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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: ->->->->->->->->->->->-> :
Acceleration for incline with slope 0.05
a = `dv / `dt; a = (vf – vo) / `dt; v0 = 0; a = vf / `dt
vAve = `ds / `dt
vAve = (vf + v0) / 2; v0 = 0; vf = 2*vAve
`ds = 10 m; `dt = 8 s
vAve = 10 m / 8 s = 1.25 m/s
vf = (2)*(1.25 m/s) = 2.5 m/s
a = (2.5 m/s) / (8 s) = 0.3125 m/s^2
Acceleration for incline with slope 0.10
a = `dv / `dt; a = (vf – vo) / `dt; v0 = 0; a = vf / `dt
vAve = `ds / `dt
vAve = (vf + v0) / 2; v0 = 0; vf = 2*vAve
`ds = 10 m; `dt = 5 s
vAve = 10 m / 5 s = 2 m/s
vf = (2)*(2 m/s) = 4 m/s
a = (4 m/s) / (5 s) = 0.8 m/s^2
Average rate = change in acceleration / change in incline slope
Change in acceleration = 0.8 m/s^2 – 0.3125 m/s^2 = 0.4875 m/s^2
Change in incline slope = 0.10 – 0.05 = 0.05
Average rate = 0.4875 m/s^2 / 0.05
= 9.75 m/s^2
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