Many students have had questions about the idea of 'flow diagrams', either because the concept is very new or because the information given in assignments has been skipped or missed.
A number of questions from students have motivated this expanded discussion, including the following question from a very capable and conscientious student answering the Query of Assignment 5:
I am really struggling with the different levels here.
I can use the given information to solve for the information we
need, but I’m having trouble making connections to the different levels.
I have tried to draw out and write out a chart from doing the work on
my own and when working through the given solution.
Is there an additional reference I can look at, read, or do some practice problems to try to get this?
I do know that the equation 'ds = (v0 + vf ) / 2 * 'dt can be used to find `ds, if we know v0, vf and `dt. Knowing `ds, I know that I can then figure out the other information. I just don't understand the flow diagram idea.
Note also that this question is very well-posed, telling me exactly what he knows and doesn't know
If well-prepared and conscientious students have either missed seeing the idea, or has trouble understanding it, then it's obvious that the entire idea needs to be addressed.
This document is therefore designed to further explain the idea to those who have difficulty with the concept, and to 'fill in the blanks' for those who have missed seeing the information.
We will develop the 'flow diagram' for reasoning out a uniform-acceleration situation in which we are given v0, vf and `dt for uniformly accelerated motion on a specific interval.
Note that assigned Class Notes and Introductory Problem Sets include further explanations and examples of the 'flow diagram' idea and its applications.
The first two levels of the diagram
The 'flow diagram' begins by listing the symbols for the given quantities v0, vf and `dt. We call these our first-level quantities. Our diagram will trace out how we use these quantities to progressively reason out the remaining quantities.
Of the seven quantities in terms of which we reason out uniformly accelerated motion we have be given three, v0, vf and `dt. Thus four remain, namely a, `ds, `dv and v_Ave.
From the initial and final velocities v0 and vf we know that we can find the change in velocity and the average velocity:
The change in a quantity during an interval is equal to the difference between its initial value and its final value, which is obtained by subtracting the initial value from the final. It is therefore clear that if we know the initial and final velocities, we can directly obtain the change in velocity.
We know that if acceleration is uniform, average velocity is equal to the average of initial and final velocities. We are assuming uniform acceleration, so knowing the initial and final velocities we can clearly determine the average velocity.
We indicate this on our diagram by writing down the symbols `dv and v_Ave for change in velocity and average velocity, and drawing lines to indicate the 'first-level' quantities from which they are determined.
We regard `dv and v_Ave in this diagram as 'second-level' quantities, since to find them we use 'first-level' quantities.
Especially when first using 'flow diagrams' we will generally list our 'first-level' quantities in a single line at the top of our diagram. Then we will list our second-level quantities on a single line, lower on the diagram. Following this pattern, our third-level quantities are listed on a line lower than the second, etc..
Our lines at this point are listed as
with the first line indicating our initial quantities and the second line the quantities we can reason out from them.
`dv is found using v0 and vf. We will therefore draw lines from v0 and vf to the symbol `dv.
v_Ave is also found from v0 and vf. We will therefore draw lines from v0 and vf to the symbol v_Ave.
Including the lines (using blue lines to indicate that v_Ave comes from v0 and vf, and green lines to indicate that `dv comes from v0 and vf), our picture now looks like the following:
Moving on to the third level; reasoning 'forward' and 'backward'
At this point of our reasoning we know v0, vf and `dt (our 'first-level' quantities), and we know how to find `dv and v_Ave (which are 'second-level' quantities).
Of the seven quantities in terms of which we understand uniformly accelerated motion, we therefore know all but two. The two we don't know are `ds and a_Ave.
We have two choices in how to proceed.
We can 'think forward', starting with what we know, and asking ourselves how our known quantities might be combined to obtain new information.
Or we can 'think backward', starting with the definitions of our remaining unknown quantities, and see if those definitions fit in with our known quantities.
To illustrate both ways of thinking, we will reason out how to find our average acceleration a_Ave by 'thinking forward', then reason out it out once again by 'thinking backward'.
We start by asking ourselves what we can find from what we know. We know v0, vf, `dt, `dv and v_Ave. What can be combined with what in this list to find new information?
If we think through the quantities we have, considering how to combine them, it might strike us that `dv and `dt, being respectively the change in velocity and the change in clock time, can be combined to find the average rate of change of velocity with respect to clock time.
Then again, this might not strike us immediately. So it's worth looking at the chain of thinking that might lead us to this conclusion:
Going back to our list v0, vf, `dt, `dv and a_Ave, we methodically begin to think through the possible combinations.
Our first-level quantities v0 and vf have already been combined to find `dv and vAve. So we have already done everything we can think of with these two quantities.
Our remaining first-level quantity is `dt. Can `dt be combined with anything on our 'known list' to find useful new information?
`dt combined with v0 to figure out how far we would go (or more precisely our displacement) if we maintained the initial velocity v0 for the entire interval; however since velocity changes that situation doesn't really occur. Or we could combine `dt with vf, but again vf represents the velocity only at the end of the interval; this velocity isn't maintained for the entire interval.
