#$&*
course phy 121
9/13 8:45
Randomized 4#$&*
course phy 121
9/10 round 3:40
An automobile traveling a straight line is at point A at clock time t = 4 sec, where it is traveling at 10 m/s, to point B at clock time t = 11 sec, where it is traveling at 20.5 m/s. Point A is 75 meters from the starting point and point B is 115 meters from the starting point. What are the average velocity and the average acceleration of the automobile during the specified time? What evidence is there that the acceleration is or is not uniform? Ave vel = change in position / change in time
Ave vel = (115 m – 75m) / (11sec – 4sec)
Ave vel = 40 m / 7sec = 5.71m/sec
Ave acc = `dv / `dt.
Ave acc = (20.5 m/s – 10 m/s) / (11sec – 4sec)
Ave acc = 10.5 m/s / 7 s = 1.5 m/s
check the units; m/s / s is not m/s. You got this in the qa; refer to those calculations and be sure you understand.
It is uniform because it is with respect to clock time.
If acceleration is uniform then the average velocity on an interval, which you calculated correctly, is equal to the average of the initial and final velocities on that interval.
What do you conclude?
&&&& I conclude that the it is not uniform because if I take the average of the initial and final velocities it becomes 20.5 m/s + 10 m/s = 30.5 m/s and then divide that by 2 to get 15.25 m/s and that does not equal the ave acc. ????I am not sure that I did this right.
You aren't comparing with the average acceleration but with the average velocity; that might well be what you meant.
Good, but check those unit calculations, and please submit a revision with an answer to the question in my last note.
&#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.
&#
#$&*
&#This looks good. See my notes. Let me know if you have any questions. &#