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Phy 121
Your 'cq_1_06.1' 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_06.1_labelMessages **
For each situation state which of the five quantities v0, vf, `ds, `dt and a are given, and give the value of each.
A ball accelerates uniformly from 10 cm/s to 20 cm/s while traveling 45 cm.
answer/question/discussion: ->->->->->->->->->->->-> :
v0 = 10cm/s, v0 is the point at which the object in question is starting to be examined in reference to its positions of interest(position or velocity/time, etc).
vf = 20cm/s, vf is the final point at which the object in question is being observed in reference to its position of interest(another way to state it than above would be the final point of interest to which acceleration is observed as a rate of change in relationship velocity vs. change in time).
`ds = 45cm, `ds is the value derived from subtracting the initial position from the final position of an object(usually located on a graphs y axis) and is considered the objects change in position.
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A ball accelerates uniformly at 10 cm/s^2 for 3 seconds, and at the end of this interval is moving at 50 cm/s.
answer/question/discussion: ->->->->->->->->->->->-> :
a = 10cm/s^2, acceleration is a rate of change associated with a change in an objects velocity vs a time interval.
`dt = change in time, or time interval, that is associated with, among other things, velocity, acceleration, etc.
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A ball travels 30 cm along an incline, starting from rest, while accelerating at 20 cm/s^2.
answer/question/discussion: ->->->->->->->->->->->-> :
`ds = 30cm, 30cm is related to a change in position.
v0 = starting from rest, or zero, this means that the first point of interest is when the object was at rest, or zero time and zero velocity.
a = 20cm/s^2, the object accelerated at a rate of 20cm/s per second along a change of position of 30cm.
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Then for each situation answer the following:
Is it possible from this information to directly determine vAve?
answer/question/discussion: ->->->->->->->->->->->-> :
Yes, it is. If I know that an object moved 30cm from rest(zero change in position or velocity and zero change in time), then I know it's initial position and it's final position which is half of the vAve equation. And I know that it changed it's velocity at a rate of 20cm/s per second, so in order to move 30cm the object needed one and a half seconds to do so. So now I know the objects change in time, 1.5s.
vAve = 30cm - 0cm / 1.5s - 0s = 20cm/s. And now that I know that (20cm/s(vf) + 0cm/s(v0)) / 2 = 10cm/s(should be vAve), and 2 * 10cm/s = 20cm/s, then I can also conclude that the relationship is linear, or uniform in nature.
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Is it possible to directly determine `dv?
answer/question/discussion: ->->->->->->->->->->->-> :
Yes, because when I'm given the acceleration between points v0 and vf, I'm told that the change in velocity to a position 30cm from rest is at a rate of 20cm/s per second. So over this positions interval of change, it changed at a rate of 20cm per second, for each second until it reached a position of 30cm. So to move 30cm at a rate of 20cm/s per second, I'd need to be moving for 1.5seconds to reach the 30cm mark in the y axis.
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30 mins
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&#Very good responses. Let me know if you have questions. &#