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Assn 7 Seed 2
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:
Find a, since `ds= v0`dt + .5 a `dt^2 =>
a= (`ds - v0) / .5 dt^2
--For slope = 0.05 and `dt = 8s,
a= (10-0) / .5*8^2=
10 / .5*64= 10 / 32 = .3125m/s/s
--For slope = 0.10 and `dt = 5s,
a= (10-0) / .5*5^2=
10 / .5*25= 10 / 12.5 = 0.8000m/s/s
=>So it seems that acceleration changes approximately 0.5 m/s/s per every 0.05 increase in slope.
Right. So the average rate of change of acceleration with respect to slope is
ave rate = change in accel / change in ramp slope = 0.5 m/s/s / (.05) = 10 m/s/s, or 10 m/s^2.
Note that the slope is a unitless quantity: since rise and run are both measured in the same units (e.g., cm), the units 'cancel' when you calculate slope = rise / run.
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&#Good responses. See my notes and let me know if you have questions. &#