I ask mainly because I am not familiar enough with newtonian mechanics and higher-level physics in general to know the repercussions of such a change, but on the simpler plane of existence, I have been given to understand that data is lost about an object's motion when you go from velocity to acceleration, due to something I will learn when I formally take Calculus (of which I have a fairly limited knowledge). For instance, in a velocity vs. time graph, one can observe the object's acceleration (a term which I would redefine in my system), its velocity, and its direction, as well as jerk and so on. Taking the first derivative gives the current definition of acceleration, which only shows acceleration (as it is currently defined, the rate of change of velocity) and jerk, etc. However, the sign of the acceleration is of nearly no consequence. In my system, the sign of the acceleration would indicate acceleration (in my terms, the rate of increase of velocity) or deceleration (a term not commonly used by physicists). Basically, I would like for acceleration to mean more than it does currently, and I am wondering if this would negatively affect high-level derivations of simple physics to the degree that it is simply incorrect.

  • $\begingroup$ Jerk is the second derivative of $v$, $\frac{d^2v}{dt^2}$. The sign of $a$ is highly significant: it determines whether it's acceleration or deceleration. $\endgroup$ – Gert Sep 8 '16 at 1:22
  • $\begingroup$ @Gert I am sorry if I am ignorant to the subject matter, and I may well just be misinformed, but this is why I disagree: acceleration is the first derivative of velocity, so if you have a negative velocity with a negative slope, although the object is accelerating (I think, at least I mean it is getting faster), the sign of acceleration would be negative. Perhaps I just don't understand the nature of acceleration, so just to clarify: if an object has negative velocity and is getting faster, is that deceleration? $\endgroup$ – Azorack Sep 8 '16 at 1:32
  • $\begingroup$ Deceleration means velocity is decreasing. Negative $a$ can also mean acceleration in the $-x$ direction though. Whichever way, the sign always matters, except in the simplest of calculations. Really the acceleration vector is $\vec{a}=\frac{d\vec{v}}{dt}$: the first derivative of the velocity vector. $\endgroup$ – Gert Sep 8 '16 at 1:42
  • $\begingroup$ @Gert Ok, I agree with everything you have said, but what I really am trying to understand is what is the significance of the sign of acceleration? $\endgroup$ – Azorack Sep 8 '16 at 1:51
  • $\begingroup$ Consider $a$ a vector. In a conventional vector space, $\vec{a}=\vec{a_x}+\vec{a_y}+\vec{a_z}$. Whether a component $(x,y,z)$ is positive or negative depends on what direction the component vector points to. Believe me, the current concept of $a$ is more than adequate to cover all aspects of Dynamics. $\endgroup$ – Gert Sep 8 '16 at 1:59

Conventions are typically used because the larger body of people found them to be the most effective way to express things. Acceleration is the first derivative of velocity because people have found that that's the most useful value to capture and give a name!

The big issue with your approach appears when you consider systems that are more than 1 dimensional. Accelerations and velocities are not always in the same direction. They may be at right angles, or at 45 degree angles, or any other direction.

This shows up quickly when you try to adapt F=ma to your system. This means that the force must also be in the direction of motion, but it should be very clear that forces are not always in the direction of motion.

More issues will crop up as you learn more calculus based physics. The equations of motion are quite simple when defined in the way people generally define them. Your system would add many extra layers of complication to those equations. You'll have to take my word for it, until you learn more calculus based physics, but hopefully the directional issue will be enough to convince you to use the definitions everyone else uses.

| cite | improve this answer | |
  • $\begingroup$ Yes, I imagined that force would create a problem, haha. This was more of a notion based on one problem I am consistently having, which is that no matter who I ask or where I search, I have yet to find the significance of the sign of acceleration. $\endgroup$ – Azorack Sep 8 '16 at 1:53
  • 1
    $\begingroup$ @Azorack The significance of the sign is defined by the coordinate system, exactly how the significance of the sign of velocity is defined by the coordinate system. However, if you think in terms of vectors, you no longer think in terms of signs, but magnitudes and directions. If you think in that respect, you can see that the direction is important as...well... the direction! When you get to vector based physics, you'll find all of those operators where sign mattered turn into vector operators which care about direction. $\endgroup$ – Cort Ammon Sep 8 '16 at 1:57
  • $\begingroup$ For example, you'll learn that work is positive if the force is in the direction of work, and negative if it is in the opposite direction. That can be rather unintuitive -- why do we have negative work?! However, if you later learn the vector based version of that equation, work is the "dot product" of Force and the movement done to the object. Once you know it's a dot product, you can go back to the definitions of dot product, and you'll see there's a very meaningful cosine term that captures the reason why work can be positive or negative. $\endgroup$ – Cort Ammon Sep 8 '16 at 1:59
  • $\begingroup$ Very interesting. I have often found myself wishing that physics, math, and sciences in general were not taught in the progression as they have been for so long, and instead taught to eliminate the "unintuitive nature" (until you get to quantum mechanics, where you must say oh well) and the "you will understand this later". $\endgroup$ – Azorack Sep 8 '16 at 2:04
  • $\begingroup$ "you will understand this later". Most kids would struggle with the vector concept. Hence young people are taught the 1D scalar version. Later, making the leap to 3D vector space isn't that hard. Worked for me, in any case. $\endgroup$ – Gert Sep 8 '16 at 2:07

Not the answer you're looking for? Browse other questions tagged or ask your own question.