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PHY 121
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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: ->->->->->->->->->->->-> :
Known quantities: `ds = 10 m; slope .05; `dt = 8 s for first run and
`ds = 10 m; slope = .10; `dt = 5 s for second run
To find the acceleration for the first run:
vAve = `ds/`dt = 10 m/8s = 1.25 m/s
vf = 2vAve -v0; vf = 2(1.25 m/s) - 0 m/s = 2.5 m/s
`dv = vf-v0 = 2.5 m/s - 0 m/s = 2.5 m/s
a = `dv/`dt = 2.5 m/s / 8 s = .3125 m/s^2
To find the acceleration for the second run:
vAve = `ds/`dt = 10 m/5s = 2 m/s
vf = 2vAve -v0; vf = 2(2 m/s) - 0 m/s = 4 m/s
`dv = vf-v0 = 4 m/s - 0 m/s = 4 m/s
a = `dv/`dt = 4 m/s / 5 s = .8 m/s^2
The change in acceleration is .8 m/s^s - .3125 m/s^ = .4875 m/s^2
The change in slope is .10 - .05 = .05
The rate is .4875 m/s^2 / .05 or 9.75 m/s^2
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