When it doesn't go faster any
faster...
In our previous discussion of
onedimensional motion we used a model in which the acceleration
was itself a varying function of time. By introducing yet another
simplification, we can get at several interesting examples of
motion without getting bogged down in complicated mathematics.
Consider the case of acceleration that is constant over time.
This case is actually what we see when objects are accelerated by
gravity near the surface of a planet such that the distance the
object covers in its travels is small compared to the distance to
the center of the planet. You know, tossed eggs, thrown rocks,
cannon balls... that sort of thing.



With acceleration being a constant, the average value which is
given by = (v_{f}  v_{i}) / (t_{f} 
t_{i}), which is the same as the instantaneous
value. The subscript _{i} refers to an initial condition
and the subscript _{f} refers to a later condition. The
initial velocity v_{i} is that measured at time
t_{i} and the final velocity v_{f} is that
measured at time t_{f}.
To simplify our handling of time, recognize that we are always
going to be dealing with changes in time, like t_{f} 
t_{i}. Now if I add any arbitrary constant to both
t_{i} and t_{f}, the difference will be
unchanged. For example let us suppose there have been 5e17
seconds since the beginning of time. (By the way if you are not
familiar with the notation, 5e17 means 5 times 10 raised to the
seventeenth power or 500,000,000,000,000,000.) Then t_{i}
might be 5e17 seconds and t_{f} might be 5e17 seconds+1.
The difference is 1 second. If I add 5e17 to both t_{i}
and t_{f} then t_{i} = 0 and t_{f} = 1.
The difference is still 1 second. So we are free to let
t_{i} = 0 as long as we let t_{f} be measured in
seconds beyond t_{i}. Now since t_{i} may be
arbitrarily set to zero, there is no reason to continue to label
the final time with an _{f}. Let's just call it t, so
=
(v_{f}  v_{i}) / (t  0).
To further clean up the notation, since the initial time is
always to be taken as zero, let's call the initial velocity
associated with it v_{0}. Also what we called
v_{f} was just the velocity measured at time
t_{f}, which we now identify as just plain t. So we can
drop subscript _{f} from velocity measured at time t,
calling it just plain v.
Now here is one of those ideas that has always
struck me as a little bit sneaky. We have been working so far
with two discrete times, initial and final. We got rid of the
initial time by agreeing to let it always be zero. Then logically
we could take the "final" designation off the other value of
time. Now the quantity t is not any specific value. It may be
taken to be any number of seconds that suits us. In other words t
may be considered a variable.
The only restriction on v is that it is the velocity
corresponding to time t. That makes v also a variable, dependent
on the independent variable t. So we have come to a relationship
between t and v similar to that we defined for x and y in our
discussion of functions .

To symbolize the functional relationship we write v = f(t), which is said, "v equals f of t", or "v is
a function of t". The symbol f stands for any particular function
we choose. In terms of our earlier talk about functions we could
have said y = f(x) This way of
identifying that two variables have a dependency relationship is
called "functional notation". It allows us to know that v is a
function of t without knowing the details of the function.
Getting back to our expression for acceleration, remember that
the instantaneous acceleration is the same as the average
acceleration so that a = (v  v_{0}) /
t. Since v = f(t), we can
rearrange the acceleration equation, a = (v 
v_{0}) / t to find out what specific f(t) v is.
Multiply both sides by t so that a * t = v
v_{0}. Then add v_{0} to both sides so
that v = v_{0} + a * t. This
means that the velocity at time t is its initial value plus the
change in velocity during time t. To see the relationship between
the velocity v and the acceleration a, run the Velocity by Algebra display. In this display
the velocity at each time is calculated from the equation we just
developed.



There is an alternative, and powerful, way to get the velocity
from the acceleration. Back in the introduction to this course, in the last few
lines, I introduced the prospect of undoing the work Newton did
in replacing millions of trivial calculations with a few complex
ones. We are going to go back to the commonsense idea that the
amount of stuff present at any time is the amount at a previous
time plus and minus any changes. This notion by the way is the
foundation of that branch of mathematics called differential
equations. Applying this to velocity, the change in velocity
between time t and time t+dt is the average acceleration over
that interval times the duration of the interval dt, where dt is
just the name we call the difference in time. Notice that the
symbol dt plays the same role as the Dt we used when talking about finite time
differences. The dt symbol is used in instances where the
duration of the time interval may become smaller without limit,
approaching zero if required. It is a subtle distinction, but
then mathematicians are like that  subtle.
The way we will handle getting velocity from acceleration
without necessarily knowing the function f in v = f(a), is to
slice time up into equal intervals of width dt and add up the
changes in velocity over each interval between time zero and time
t.
Look at the Velocity by Time
Slicing display to see the velocity curve developed in this
way. Notice that the change in velocity during any time interval
is just the area of that little rectangle of width dt and height
a. The value of velocity at any time is just the sum of the area
of all those little time slices to the left of that time. In
other words velocity at any time is the area under the
acceleration curve up to that time. Of course since we are
working with constant acceleration, the area of each time slice
is exactly equal to dt times the a at any point in the slice.


