#$&*
Phy 121
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 **
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: ->->->->->->->->->->->-> :
Incline 1:
`ds = 10m
`dt = 8s
V0 = 0m/s
Slope = .05
Incline 2:
`ds = 10m
`dt = 5s
V0 = 0m/s
Slope = .1
Acceleration for slope 1:
`ds = (v0+vf)/2 *`dt
Vf = (2 *`ds - v0) / `dt
Vf = (2 * 10m)/8s = 2.5m/s
Vf = v0 + a *`dt
A = (vf - v0)/`dt
A = 2.5m/s / 8s = .3125m/s/s
Acceleration for slope 2:
Vf = ((2 *`ds - v0) / `dt)
Vf=(2*10m)/5s = 4m/s
A = (vf - v0)/`dt
A = 4m/s / 5s = .8m/s/s
So, the average rate at which the acceleration is changing with respect to slope is:
`da / `dslope
=(.8m/s/s - .3125m/s/s)/(.1-.05)
=(.4785m/s/s)/(.05) = 9.5cm/s/s
@&
This would be 9.5 m/s^2.
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&#Very good responses. Let me know if you have questions. &#