Phy 231
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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: ->->->->->->->->->->->-> :
`ds= 10m
`dt= 8s
V0=0m/s
The 1st ramp with slope of .05:
10m/8s= 1.25m/s= vAve
1.25m/s= (v0+vf)/2, so 2(1.25)= 2.5m/s as vf and `dv
2.5m/s/8s= 0.31m/s^2= acceleration
2nd ramp:
`dt= 5s
`ds=10m
V0=0m/s
10m/2= 2m/s= vAve and 2*2= 4m/s and vf and `dv
4m/s / 5s= 0.8m/s^2
The average rate of change is (0.8m/s^2-0.31m/s^2)/ (.10-.05)= .49/.05= 9.8m/s^2
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15min
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&#Very good work. Let me know if you have questions. &#