Just to revisit the idea of a parameter being related to its
rate of change by the area under the rate of change curve, look
at the development of displacement from the area under the
velocity curve. In this case we need to pay more attention to how
we calculate the area. In calculating the area under the
acceleration curve previously there was never a question about
what height to use for our time slices. They were the same height
everywhere. Since the velocity curve is not flat, each time slice
has a low side and a high side. If you use the high side
dimension the calculated area of each time slice will be too
large. If the low side is used the area will be too small. Look
at the next display, Coarse Velocity Time
Slice for an illustration.

Now suppose we increase the number of time slices as
illustrated in the Fine Velocity Time
Slice display. The images above and at the right illustrate
the total error in area under the velocity curve for two dt
values, one large and the other smaller. If you count up the
little triangles stacked in the red boxes you will see in each
instance that the total error is the same as the area of a single
time slice, centered along the taxis. The altitude at the
midpoint of the velocity curve fixes the height of such a time
slice. The width of that time slice is just dt. As we make more
time slices by reducing dt, the error area must decrease. In fact
we can reduce the error between the actual value of v at any t
and the total area under the curve up to that t, to less than any
arbitrarily small value by choosing dt small enough.



Finally take a look at the Really Fine
Time Slice display to see how closely the area under the
velocity curve fits the displacement curve calculated from the
formula we developed. You can see that with 100 time slices, the
error is about as small as the screen resolution. The computer
could easily handle 1000 or 10,000 time slices, which is fine
enough for all practical purposes.
Are there any questions?

Here might be a good place to relax for a few minutes, like
these guys, and look back over the trail that brought us to this
point. Working in 1 dimension, we defined the idea of the
position of a particle as a distance from the origin of a
reference frame. Then we defined velocity as the rate of change
of position with respect to time. Next we defined acceleration as
the rate of change of velocity with respect to time. Armed with
those definitions you will be able to determine average values of
velocity if you have a plot of position vs. time, and average
values of acceleration if you have a plot of velocity vs. time.
By taking smaller and smaller dt, you can make the average values
as close to the instantaneous values as you like. In the limit as
dt approaches zero, the average value approaches the derivative
of the function, dx/dt, which is the slope of the curve.
Frequently it is the position as a function of time that is
unknown and the acceleration that is known. In the case where the
acceleration is constant we derived an expression for the
velocity and position as functions of time using the fact that
the average acceleration and instantaneous acceleration have the
same values. Then we made the case that a general way to get the
value, at time=t, of a variable from its rate of change was to
calculate the area under the rate of change curve from zero to t.
And, that we could calculate that area by summing up the areas of
time slices, reaching any required precision by taking small
enough dt. This last result is of profound importance.



It was the realization by Isaac Newton that the sum of tiny
rectangular slices could be made to approximate the area under
any reasonable curve that led to the invention of integral
calculus. In principle Newton and his contemporaries could carry
out the simple multiplication and summing process for thousands
or millions of little rectangles. As a practical matter however
life is too short to get very far that way using a quill pen and
parchment. So they had to come up with a trick to replace the
millions of trivial calculations with a few complex ones. If they
had had a personal computer, integral calculus might never have
been invented. We have one, so we will use it to bypass most of
the complications presented by calculus.
While we are philosophizing here I will explain one of the
differences between this program and a standard college physics
course. I make a distinction between knowledge and understanding.
Knowledge involves gathering, remembering and organizing facts to
get a result. Understanding involves gathering, remembering and
organizing basic principles that may be used to generate facts in
a wide variety of situations. The more you understand, the less
you need to know.
Most physics courses are centered on exams and exams are
centered on problem solving. There is nothing wrong with that by
the way. You need those skills. But one of the consequences of
this focus on high speed problem solving is that there is a great
temptation to extract certain equations from the explanations,
like x  x_{0} = v_{0} * t + 1/2
* a * t^{2} and commit them to memory. Then at
exam time select one or more of the equations and plug in the
known quantities to get the unknowns. This is an efficient way to
take exams but requires you to know a lot of information.
My purpose here is to strengthen your understanding of this
material. We will look at the information we cover this way and
that, turning it inside out sometimes, and finding links among
the topics so that you develop real understanding. In this
playing with the ideas there are few problems to solve and no
exams but your ability to solve problems and take exams will be
better for having the understanding. Beyond that, in the "real
world", that place you go after college, you will have a much
better shot with this understanding at solving problems no one
has thought of before.

