PHY 121
Your 'cq_1_05.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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The problem:
A ball accelerates at 8 cm/s^2 for 3 seconds, starting with velocity 12 cm/s.
• What will be its velocity after the 3 seconds has elapsed?
answer/question/discussion: ->->->->->->->->->->->-> :
Final velocity= initial velocity + acceleration x time
final velocity= 12 cm/s + 8 cm/s^2(3s)
=12cm/s +24cm/s
=36 cm/s
Final Velocity= 36 cm/s
• Assuming that acceleration is constant, what will be its average velocity during this interval?
answer/question/discussion: ->->->->->->->->->->->-> :
12 +36 = 48
48/2= 24
• How far will it travel during this interval?
answer/question/discussion: ->->->->->->->->->->->-> :
Distance= Rate x Time
Distance= 24 cm/s x 3 sec
Distance= 72 cm
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10 min
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When determining the distance traveled should we use the average velocity?
&#This looks very good. Let me know if you have any questions. &#
The information below has been recently edited into a version of a problem you have already completed.
You should review this discussion, which addresses the limitations to the 'distance = rate * time' idea. It's important to see how this idea generalized into the overall idea of rate. I believe you understand, but the discussion below can help clarify what you know and might serve you as a useful reference.
COMMENTS ON GOOD STUDENT SOLUTION
The student's solution:
r=d/t
d=10cm/s - 40cm/s = - 30cm/s
t=3s
r=30cm/s / 3s
r=10cm/s^2.
INSTRUCTOR COMMENTARY
Your calculation is correct, as are your units, so you did very well on the question.
However you need to move your thinking beyond the formula r = d / t.
This formula, while useful in lower-level courses, has serious limitations in this context.
The formula r = d / t is fine for teaching the very important concept of rate in courses below the level of algebra, but it isn't specific enough for applications in higher-level courses.
Specifically for this problem, the use of d for the change in velocity is likely to cause confusion.
Instead of
r = d / t,
you should be thinking
ave rate = change in A / change in B.
The change in A is sometimes a distance, in which d could be an appropriate letter, but more often it is something else, as in this example.
Another problem with the letter 'd' is that it is commonly used for 'distance'.
'Distance' is not the most commonly used quantity ussed in analyzing motion; rather displacement is the more appropriate quantity.
Since the letter 'd' is more likely to invoke the word 'distance', it is likely to be misleading.
Still another problem with the letter 'd' is that it has a universal and specific meaning in mathematical notation, and to avoid
confusing notation the letter 'd' needs to be reserved for that meaning.
As you have seen by now, we use the letter 'd' in the prefix 'change in'. Thus when analyzing a quantity x, the symbol `d
(which in our notation stands for the Greek Delta), `dx means 'change in x'.
So for example the expression `dy / `dx stands for 'change in y / change in x', which by definition of average rate of change stands for the average rate of change of y with respect to x.
At the level of Principles of Physics and even more so in General College Physics we consider the concept of the instantaneous rate of change, which occurs when `dx gets smaller and smaller, approaching zero. The limiting value of `dy / `dx, and `dx shrinks to zero, is universally represented by the expression dy / dx.
In University Physics we take this a step further with the concept of the derivative function dy/dx.
The equation r = d / t is also confusing even in the context of distance and time.
The symbol t generally stands for the 'running time' on a timekeeping device (e.g., a clock), and not for the interval between to events that occur at different clock times.
To use the same symbol t for the duration of the time interval during which the distance is covered is inherently confusing.
The use of r for the rate is pretty much OK, but if there are a number of different types of rate involved in a situation, as in the present situation, the use of a single letter for 'rate' can be confusing.
The formula r = d / t should therefore be replaced in our minds by a more general formula like
r = `dx / `dt,
where x is position and t is clock time.
`dx is literally read as 'delta x' or 'change in x'.
If x represents position, then `dx represents change in position.
`dt is read as 'change in t', so if t represents clock time, `dt represents change in clock time.
Thus r in this context would stand for 'change in position / change in clock time', which by definition of
average rate is the average rate of change of position with respect to clock time.
