The following is a solution to the given problem.

Please compare the given solution with your solution and submit a self-critique of any error(s) in your solutions, and/or and additional questions or comments you might have. 

Simply copy your posted document into a text editor and insert revisions, questions, and/or self-critiques, marking your insertions with ####.  Submit using the Submit Work Form.

Among other things, this problem should illuminate

• the definition of average rate of change

• the difference between velocity and acceleration

• the use of units, including the the algebra of units

• the units of velocity and acceleration

These problems are based on the definition of rate of change:

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)

********************

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?

We solve this by applying the definition of average rate of change.

• 10 cm and 20 cm are positions, so quantity A is position and quantity B is change in clock time.

• The average rate of change of position with respect to clock time is therefore

ave rate = change in position / change in clock time = 10 cm / (5 s) = 2 cm/s.

The average rate of change of position with respect to clock time is by definition the average velocity.

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?

Common erroneous answers include 10 cm, 10 cm/s, 90 cm, and 90 cm/s.

The correct answer is 10 cm / s^2.  Be sure you understand how this answer is obtained:

We solve this by carefully applying the definition of average rate of change.

• 10 cm / sec and 40 cm / sec are velocities, so quantity A is velocity and quantity B is change in clock time.

• The average rate of change of velocity with respect to clock time is therefore

ave rate = change in velocity / change in clock time =

(30 cm / s) / (3 s) =

10 (cm/s) / (s) =

10 (cm/s) * (1/s) =

10 cm/s^2.

• Note that this result is in units of cm/s^2, not cm/s.

Be sure you are aware of the difference between the definition of average velocity and the definition of average acceleration, and of the difference in the units of velocity and acceleration.

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.

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?

The average rate of change of position with respect to clock time is

• ave rate = (change in position) / (change in clock time).

It follows that

• change in position = ave rate * change in clock time,

or more specifically

• change in position = ave rate of change of position with respect to clock time * change in clock time.

Thus for this example,

• change in position = 5 cm/s * 10 s = 50 cm.

Some erroneous solutions:

STUDENT SOLUTION:  5cm/sec/10sec=.5cm/sec^2
 

INSTRUCTOR COMMENT:  If 5 cm/s was the change in velocity, then 5 cm/s / (10 s) = .5 cm/s^2 would be the rate of change of velocity with respect to clock time. 

However 5 cm/s is not the change in velocity, it is the average rate of change of position with respect to clock time, i.e., the average velocity.  This is not the same thing as the change in velocity. 

• vAve and `dv are easy to confuse, but they are very different quantities with very different meanings.

INFORMATION AND QUESTION TO BE ANSWERED: 

Ave. rate of change of position with respect to clock time = change in position / change in clock time.

Abbreviate this by ave rate = `dx / `dt.

You are given the ave rate and the change in clock time. You need to find the change in position.

In the abbreviated notation, you are given ave rate and `dt. You want to find `dx.

• The Question:  Given this information, how do you find `dx?

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.

Good student response: The first thing to remember is that the rate of change definition is the change of A divided by the change of B. I will make a note of this to report back to in the future.

Some instructive student errors are included below:

Instructor Note #1:  It's very important to label the A and the B quantities carefully and correctly.  If you get in the habit of labeling these quantities with the correct terms, you will more quickly expand your understanding of this important definition and your ability to apply it correctly in the many different situations you will encounter in this course.  It is very important in all fields to apply and understand the words and phrases used in definitions and descriptions. 

Many of the examples given below illustrate basically good student work in which the student nevertheless labeled some of the quantities incorrectly, in a way that could lead to later confusion.

Responses to first question:

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?

STUDENT RESPONSE TO FIRST QUESTION:   “A” is the rate of change in distance (10cm) and “B” is the rate of change of the clock (5sec). Average rate of change is 2cm/1sec.< /p>

INSTRUCTOR CRITIQUE:

Your final result is correct and your steps are correct, but you didn't correctly label the A and the B quantities:

• The A quantity is position, and the B quantity is clock time. 

The definition says that

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

If A is the rate of change in position and B the rate of change in clock time, then by this definition we would have

• average rate of change of rate of change in position with respect to rate of change of clock time = (change in rate of change of position) / (change in rate of change in clock time).

It should be clear to you that this contains way too many words, and that it describes something more complicated that the requested quantity.

What we want here is the average rate of change of position with respect to clock time. So the A quantity is simply position, and the B quantity is simply clock time.

Responses to second question: 

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?

STUDENT RESPONSE:  Quantity A is 30cm/second and B is 3 seconds. So, the ball is gaining a velocity of 10cm/sec for each second that it is rolling.

INSTRUCTOR CRITIQUE:  Your final answer is correct, and your reasoning process is good, but you have not correctly labeled all the quantities.

Quantity A is velocity, and quantity B is clock time.

30 cm/s is the change in velocity, which you have correctly obtained by subtracting 10 cm/s from 40 cm/s.

However quantity A being velocity, and 30 cm/s being change in velocity, we cannot say that quantity A is 30 cm/s.

30 cm/s is the change in A, not the A quantity itself.

3 seconds is a time interval, which is an interval between two clock times, or a change in clock time. It is not a clock time, so 3 seconds is not the B quantity.

3 seconds is the change in the B quantity, not the B quantity itself. </h3>