Now in spite of what I said about problem solving, let's
return to the thread of the story with some examples of the use
of the constant acceleration, 1 dimensional motion equations. We
know from experiment that the acceleration due to gravity on a
body is very nearly constant. The value of that constant near the
surface of the Earth is approximately 9.8 meters per second per
second.
In the first case consider simply dropping the ball from a
window 10 meters above the ground. How long does it take to reach
the ground? In solving problems the trick is in translating the
words into quantities about which we know something. It looks
like "dropping", "10 meters" and "reach the ground" are key bits
of information. I interpret "dropping" to mean that v_{0}
is zero. If we let ground level be the origin of our reference
frame, then x_{0} is +10 meters. Since there is no
mention of any other force on the ball, we take gravity to be the
only motive force so acceleration is 9.8 meters per second per
second. The minus sign because the acceleration is directed
toward smaller displacement numbers. Finally "reach the ground"
means that at the time we are solving for, x=0. Next, scan around
for some relationship that relates v_{0}, x_{0},
a, x and t.
The formula x  x_{0} = v_{0}
* t + 1/2 * a * t^{2} seems to fit the bill. If we
just plug in the values we are given, we get 0
10 m = 0 * t s +1/2 * (9.8 m/s^{2}) * t
s^{2}. This reduces to 10 m =
4.9 m/s^{2} * t s^{2}. Notice that we are
still OK with the units. So 10 m/(4.9
m/s^{2}) = t s^{2}, or t = ((10/4.9)
s^{2})^{.5} = 1.43 s to 3 significant figures. We
reject the negative square root as not having any physical
significance.
The process we just went through will work for any initial
conditions of position or velocity, so balls thrown from the
ground up, from a height down or from a height up, where the
question is "how long to reach any condition?", work the same way
just with different numbers. Physics professors are usually not
content to ask questions in such a straightforward fashion. They
like to ask things backwards, or sometimes even sideways.



Consider a ball tossed upward from our 10 meter window with an
initial velocity of +20 meters per second (m/s). To what height
does it rise above the ground, and what is its speed just before
it hits the ground? Again extract the given data from the words.
We get v_{0}=+20 m/s, x_{0}=+10 m and a=9.8
m/s^{2}. Now we have to look in the question for
additional information. There are two parts to the question. In
the first part we want to know "to what height?" In the second
part we want to know "how fast". Two part questions are usually
best worked separately but you may have to use the answer to one
part to get the other part figured out. If the questioner is
particularly nasty, she may ask them in reverse order so you need
to work the second part first.
So what do we know about the ball at the time of
maximum height. Well, we know it isn't going any higher and will
soon be going lower so the magnitude of its velocity must pass
through zero at that instant. This is one of those statements
that is obvious after you hear it but unless you have an
intuition about the physics of a situation, you might get hung up
for a long time trying to figure out that in the first part of
the question, v=0 at the maximum height. Then you have to
recognize that knowing that is worthwhile. The question asks
"what height?", not "how long?". Here is where you need take what
you can get if what you want is not readily available. Any
information is preferable to none.
Since v = v_{0} + a * t we can
now find the time of maximum height, t = (v 
v_{0}) / a = (0  20 m/s) / 9.8 m/s^{2} = 2.04
s. Now we can use x  x_{0} =
v_{0} * t + 1/2 * a * t^{2}, which after
plugging in everything we know gives us x = 10 m
+ 20 m/s * 2.04S 4.9 m/s^{2} *
2.04^{2}S^{2} = 10 m + 40.8 m  20.4 m = 30.4
m.
Now that we know the height from which the ball is falling, we
can use x = 1/2 * a * t^{2} to
get the time of the fall to be t = (2 * x /
a)^{.5} = (2 * (30.4) / (9.8))^{.5} = 2.49
s. Then using v = a * t we can get
the velocity at impact of v = 9.8
m/s^{2} * 2.49 s = 24.2 m/s
I can
sense that you are growing weary of all this algebra in the name
of better understanding. I know I am. Wouldn't it be nice to have
a mathematical model of the constant acceleration problem into
which you could put initial conditions and out of which would
come the answer to any required degree of precision. Well we just
happen to have such a homework engine in our "Physics1" program.
If you are curious you may jump to the Physics1 download site.
You will still have
to practice a lot of problem solving to be successful, but all
that work gets in the way of understanding. In this program we
will be looking to minimize the tedium.
As an alternative to the "Physics1" program, we offer in this
online edition a limited scope version of the constant
acceleration model I mentioned above. In the Free Body Diagram
display you may enter initial position, velocity and
acceleration, and see the results displayed as a plot of position
and velocity vs. time. Then using the cursor, you may get
approximate values for many parameters. This tool will be
available to you after we have a few more lessons under our
belt.
Are there any questions?
In the next section we are going to cover some vector
arithmetic to prepare ourselves to handle more than
onedimensional situations.

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