Moving on to our second-level quantities, `dt (change in clock time) combined with `dv (change in velocity) can clearly be used to find an average rate of change. We could apply the definition of average rate of change to find either the average rate of change of clock time with respect to velocity, or the average rate of change of velocity with respect to clock time. We should recognize the latter as the average acceleration.
Thus `dv and `dt can be combined to find the average rate of change of velocity with respect to clock time, which is the average acceleration a_Ave.
Thus by combining `dv and `dt we have found a new quantity, a_Ave. This is a third-level quantity, since to find it we had to use the second-level quantity `dv.
This is an example of 'thinking forward'. Whether immediately or through methodically thinking through possible combinations, we found that `dv and `dt can be used to find a_Ave.
We will place a_Ave at the 'third level' and connect it with lines to `dv and `dt, the quantities from which is is calculated.
We could have 'thought backward' to reason out a_Ave. We would start with the definition of a_Ave, then we would look to our known quantities to see if anything fits the definition.
a_Ave is the average acceleration, defined to be the average rate of change of velocity with respect to clock time. By the definition of average rate, this is (change in velocity) / (change in clock time), or `dv / `dt.
Thus we can find a_Ave if we know `dv and `dt.
Looking at our known quantities, we find that `dv and `dt are among them.
Just as before we conclude that a_Ave can be reasoned out from our known quantities `dv and `dt, so we place a_Ave at the 'third level' and connect it with lines to `dv and `dt, the quantities from which is is calculated.
It might seem from above that 'thinking backward' is easier. It's certainly shorter. However note that you didn't see the 'thinking-backward' solution until after you had seen the 'forward' reasoning. So you might not be in a position to judge which is the easier.
In general it turns out there is no clear consistent choice for which 'reasoning direction', forward or backward, is easier. It depends on the situation, and it often depends even more on how your mind works, on how your brain happens to be 'wired' and how your previous experiences have influenced your ways of thinking.
We still have `ds to reason out. We can do so by 'thinking forward' or 'thinking backward'.
It's probably more difficult to reason out `ds by 'thinking backward'. This is because, unlike a_Ave in the preceding, `ds is not defined in terms of the other quantities. So you don't have an obvious definition to use.
However this doesn't mean you don't have any definitions available. `ds is certainly involved in at least one other definition.
`ds represents change in position. One of our most basic definitions involves change in position: our definition of average rate of change of position with respect to clock time, i.e., average velocity. Thus v_Ave = `ds / `dt.
Knowing that v_Ave = `ds / `dt, we can look back at our known quantities v0, vf, `dt, v_Ave, a_Ave and `dv and quickly see that we know v_Ave and `dt. Thus we will be able to solve v_Ave = `ds / `dt for `ds, obtaining
`ds = v_Ave / `dt.
To find `ds by 'thinking forward', we look at what we know and think about how we can 'put the pieces together' to get new information:
Looking at our known quantities v0, vf, `dt, v_Ave, `dv and a_Ave, if we systematically 'pair up' quantities we will soon be lead to consider what we can find from `dt and v_Ave.
When we think about v_Ave, we should recall that its definition leads us to understand that v_Ave = `ds / `dt. Now, knowing v_Ave and `dt, we can solve for `ds, again obtaining
`ds = v_Ave / `dt.
The difference, in this example, between 'thinking forward' and 'thinking backwards' might be summarized in the following statements:
When we 'think forward', v_Ave is on our list of known quantities, so we are logically led to consider v_Ave.
When we 'think backward' we have to think of the definition, but without a list of known quantities to guide us.
The diagram at the third level
With `ds and a_Ave on the 'third level', our diagram is now as follows:
Adding lines to indicate the new relationships, using 'red' lines to indicate the reasoning of `ds and 'purple' lines for a_Ave, we have our final diagram:
Summarizing, we see that the above diagram is on three levels:
The first level indicates our given information v0, vf and `dt.
The second level indicates the quantities v_Ave and `dv, obtained using first level quantities.
The third level indicates the quantities `ds and a_Ave, both of which require second-level quantities (as well as a first-level quantity).
The 'blue' lines indicate the quantities used to find v_Ave.
The 'green' lines indicate the quantities used to find v_Ave.
The 'red' lines indicate the quantities used to find v_Ave.
The 'purple' lines indicate the quantities used to find v_Ave.
Once we understand the relationships ... :
Once we really understand the relationships among the seven quantities v0, vf, `dt, v_Ave, `dv, a_Ave and `ds, we will spontaneously be able to think about a given situation in whichever 'direction' has become more natural to us.
For example, we will get used to the idea that once we know two of the three quantities v_Ave, `ds and `dt, we can find the other, and it won't matter much to us whether we think of this from the 'forward' or 'backward' perspective.
When we are still 'sorting out' these ideas, though, it's a good idea to consciously approach a question from both directions.
Sometimes we really do need the equations:
In some cases, reasoning in the way illustrated here turns out to be impossible and we must rely on equations (this is the case if we are given a, `dt and either v0 or vf; the definitions alone aren't sufficient, at least without using more mathematics than we wish). However in all other cases, if we know three of the five quantities v0, vf, `dt, a and `ds, we can reason out the values of the seven quantities v0, vf, `dt, a, `ds, `dv and v_Ave.
