A study of Variable Acceleration by Miles Mathis

A STUDY OF
VARIABLE ACCELERATION

by Miles Mathis

First posted December 20, 2009

Abstract: I will analyze a textbook solution of variable acceleration, showing that it is incorrect in both method and answer. It is incorrect because it is solved improperly with integration when it would be easier and more transparent to solve with the second and third derivatives. Once again, I will show that the calculus is taught upside down.

[Apparently many readers have been mystified by this paper. They do not comprehend my method, and assume I wrote it standing on my head. But I stand by it. In my opinion, it is far easier and far more transparent to solve for distance using the third derivative here than by integrating. I honestly believe that if you penetrate this paper, some important scales will fall from your eyes.]

In previous papers I have shown many problems with the modern calculus. In this paper I will show problems in applying the calculus to variable acceleration. To do this, I will follow a physics textbook solution line for line.

To start, we must ask what we mean by a variable acceleration. It could mean two things. One, it could mean that we were speeding up and slowing down, so that our change in velocity was not constant. That is not what I mean here. What I mean is an acceleration represented by a power of 3 or more, as in the curve equation x = t³. That means that you take a constant acceleration and then accelerate it. For example, you take your car out on the highway and press down on the gas at a constant rate. If your foot and engine work like they should, you will have created a constant acceleration. Now, take that whole stretch of highway, suck it up into a huge alien spacecraft, and accelerate the spacecraft out of orbit, in the same direction the car is going. The motion of the car relative to the earth or to space is now the compound of two separate accelerations, both of which are represented by t². So the total acceleration would be constant, not variable, but it would be represented by t⁴. This is what I am calling a “variable acceleration” here. It is not really variable, it is just a higher order of change.

The acceleration would be represented by t³ if the alien spacecraft had a constant velocity instead of a constant acceleration. An acceleration is two velocities over one interval, so t³ is three velocities over one interval. Or, it is three changes in x over one defined interval, say one second. We can write that as either three x's or three t's, but it is common usage to use three t's in the denominator instead of three x's in the numerator.

The cubed acceleration can also be created in a car, by increasing your pressure on the gas pedal at a constant rate of increase. This will cause a cubed acceleration in the first few seconds.

In engineering, a higher order acceleration like this is called a “jerk” (though it is usually applied to a negative acceleration, as in a jerk to a stop). I call the positive acceleration a cition in my first paper on the derivative, from the Latin “citius”. As in the Olympics motto “citius, altius, fortius”: faster, higher, stronger.

Because this sort of acceleration is often called a variable acceleration in physics textbooks, most people seem to think it isn't constant, and therefore can't be averaged like the squared variable. But higher powers can be constant, if they are created by a constant process like the one I proposed above with the car and the spacecraft. If the car and spacecraft are both accelerating at a constant rate, the higher power total acceleration will also be constant. Just because an acceleration has a power greater than two does not mean it isn't constant. We will see how important this is below.

When I say that the acceleration is constant, I do not mean that we can average the velocity. Yes, the velocity is accelerating itself, so we cannot find the velocity at a given time by averaging. We have to take the second derivative, not the first. When I say the acceleration is constant, I mean that it increases at a consistent rate. It is not fluctuating.

Now let us look at all the problems encountered by modern mathematicians in trying to analyze this situation. In physics textbooks, the chapter on velocity and acceleration normally comes very early. In my textbook¹, it comes in chapter 2. You don't need calculus for constant velocity, but for “instantaneous velocity” you do, so we get an entire subsection for that. To begin, we get a graph plotting x against t and are given a curve (but no curve equation).

In the next section, constant acceleration is covered, and we are given a graph that plots v against t, with a similar curve. And in the section after that, we find “variable” acceleration. We are given a graph that again plots v against t, with a curve.

We should already have several questions. Since we are measuring the curvature of these curves with the graph, and finding tangents and areas beneath them, shouldn't our methods be analogous as we go from velocity to acceleration to variable acceleration? In other words, if we plot x against t in the first graph, shouldn't we plot x against t in all the graphs? Or, by another method, we would plot x against t in the first graph, v against t in the second, and a against t in the third. That would keep our method even and unchanged as we moved from one rate of change to the next.

