#$&*
MTH 279
@&
I've responded by email. Let me know if you haven't received my response.
*@
Phy 201
Your 'cq_1_07.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_07.2_labelMessages **
Seed 7 2
An automobile rolls 10 meters down a constant incline with slope .05 in 8 seconds, starting from rest. The same automobile requires 5 seconds to roll the same distance down an incline with slope .10, again starting from rest.
• At what average rate is the automobile's acceleration changing with respect to the slope of the incline?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
m=.05
ds=10 m
dt= 8s
vAve=1.25m/s
v0=0
vf= 2.5 m/s
dv= 2.5 m/s
aAve= 2.5 m/s/ 8s
=.312 m/s/s
m= .1
ds= 10 m
dt= 5 s
vAve=2 m/s
v0=0
vf= 4 m/s
dv= 4 m/s
aAve= .5 m/s/s
@&
Good to here.
*@
the acceleration is changing at an avg rate of .40625 m/s/s when the slope increases from .05 to .1.
&&&&&&&&7
(avg dv)/ (avg dt)= aAve
3.25/6.5= .5m/s/s
This is my best guess im not entirely sure what to do.
@&
The question asks for the average rate of change of acceleration with respect to ramp slope.
Confirm for your self that by the definition of average rate, this would be
(change in acceleration) / (change in ramp slope).
The change in ramp slope is from .05 to .10, a change of .05. The change in acceleration is from .312 m/s^2 to .8 m/s^2, a change of .488 m/s^2.
So the requested rate is
.488 m/s^2 / (.05 ) = 9.76 m/s^2.
*@
@&
You don't say how you got your average rate of change. Your result doesn't appear to agree with the result that would be obtained from your previous results.
You do need to apply the definition of average rate very carefully.
*@
#$&*
Seed 7 2
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: ->->->->->->->->->->->-> :
m=.05
ds=10 m
dt= 8s
vAve=1.25m/s
v0=0
vf= 2.5 m/s
dv= 2.5 m/s
aAve= 2.5 m/s/ 8s
=.312 m/s/s
m= .1
ds= 10 m
dt= 5 s
vAve=2 m/s
v0=0
vf= 4 m/s
dv= 4 m/s
aAve= .5 m/s/s
@&
My previous note indicated that this result was good, but 4 m/s / (5 s) = .8 m/s^2, not .5 m/s^2.
*@
@&
Good to here.
*@
the acceleration is changing at an avg rate of .40625 m/s/s when the slope increases from .05 to .1.
&&&&&&&&7
(avg dv)/ (avg dt)= aAve
3.25/6.5= .5m/s/s
This is my best guess im not entirely sure what to do.
@&
The question asks for the average rate of change of acceleration with respect to ramp slope.
Confirm for your self that by the definition of average rate, this would be
(change in acceleration) / (change in ramp slope).
The change in ramp slope is from .05 to .10, a change of .05. The change in acceleration is from .312 m/s^2 to .8 m/s^2, a change of .488 m/s^2.
So the requested rate is
.488 m/s^2 / (.05 ) = 9.76 m/s^2.
*@
@&
You don't say how you got your average rate of change. Your result doesn't appear to agree with the result that would be obtained from your previous results.
You do need to apply the definition of average rate very carefully.
*@
#$&*
** **
10 minutes
** **
@&
Good start, but the last step needs to be revised as indicated.
&#Please see my notes and submit a copy of this document with revisions, comments 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.
&#
*@