#$&*
Phy 201
Your 'cq_1_07.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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An automobile rolls 10 meters down a constant incline with slope .05 in 8 seconds, starting from rest. The same automobile requires 5 seconds to roll the same distance down an incline with slope .10, again starting from rest.
• At what average rate is the automobile's acceleration changing with respect to the slope of the incline?
answer/question/discussion: ->->->->->->->->->->->-> :
the acceleration for the first incline is calculated like this:
d=v0*t+1/2 *a*t^2
10m=1/2 *a*(8s)^2
10m= ½*a*64s^2
20m=a*64s^2
a=0.3125m/s^2
The acceleration for the second incline is calculated like this:
d=v0*t+1/2 *a*t^2
10m=1/2*a*(5s)^2
10m=1/2*a*25s^2
20m= a*25s^2
a=0.8m/s^2
Then to get the average rate of change of the acceleration, then you would first find the difference between the accelerations: 0.8m/s^2 - 0.3125m/s^2 = 0.4875m/s^2
Then you should find the difference in the times of the two inclines: 8s- 5s = 3s
Then you divide the difference in acceleration by the difference in time to ger:
0.4875m/s^2 / 3s = 0.1625m/s^3
#$&*
@& You would divide by the difference in the slopes. This is rate of change of acceleration with respect to slope.
Otherwise very good.*@