Instead, we find the textbook plotting v against t when solving for a variable acceleration. This is not a quibble: it must be important, because the curve determines the tangent and the area under the curve. If you have a different curve, the tangent and the area are different, too. Well, plotting x against t will not give you the same curvature as plotting v against t or “instantaneous a” against t, will it? If we are going to differentiate or integrate, shouldn't we be careful to get the right curve?

Another problem. All textbooks I have seen solve problems of variable acceleration with integration. But that is upside down. When solving with respect to t, you should differentiate down and integrate up. In other words, if you are given an acceleration and you want to find a velocity, you differentiate. The derivative of t² is 2t, where t² is the acceleration and 2t is the velocity. Conversely, to go from a velocity to an acceleration, you integrate. The integral of 2t is t². But in textbooks, they integrate from a velocity graph, and find a velocity from a variable acceleration by integrating only once. Since a velocity is two steps from a variable acceleration, they should be seeking the second derivative, not the first integral.

To be even more specific, let me quote from the textbook:

If x is given by x = At³ + Bt, then v = dx/dt = 3At² + B. . .Then, since a = dv/dt, a = 6At.

What the authors are doing here is preparing you to integrate. They are showing you how differentiating works, and then preparing you to reverse it in the upcoming problem. You probably don't see anything wrong there, but as I pointed out in my paper on the exponential derivative, modern calculus is a jumble. What the textbook has done is take the first derivative and then the second, as you see, but they have called the first derivative of a variable acceleration a velocity and the second derivative a constant acceleration. That is backward. Just look at the equations: v = 3At² + B? Since when can you write a velocity as a squared variable? Or, a = 6At? Isn't 6t a straight line on a graph? That isn't an acceleration.

My initial reaction was that this textbook author is just a nut, but by looking around me I have found that all of calculus now “works” this way. According to current wisdom, velocity is always supposed to be the derivative of distance, and acceleration the derivative of velocity, and that is what causes this horrible confusion. As I showed in previous papers, Wikipedia and most modern sources define the derivative like this, which is enough to raise Newton from the grave. He would tell you that when you are looking at the time variable, velocity is the derivative of acceleration, not the reverse. When manipulating t, you differentiate down and integrate up. When applying the calculus with respect to t, you differentiate down and integrate up.

Some will not see my point. They will think I am the nut. They will say, “What in the devil are you talking about? The first equation x = At³ + Bt is the distance. That is what the book means by 'x is given by'. That is what 'x equals' means, you dope!” But no, I am not the dope, so pay attention here. This is where it all comes out in the wash. The standard model and standard reply is wrong again, since the equation x = At³ + Bt is the curve equation on the graph, and it represents a variable acceleration. That equation is not x, that equation is the variable acceleration. You have been fooled by the “x=”.

Bear with me, please. Look closely at the equation x = At³ + Bt. The physical displacement x is not given by that equation. That equation applies to the graph only. It is telling you an x-distance from the y-axis at time t. The x in that equation tells you what x you are at, at the given value of t, but it does not tell you the distance traveled on the curve, since the curve is curving. To say it another way, x in a curve equation does not equal x in a physics equation. So x = At³ + Bt will not tell you a value for total distance traveled after t. If it did, we wouldn't need calculus at all, we could just read the value for x right off the graph, for any and all curves. But no, to find x you have to use physics equations, not curve equations. The equation x = At³ + Bt is a curve equation, and because it has a t³ in it, it must stand for a variable acceleration.

[Clarification, June 2015. I have finally understood that my claiming that distance is the derivative of velocity is what has shut most readers down when trying to understand this paper. We are all taught that velocity is the rate of change of distance, and in other papers I even admit that is true. So they don't understand what I am saying here, by turning that upside down. They think I am just crazy, or at least wildly inconsistent. But what I thought was clear from the beginning is that I am trying to show that when you are doing your calculus operations on the variable t, you actually differentiate down from velocity to distance. You take the derivative of t in the velocity equation to find the distance—which means that in this case, the distance is derivative of the velocity. If you follow my actual equations, you will see exactly what I mean, and why I am right.

