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PHY 201
Your 'cq_1_07.1' 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_07.1_labelMessages **
Asst_7_Seed_7.1
A ball falls freely from rest at a height of 2 meters. Observations indicate that the ball reaches the ground in .64 seconds.
Based on this information what is its acceleration?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
Acceleration is the change in velocity over the change in clocktime.
v0=0m/s
vf=2m/.64sec = 3.125m/s
(`dv= 3.125m/s - 0m/s) = 3.125m/s
a=3.125m/s / .64s = 4.88m/s^2
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Is this consistent with an observation which concludes that a ball dropped from a height of 5 meters reaches the ground in 1.05 seconds?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
v0 = 0m/s
vf= 5m / 1.05s = 4.76m/s
`dv = 4.76m/s
a= 4.76m/s / 1.05 s = 4.53m/s^2
The acceleration of the second ball is slightly slower, about 5% the rate of the first example
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Are these observations consistent with the accepted value of the acceleration of gravity, which is 9.8 m / s^2?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
I do not believe these observations are consistent as in the second observation, the ball had less than the accepted value for acceleration of gravity, 9.8m, it had 5 meters to travel with more than one second. Based upon these calculations, I don't find the observations to be consistent with the given value for acceleration of gravity.
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15mins.
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You have used vAve for `dv in these solutions. This is a common source of error, so be sure you use the definitions carefully.
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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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Asst_7_Seed7.2
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: ->->->->->->->->->->->-> scussion:
When the slope equaled .10, `dt= 5seconds and `ds remained 10 meters.
Using the equation, `ds = [(v0 + vf)/2] * `dt, and solving for vf, we get:
2*(`ds / `dt)= `vf ; substituting the known values:
2*(10m / 5s) = vf
2*(2m/s) = vf
4m/s = vf
Now using the equation vf = v0 + a * `dt and solving for a:
vf - v0 / `dt = a, and substituting known values:
4m/s - 0m/s / 5 sec = a
.8m/s^2 = a(.10)
For slope = .05, we can use the same equations for figuring acceleration:
vf = 2(10m / 8s)
vf= 2(1.25m/s)
vf= 2.5m/s
a= 2.5m/s / 8s
a(.05)= .3125m/s^2
Acceleration changes with a quicker rate as slope is doubled.
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Good so far, but you still have to find the average rate of change of the automobile's acceleration with respect to the slope of the incline.
To understand what this is asking you will want to return to the definition of an average rate of change.
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Be sure to include the entire document, including my notes. &#
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