cq_1_011

phy201

Your 'cq_1_01.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

The problem:

Here is the definition of rate of change of one quantity with respect to another:

The average rate of change of A with respect to B on an interval is

average rate of change of A with respect to B = (change in A) / (change in B)

Apply the above definition of average rate of change of A with respect to B to each of the following. Be sure to identify the quantity A, the quantity

B and the requested average rate.

If the position of a ball rolling along a track changes from 10 cm to 20 cm while the clock time changes from 4 seconds to 9 seconds, what is the

average rate of change of its position with respect to clock time during this interval?

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

D=r * t

D=10cm-20cm=10cm

t=4s-9s=5s

r= d/t

r=10cm/5s

r=2cm/s

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If the velocity of a ball rolling along a track changes from 10 cm / second to 40 cm / second during an interval during which the clock time changes by

3 seconds, then what is the average rate of change of its velocity with respect to clock time during this interval?

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

r=d/t

d=10cm/s - 40cm/s = - 30cm/s

you had 30 cm/s; however 10 - 40 = -30, not 30. I've made the correction in your otherwise correct solution.

t=3s

r= - 30cm/s / 3s

r= - 10cm/s^2

You have a good solution (but see my note on the sign error), and we can use this solution as a basis for the deeper understanding we will be developing in this course. Read on.

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.

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#$&*

If the average rate at which position changes with respect to clock time is 5 cm / second, and if the clock time changes by 10 seconds, by how much does

the position change?

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

D=r * t

D= 5cm/s * 10s

D= 50 cm/s^2

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You will be expected hereafter to know and apply, in a variety of contexts, the definition given in this question. You need to know this definition word

for word. If you try to apply the definition without using all the words it is going to cost you time and it will very likely diminish your

performance. Briefly explain how you will ensure that you remember this definition.

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

You must know the definition of the terms word for word and know how to apply it to different contexts, this will save you time and increase your

performance

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You are asked in this exercise to apply the definition, and given a general procedure for doing so. Briefly outline the procedure for applying this

definition, and briefly explain how you will remember to apply this procedure.

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

in this procerdure rate of change of an object is defined by the change of its distance divided by its change of time. I use the formula D=r * t to

find the solution and manipulate the formula to find the answer I am looking for.

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#$&*

20 minutes

:-)

Your solutions are excellent. However you need to move to a more specific notation. Though you've adapted it well, the formula r = d / t has serious limitations at this level.

I've used one of your solutions as a basis for an extensive edit of the 'feedback' document for this series of problems, and I've inserted that edit above. I believe you will understand everything very well; please review my edit and let me know if you have questions.

cq_1_011

Your 'cq_1_01.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

** **

you had 30 cm/s; however 10 - 40 = -30, not 30. I've made the correction in your otherwise correct solution.

You have a good solution (but see my note on the sign error), and we can use this solution as a basis for the deeper understanding we will be developing in this course. Read on.

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