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course phy 121
9/11 around 9:00
Randomized 2-1 revised#$&*
course Phy 121
9/9 3:35
Randomized 2-1#$&*
course Phy 121
9/8 around 7:00
If the velocity of the object changes from 3 cm / sec to 11 cm / sec in 9 seconds, then at what average rate is the velocity changing?Ave rate = change in distance / change in time
distance can't be negative; change in position can, so think in terms of change in position / change in clock time
8cm/sec / 9sec = 0.89 cm/sec
A ball rolling from rest down a constant incline requires 8.6 seconds to roll the 68 centimeter length of the incline.
• What is its average velocity?
Ave velocity = distance / time
vAve = 68cm / 8.6sec = 7.91 cm/sec
An object which accelerates uniformly from rest will attain a final velocity which is double its average velocity.
• What therefore is the final velocity of this ball?
&&&& final vel = ave vel * 2
= 7.91 cm/sec * 2
= 15.82 cm/sec
you've& figured out the average velocity of this ball. The preceding statement tells you how to find the final velocity. This should help.
• At what average rate is the velocity of the ball therefore changing?
??????????I am not really sure how to go about answering these questions.
&&&& the ave rate of change = change of A / change of B
Ave rate = change in distance / change in time
Ave rate = 11 cm/s – 3 cm/s / 9 sec
Aver rate = 8 cm/s / 9 sec
Ave rate = 0.89 cm
(cm/s) / s = cm/s^2, not cm.
You are no longer on the first problem, so 11 cm/s and 3 cm/s are not relevant here.
You have found that average velocity is 7.91 cm/s and final velocity is 15.82 cm/s.
You now need to find the average rate of change of the velocity with respect to clock time.
You have You've gone back to the velocities given for a preceding situation. In the present case the velocity goes from 0 to 15.82 cm/s in 8.6 sec.
what does the definition of average rate of change of A with respect to B tell you about the average rate of change of velocity with respect to clock time?
For the given situation, what do you conclude that average rate is for the ball?
An automobile accelerates uniformly down a constant incline, starting from rest. It requires 15 seconds to cover a distance of 172 meters. At what average rate is the velocity of the automobile therefore changing?
Ave velocity = distance / time
vAve = 172 m / 15 sec = 11.47 m/sec
???? How would I find the rate of change if I am only given this information… there is not a way to determine the change in time and distance?
&&&&Ave rate = change in distance / change in time
Ave rate = 11.47 m/sec / 15 sec
Ave rate = 0.76 m/cm^2
That is vAve / `dt, not `dv / `dt.
You have the average velocity. The initial velocity was zero.
What therefore is the final velocity?
What is the change in velocity, i.e., the change from initial to final velocity?
Now what do you get for the average acceleration?
It's very good to notice that you're missing something. Most students miss this and never realize they have.
See my notes and spend up to 15 minutes trying to construct a solution, and/or inserting questions. Don't get hung up on it for longer than that, and submit what you have at that point. I want to see if you can get it from my notes, but I don't want to tie you up for hours. If you can't get it in 15 minutes, I'll give you further hints.
&#Please see my notes and submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
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See my note above.
Now see if you can figure out the average rate of change of velocity with respect to clock time for that last situation. You already have vAve = 11.47 m/s. The sequence of questions before the last shows how to reason this out.
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Good, but you still have a little ways to go.
Please submit another revision. This time mark your insertions with ####, so I can distinguish them from previous insertions, and be sure to put that mark after as well as before each insertion so I can tell immediately where the inserted information begins and ends. As before, limit yourself 15 minutes then if necessary insert questions.
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