Your 'cq_1_7.2' 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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Seed quest 7.2
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 the accel of both
Ds = v0*dt + .5a * dt^2
10m = 0 * 8s + .5a * 8^2
10m = .5a * 64
0.15625 = .5a
.3125m/s/s = a
Ds = v0 *dt + .5a * dt^2
10m = 0 * 5 + .5a * 5^2
10m = .5a * 25
.4 = .5a
.8m/s/s = a
.3125m/s/s
.8m/s/s
Find difference in accel
.8 - .3125 = .4875m/s/s
Find difference in slope
.1 - 0.05 = 0.05
Divide diff accel by diff slope
.4875/.05 = 9.75
I don’t think that is right…
It's right. The ramp slope is unitless; acceleration has units m/s^2. So this rate of change ends up with units of m/s^2, identical to the units of acceleration.
Find the average of the two accelerations
(0.8 + 0.3125)/2 = .55625
Find average of the two slopes
(.1 + .05)/2 = .075
Divide ave accel by ave slope
.55625/.075 = 7.416
I don’t think that is right either…
I don’t know how to do it.
This is not right. The former was right. The former approach was a direct application of the definition of rate of change.
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25 mins
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Can you explain what I was supposed to do? I think I was on the right track, but got off somewhere.
Your first attempt was a correct application of the definition of rate of change, and you got a correct result.
You're doing great.