#$&*
course Phy 121
3.21.11 at 8:00pm
Resubmit 72#$&*
course PHY121
3.12.11 at 6:17pm
cq_1_072#$&*
PHY121
Your 'cq_1_07.2' report has been received. Scroll down through the document to see any
comments I might have inserted, and my final comment at the end.
** **
An automobile rolls 10 meters down a constant incline with slope .05 in 8 seconds, starting
from rest. The same automobile requires 5 seconds to roll the same distance down an incline
with slope .10, again starting from rest.
At what average rate is the automobile's acceleration changing with respect to the slope of
the incline?
answer/question/discussion: ->->->->->->->->->->->-> :
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0.75m/0.05 = 15
10m/8sec = 1.25m/sec
10m/5sec = 2m/sec
2m/sec-1.25m/sec = 0.75m
Slope 0.10 - slope 0.05 = 0.05 difference
Not sure what untis would go with the 15 on line 1?
####What has me confused, I think, is this: if v0=0, isnt vAve and acceleration the same figure?####
@& vAve is the average rate of change of position with respect to clock time.
average acceleration, or aAve, is average rate of change of velocity with respect to clock time.
Average velocity, final velocity, change in velocity, and average acceleration are all different things. In some cases two or more of those velocities might be equal, but generally this won't be the case. However acceleration is a completely different quantity than velocity. No acceleration is ever equal to a velocity; among other things acceleration and velocity have different units.
Concentrate on that first car for the moment. Its average velocity is 1.25 m/s, as you say. What therefore are its final velocity and the change in its velocity? What is the resulting average acceleration?
Then reason out the motion on the second incline.*@
** **
10 minutes
** **
@& You haven't calculated accelerations, but you have done most of what would be required to
find the average rate of change of average velocity with respect to slope.
If you divide the change in average velocity by the change in slope, you get the average
rate of change of average velocity with respect to slope.
Now find the accelerations on the two slopes, and follow the same reasoning to get the
requested result.
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#$&*
@& See my note and see if you can do another revision, using &&&& this time.
*@"
@&
&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
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Solution
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*@
@& vAve and acceleration are completely different quantities.
vAve is average rate of change of position with respect to clock time, and has units like cm/s or m/s.
average acceleration is average rate of change of velocity (not position) with respect to clock time and has units like cm/s/s or m/s/s, equivalently expressed in cm/s^2 or m/s^2.*@