Constant Acceleration in 1 Dimension

In our previous discussion of 1 dimensional 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 which 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 balls, thrown rocks, artillery shells... that sort of thing.

With acceleration being a constant, the average value which is given by

= (v_{f} - v_{i}) / (t_{f} - t_{i}) ,

is the same as the instantaneous value. The subscript 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 which 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 that 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 vv = 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 from Constant Acceleration
display.
Look at the Velocity from Time Slices display to see the velocity curve developed as the area under the acceleration curve.

Next we will work on the displacement of our particle undergoing constant acceleration. Because the acceleration is constant, the average velocity over the interval of time between 0 and t is

= (v_{0} + v) / 2

using the normal averaging formula. If the rate of change of
velocity were not uniform, ie acceleration a constant, the
average would have to be some sort of weighted average. We could
not get it from just the initial and current velocities.
If the position of the particle is x_{0} at t=0 then
the position x at time t is given by

x = x_{0} + * t .

This follows from the definition of an average velocity. But is 1/2*(vx = x_{0} + 1/2 * (v_{0} + v) * t .

Or, if we choose the origin of our reference frame such that x0=0, x = 1/2 * (v_{0} + v) * t = 1/2 *
v_{0} * t + 1/2 * v * t

We may substitute for v in the preceding equation,
v = v_{0} + a * t ,

giving this result
x = 1/2 * v_{0} * t + 1/2 *
v_{0} * t + 1/2 * a * t^{2} .

Or
x = v_{0} * t + 1/2 * a *
t^{2} .

Now we may add the displacement x to our graph of the parameters of the motion of a particle under constant acceleration. Look at the Displacement From Constant Acceleration display to see this.

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.

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.

From the examples shown here we might reasonably conclude that we could reduce the error in calculating any variable from its rate of change, as small as we wish by taking smaller and smaller time slices. And that turns out to be the case.

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 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 which 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, rembering and organizing basic principles which 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 get0 -10M = 0 * tS +1/2 * (-9.8M/S^{2})
* tS^{2} .

This reduces to
-10M = -4.9M/S^{2} * tS^{2}
.

Notice that we are still OK with the units. So
-10M/(-4.9M/S^{2}) = tS^{2} , or t = ((10/4.9)S^{2})^{.5}
= 1.43S

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 (MPS). 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}=+20MPS, x_{0}=+10M and
a=-9.8M/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 second 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 - 20MPS) /
-9.8M/S^{2} = 2.04S .

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 = 10M + 20M/S * 2.04S -4.9M/S^{2} *
2.04^{2}S^{2} = 10M + 40.8M - 20.4M = 30.4M
.

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} = 2.49S .

Then using
v = a * t

we can get the velocity at impact of
v = -9.8M/S^{2} * 2.49S =
-24.2M/S

I can sense that you are growing weary of all this algebra in the name of better understanding. I know I am.

In the next section we are going to cover some vector arithmetic to prepare ourselves to handle more than 1 dimensional situations.