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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: ->->->->->->->->->->->-> :
Given situation 1: ‘ds=10m, slope=.05, ‘dt=8s, v0=0
Given situation 2: ‘ds=10m, slope=.10, ‘dt=5s, v0=0
Find: a for each situation and average rate of change of a with respect to change in slope.
situation 1: ‘ds=10m, slope=.05, ‘dt=8s, v0=0
find a
‘ds = v0*‘dt + .5*a*’dt^2, solve for a
‘ds - v0*’dt = .5 * a * ‘dt^2
(ds - (v0*’dt))/(.5*’dt^2) = a
(10m - (0m/s*8s))/(.5*(8s)^2) = .313m/s^2
situation 2: ‘ds=10m, slope=.10, ‘dt=5s, v0=0
find a
‘ds = v0*‘dt + .5*a*’dt^2, solve for a
‘ds - v0*’dt = .5 * a * ‘dt^2
(ds - (v0*’dt))/(.5*’dt^2) = a
(10m - (0m/s*5s))/(.5*(5s)^2) = .8m/s^2
Situation 1 a=.313m/s^2, situation 2 a=.8m/s^2
Change in quantity A(accel) with respect to change in quantity B(slope)
(.8m/s^2 - .313m/s^2)/(.10 - .05)=9.74m/s^2
????Not sure why the slope doesn’t have units in the problem????
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Slope is rise / run.
Both rise and run are distances.
If they are expressed in the same units, which is usually the only way that makes good sense, then the units will divide out, leaving the quantity unitless.
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&#Good responses. See my notes and let me know if you have questions. &#