Phy 231
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.
** **
Copy the problem below into a text editor or word processor.
• This form accepts only text so a text editor such as Notepad is fine.
• You might prefer for your own reasons to use a word processor (for example the formatting features might help you organize your answer and explanations), but note that formatting will be lost when you submit your work through the form.
• If you use a word processor avoid using special characters or symbols, which would require more of your time to create and will not be represented correctly by the form.
• As you will see within the first few assignments, there is an easily-learned keyboard-based shorthand that doesn't look quite as pretty as word-processor symbols, but which gets the job done much more efficiently.
You should enter your answers using the text editor or word processor. You will then copy-and-paste it into the box below, and submit.
________________________________________
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 at the midpoint would be 5+13/2 which equals 9 seconds.
• What is the velocity at the midpoint of this interval?
->->->->->->->->->->->-> answer/question/discussion:
The velocity at the midpoint would be 16 + 40 over 2 which equals 28cm/s
• How far do you think the object travels during this interval?
->->->->->->->->->->->-> answer/question/discussion:
Using the average velocity of 28cm/s equals Distance over 9 seconds. Solving algebraically the distance is equal to 252cm.
• By how much does the clock time change during this interval?
->->->->->->->->->->->-> answer/question/discussion:
The clock time changes by 9 seconds because 13 – 5 is equal to 9 seconds.
• By how much does velocity change during this interval?
->->->->->->->->->->->-> answer/question/discussion:
velocity changes by 40 – 16 which is equal to 24cm/s.
• What is the average rate of change of velocity with respect to clock time on this interval?
->->->->->->->->->->->-> answer/question/discussion:
the average rate of change would 40cm/s – 16cm/s over 9 seconds which is equal to 2.67cm/s/s
• What is the rise of the graph between these points?
->->->->->->->->->->->-> answer/question/discussion:
the rise of the graph between the two is equal to 40-16 which is 24cm/s
• What is the run of the graph between these points?
->->->->->->->->->->->-> answer/question/discussion:
the run is equal to 13 – 5 which is equal to 9 seconds.
• What is the slope of the graph between these points?
->->->->->->->->->->->-> answer/question/discussion:
The slope of the graph is equal to 24 over 9 which is equal to 2.67cm/s/s
• What does the slope of the graph tell you about the motion of the object during this interval?
->->->->->->->->->->->-> answer/question/discussion:
The slope of this graph tells us the acceleration of this object
• What is the average rate of change of the object's velocity with respect to clock time during this interval?
->->->->->->->->->->->-> answer/question/discussion:
the average rate of change would 40cm/s – 16cm/s over 9 seconds which is equal to 2.67cm/s/s
** **
20 min
** **
&#This looks very good. Let me know if you have any questions. &#