The current question does not deal with change in position, but with change in velocity. In this case it would be more
appropriate to use v for velocity, so that our desired rate will be
r = `dv / `dt.
We said above that the use of r for rate is 'pretty much' OK. However we are already running into a problem with the use of
just plain r.
We previously used r for the average rate of change of position with respect to clock time, r = `dx / `dt.
Now we are using r for the average rate of change of velocity with respect to clock time, r = `dv / `dt.
For the present we'll just make note of this ambiguity. Shortly we'll resolve it.
Now let's return to your solution.
Your calculation for this problem, which I repeat arrived at the correct final result using correct reasoning and including correct units (when 80% of students in your course at this stage at least fail to do the units calculation correctly, and most do not spell out their reasoning) was presented as follows:
r=d/t
d=10cm/s - 40cm/s = - 30cm/s
t=3s
r=30cm/s / 3s
r=10cm/s^2.
This is a very good solution but would be clearer if we used different symbols. The same solution, using `dv instead of d
for change in velocity and `dt instead of t for change in time, would read
r =`dv / `dt
`dv = 10cm/s - 40cm/s = -30cm/s
`dt = 3s
r = - 30cm/s / 3s
r = - 10cm/s^2.
It should be clear why the notation used in this revision is more specific and clearer than the notation of the r = d / t formula.
Now let's consider the ambiguity in the use of the letter r.
The quantity we calculate here is `dv / `dt, the average rate of change of velocity with respect to clock time.
We could use the abbreviation 'ave roc of v wrt t' instead of just r. With this notation the solution would read
ave roc of v wrt t =`dv / `dt
`dv = 10cm/s - 40cm/s = -30cm/s
`dt = 3s
ave roc of v wrt t = - 30cm/s / 3s
ave roc of v wrt t = - 10cm/s^2.
This clearly defines the nature of the average rate we are calculating. With this notation we have a complete and specific solution to the problem.
Since we've come this far, we have a good opportunity to go a little further and use a name for the average rate of change of velocity with respect to clock time.
We will call this quantity the 'average acceleration'. Average acceleration is defined to be the average rate of change of velocity with respect to clock time.
Using the notation a_Ave for this quantity your solution becomes
a_ave =`dv / `dt
`dv = 10cm/s - 40cm/s = -30cm/s
`dt = 3s
a_Ave = - 30cm/s / 3s
a_Ave = - 10cm/s^2.
Since we've gone this deeply into the discussion, one other idea worth noting is that in most of the situations we encounter in the early part of a first-semester physics course, we expect acceleration to be uniform, unchanging, for the duration of some extended time interval.
If this is the case, then the average acceleration is the same as the initial acceleration or the final acceleration, or the acceleration at any other instant during that interval. In this case we can simply drop the subscript 'Ave' and use the letter a for 'the constant acceleration'.
In your problem it was not specified that the acceleration is constant. So the best you can say from the given information is that a_Ave = - 10 cm/s^2.
Had it been specified that acceleration is constant, then your solution could have read
acceleration is constant, so a_Ave can be represented simply by the letter a
a_ave = `dv / `dt, so
a = `dv / `dt.
`dv = 10cm/s - 40cm/s = -30cm/s
`dt = 3s
a = - 30cm/s / 3s
a = - 10cm/s^2.
One final thing it worth noting:
The average slope of the v vs. t graph, between two graph point, represents an average acceleration. If acceleration is constant, the v vs. t graph, which represents velocity vs. clock time, has constant slope so the graph is a straight line. The converse is also true: If the v vs. t graph is a straight line, then acceleration is constant.
All these ideas will be developed in this assignment and in the next few assignments. This discussion should be a worthwhile reference as you continue to sort through these ideas.
Your work on the assignment is very good.
The information above has been recently edited into a version of a problem you have already completed.
You should review this discussion, which addresses the limitations to the 'distance = rate * time' idea. It's important to see how this idea generalized into the overall idea of rate. I believe you understand, but the discussion below can help clarify what you know and might serve you as a useful reference.