 24 cm/s ###
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• What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
The run of this graph was determined by completing the following: 40-16= 24.
### The run of the graph is 8 sec. ####
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• What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
The rise on the graph between these points if 0.375. This was discovered by taking 13-6 = 9 and 40-16=24. Once obtaining these results, I divided 9/24 = .375
### The slope is determined by rise/run. So based on our numbers slope = 24cm/s / 8 sec= 3 cm/s / s = c cm/s * 1s = 3 cm/s ^2 ####
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• What does the slope of the graph tell you about the motion of the object during this interval?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
It is increasing.
#### The slope is the same as the average rate of change. ####
More correctly the average rate of change of velocity with respect to clock time.
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• What is the average rate of change of the object's velocity with respect to clock time during this interval?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
The average rate of change for this objects velocity in respect to clock time during this interval is found by the change in a/change in b. So 24/9= 2.6
### The average rate of change is = change in a/change in b
So 24 cm/s / 8 seconds = 3 cm/s / s = 3cmcm/s * 1s = 3 cm/s^2 ####
### With determining the slope and average rate of change for velocity, how come the answer is in cm/s ^2??? Shouldn’t the seconds cancel each other out and simply be cm/s??? I found this a little tricky.
This is tricky, but it just follows from the rules of arithmetic for fractions.
(cm / s) / s means (cm / s) * (1 / s), which means cm * 1 / (s * s) = cm / s^2. ####
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This is a revised version of my first submission for this assignment.
Your revisions are very good, though there is one significant error and a couple of smaller errors which will be easy to correct.
See my notes and let me know if you have questions.