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If I answered problems 3 and 4 incorrectly, could you please explain how to find the answers. I thought since the needed quantities were not given that I was supposed to simply explain how to do the problem.What exactly are constant and straight inclines? Are they both simply level inclines?
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`dv = vf -vi = 15cm/s - 3cm/s = 12cm/s
Avg rate of `dv = `dv/`dt = (12cm/s)/10s = 1.2cm/s
2.A ball rolling from rest down a constant incline requires 4.4 seconds to roll the 86 centimeter length of the incline.
What is its average velocity?
vAve = `ds/`dt
`dt = 4.4s
`ds = 86cm
vAve = 86cm/4.4s = 19.55cm/s = 20cm/s (rounded)
3.An object which accelerates uniformly from rest will attain a final velocity which is double its average velocity.
What therefore is the final velocity of this ball?
What average rate is the velocity of the ball therefore changing?
Since we have not atual quantities for this problem, we cannot give an actual number as an answer. We can, however, provide a method or formula for solving it.
Final velocity, vf = 2vi
`dv = vf - vi = 2vi -vi = vi
So change in velocity is equal to the initial velocity.
Then the average rate of change of velocity is equal to the initial velocity over the time interval. We do not know what the time interval is, so all we can say is:
avg rate of `dv = vi/`dt
4.An automobile accelerates uniformly down a constant incline, starting from rest. It requires 10 seconds to cover a distance of 138 meters. At what average rate is the velocity of the automobile therefore changing?
`dv = distance change / time interval = 138m / 10s = 13.8m/s
We know from reading the problem that the velocity at the bottom of the ramp will be faster than the velocity at the top. But in order to find the average rate at which velocity changes, we must have another set of coordinates. We need the distance covered over a different time interval. Then we could find the change in velocity betweent the two time intervals. Then to find the rate at which velocity changes, we divide the change in velocity by the time interval:
avg rate of `dv = `dv/`dt
It would look something like this:
avg rate of `dv = (vf - 13.8m/s)/(tf - 10s)"
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Good work. Let me know if you have questions.