What I didn't originally make clear is why this is so. It turns out it is because t is normally in the denominator. Being in the denominator in all equations of motion, t naturally acts upside-down to x as a matter of differentiation or integration. Not realizing this, the mainstream has butchered many of these manipulations when they are monitoring t. Any time they are finding anything “with respect to t” they are failing to take this into consideration.]

The entire modern interpretation of the calculus is upside down in this regard! To show this, let us look at the textbook solution of a specific problem:

Given a = 7m/s³ and 2s, find v final from rest.

v = ∫(7m/s³)tdt = (7m/s³)t²/2 = (3.5m/s³)t² = 14m/s.

That solution looks like a fudge to me, from the start. If the moderns don't understand the foundations of the calculus, or how it works, it is unlikely they will be able to apply it in a logical and correct manner. In fact, the solution can't be right, because in the math they have taken one integral of time, to convert a variable acceleration to a velocity. Since the velocity is two steps of differentiation or integration away from the variable acceleration, in the differential table³ or in real life, that solution can work only by some sort of accident or push or other miracle. But I will show that it doesn't work. That solution is not correct.

But first let us see why the textbook is integrating. We only have to look at its own explanation [this follows the quote above, explaining differentiation]:

The reverse process is also possible. If we are given the acceleration as a function of time, we can determine v as a function of time; and given v as a function of time, we can obtain the displacement, x.

The textbook integrates because it believes it is reversing the above process. But because I have just shown their first process was upside down, this can't work. They thought they were going x→v→a with differentiation, so now they think they are going a→v→x with integration. But regarding the variable t, differentiation is the process a→v→x. Differentiation goes down, and integration goes up. They are trying to differentiate up and integrate down, when with t you have to do the reverse.

One more time, for good measure. We are given a curve equation, say y = x³. That is a curve equation, so it must stand for a curve. It does not stand for the point y or the distance y, since a point or distance y cannot curve. The only “y” it gives us is some vertical distance from the horizontal axis at some value of x. But that is not the solution for the distance traveled along the curve. Therefore it is not the solution to any physics problem. The equation y = x³ is not telling us a displacement, given an acceleration. It is telling us the acceleration.

Now let us solve the problem, without using integrals. We will start with the velocity. As I said, we need to find the second derivative, since velocity is the second derivative of a variable acceleration. The second derivative of t³ is 6t, so while the time is changing by the cube, the velocity will be changing by 6's. You can see this clearly by taking the lines out of my differential tables:

Δx³ = 1, 8, 27, 64, 125, 216, 343
ΔΔx³ = 1, 7, 19, 37, 61, 91, 127
ΔΔΔx³ = 6, 12, 18, 24, 30, 36, 42
ΔΔΔΔx³ = 6, 6, 6, 6, 6, 6

The first line is the cubed acceleration, the second is the first rate of change of that acceleration, and the third is the second rate. The second line is (a sort of) first derivative of the first line, and the third line is (a sort of) second derivative. We are straightening out the curve. So the third line gives us a velocity. You can see that it is changing the same amount in between numbers. The differential is constant. That is the definition of a velocity. You can see that the velocity is changing by 6's. Its rate of change is 6. In our current problem, its rate of change is 6t, and t is 2, so at t=2, its rate is 12. Again, you can see that right from the table. The second entry is 12. But we have an acceleration of 7, not 1, so we multiply by 7 and divide by 2 (to take into account the first halved interval). This gives us v = 42m/s.

v = ad²(t³)/2 = 3at

That is the new equation for velocity, given a cubed acceleration. This is logical since we can derive the current equation for normal (squared) acceleration in the same way. The current equation is v = at. Current textbooks don't derive that equation with calculus, they just take it as given or derive it from the classical equation a = v/t. But we can now expand it showing the derivative:

v = ad(t²)/2 = at

Since the derivative of t² is 2t, we get the current equation. This means we can intuit the velocity equation for an acceleration of t⁴ as

v = ad³(t⁴)/2 = 24at/2 = 12at

And v = ad⁴(t⁵)/2 = 120at/2 = 60at

I have shown a simple method for taking higher order acceleration equations straight from my table of differentials. No one has ever done this before, that I know of. It is certainly not done presently, because, as I showed you, current textbooks solve with integration.

Let us look at the textbook's solution. They found v=14m/s, remember? I found 42m/s. I bet you think they are right and I am wrong, but I can prove they are wrong very easily. An acceleration of 7m/s³ must be greater than an acceleration of 7m/s², right? A cubed acceleration is the motion of an acceleration, so the distance traveled has to be greater. So let us solve the same problem for a = 7m/s² instead of 7m/s³. Using current equations for constant acceleration, we find

v = at = 14m/s.

They found the same final velocity for 7m/s³ and 7m/s². That is impossible. An object accelerated to a cube must be going faster at all t's than an object accelerated to a square. That much is clear to anyone, I hope. So the textbook solution is a blatant fudge, one that doesn't even get the right answer.

We can also use the differential table to find the distance here. But first let use my velocity 42m/s to find a solution. Because our acceleration is constant (or consistent), we can tweek the old equations.

x = v_ft/2 = (3at)t/2 = 3at²/2 = 42m.

Going to the table, we see that the object is moving 6 during each interval of 1. That is what 6, 6, 6, 6 means. Since our acceleration is 7, we just multiply. In doing this, we are using the third derivative, like so:

x = a(d³t³)t²/4 = 42m

To find a distance from a cubed acceleration, we take the third derivative. We differentiate down three times.

Let me clarify that. Some have not understood what I am doing here. They have complained that I am treating the acceleration as a motion constant and thereby trying to average the velocity or distance over the elapsed time. That is not what is happening. When I say that the object is moving 6 during each interval, I should say subinterval. I do not mean that the object is traveling 6 during each 1/7 of a second or something, the same distance over each equal time. No, my third derivative is telling us that the object is moving 6 for each constituent velocity, and a cubed acceleration is made up of three of those. You really have to study the tables to see what I am doing, and no one has done that in centuries. The calculus hasn't been taught like that, so my simple manipulations seem mysterious.

Let us see what the textbook got:

To get the displacement, we use x₂ = ∫ v(t)dt with v₁ = 0, v₂ = 14m/s, and t₂ = 2s.

x₂ = ∫(3.5m/s³)t²dt = (3.5m/s³)t³/3 = 9.33m

Again, let us check that by comparing it to the solution for the distance traveled after 2 seconds at 7m/s². Everyone agrees that the equation x = at²/2 works for constant acceleration, so we find 14m. The textbook found a number less than that, therefore the textbook cannot possibly be correct. A cubed acceleration must give us more displacement after any amount of time than a squared acceleration.

I have just proven that the textbook solution is a fudge, in both method and answer.

Conclusion: I have shown in a direct manner that modern physicists and mathematicians do not know how to use the calculus. Anytime you see a scientist integrating down from accelerations to distances in this way, you know that madness is afoot. In other papers I have shown that the calculus is misdefined, and now I have shown that it is misused, even in simple problems. Nor is this an isolated incident, since I have shown that Wikpedia, as the mouthpiece of common wisdom, defines the derivative up instead of down. Students are currently taught to differentiate a distance to get a velocity and differentiate a velocity to get an acceleration, when that is upside down. Since all are equations of motion, motion is defined by time, and the calculus is normally applied to the time variable (with respect to t), we have to reverse that process. In most real operations, we must differentiate a velocity to find a distance.

Given this fundamental misunderstanding, we can now see why scientists and mathematicians hide away in esoteric problems and esoteric maths. They can't do simple math, either algebra or basic calculus, so they must take cover under slippery operators in slippery fields. If you had thought that the math in places like Physical Review Letters was a big con game, you are right. Most math is a con game, and that includes the simple maths you were taught in high school and college. If the math in chapter 1 and 2 is false, you know the math in chapter 30 is false.

[For more on this, you may now read my newest paper called Calculus is Corrupt, where I show a major fudge first used by Lagrange in the 1700's. It is directly related to the upside-down calculus.]

¹General Physics, Douglas C. Giancoli, 1984.
³http://milesmathis.com/are.html

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