In order to be accurate, the title of this section should be "One Dimensional Equations of Motion for Constant Acceleration". Given that such a title would be a stylistic nightmare, let me begin this section with the following qualification. The equations of motion are valid only when acceleration is constant and motion is constrained to a straight line.

Given that we live in a three dimensional universe in which the only constant is change, you may be tempted to dismiss this section outright. It would be correct to say that no object has ever traveled in a straight line with constant acceleration anywhere in the universe at any time — not today, not yesterday, not tomorrow, not five billion years ago, not thirty billion years in the future, never. This I can say with absolute metaphysical certainty.

So what good is this section then? Well, in many instances, it is useful to assume that an object did or will travel along a path that is essentially straight and with an acceleration that is nearly constant. That is, any deviation from the ideal motion can be essentially ignored. Motion along a curved path may also be effectively one-dimensional if there is only one degree of freedom for the objects involved. A road might twist and turn and explore all sorts of directions, but the cars driving on it have only one degree of freedom — the freedom to drive in one direction or the opposite direction. (You can't drive diagonally on a road and hope to stay on it for very long.) In this regard, it is not unlike motion restricted to a straight line. Approximating real situations with models based on ideal situations is not considered cheating. This is the way things get done in physics. It is such a useful technique that we will use it over and over again.

Our goal in this section, is to derive new equations that can be used to describe the motion of an object in terms of its three kinematic variables: velocity, displacement, and time. There are three ways to pair them up: velocity-time, displacement-time, and velocity-displacement. In this order, they are also often called the first, second, and third equations of motion, but there is no compelling reason to learn these names. Since we are dealing with motion in a straight line, the symbol *x* will be used for displacement. Direction will be indicated by the sign (positive quantities point in +*x* direction, while negative quantities point in the −*x* direction). Determining which direction is positive and which negative is entirely arbitrary. The laws of physics are isotropic; that is, they are independent of the orientation of the coordinate system. As long as you are consistent, it doesn't matter. Some problems are easier to understand and solve, however, when one direction is chosen positive over another.

Calculus is an advanced math topic, but it makes deriving two of the three equations of motion much simpler. By definition, acceleration is the first derivative of velocity with respect to time. Take the operation in that definition and reverse it. Instead of differentiating velocity to find acceleration, integrate acceleration to find velocity. This gives us the velocity–time equation. If we assume acceleration is constant, we get the so-called first equation of motion [1].

a = |
dv |
||||

dt |
|||||

dv = |
a dt |
||||

v |
Δt |
||||

⌠ ⌡ |
dv = |
⌠ ⌡ |
a dt |
||

v_{0} |
0 | ||||

v − v_{0} = |
aΔt |
||||

v = |
v_{0} + aΔt |
[1] |

Again by definition, velocity is the first derivative of displacement with respect to time. Reverse the operation in the definition. Instead of differentiating displacement to find velocity, integrate velocity to find displacement. This gives us the displacement–time equation for constant acceleration, also known as the second equation of motion [2].

v = |
dx |
|||||

dt |
||||||

dx = |
v dt = (v_{0} + at) dt |
|||||

x |
Δt |
|||||

⌠ ⌡ |
dx = |
⌠ ⌡ |
(v_{0} + at) dt |
|||

x_{0} |
0 | |||||

x − x_{0} = |
v_{0}Δt + ½aΔt^{2} |
|||||

x = |
x_{0} + v_{0}Δt + ½aΔt^{2} |
[2] |

Unlike the first and second equations of motion, there is no obvious way to derive the third equation of motion (the one that relates velocity to displacement) using calculus. We can't just reverse engineer the definitions. We need to play a rather sophisticated trick.

The first equation of motion relates velocity to time. We essentially derived it from this derivative…

dv |
= a |

dt |

The second equation of motion relates displacement to time. It came from this derivative…

d^{2}x |
= a |

dt^{2} |

The third equation of motion relates velocity to displacement. By logical extension, it should come from a derivative that looks like this…

dv |
= ?? |

dx |

But what does this equal? Well nothing by definition, but like all quantities it does equal itself. It also equals itself multiplied by 1. We'll use a special version of 1, *dt*/*dt*, and then do a bit of special algebra — algebra with infinitesimals. Look what happens when we do this. We get a derivative equal to acceleration and another equal to the inverse of velocity.

dv |
= | dv |
· | dt |
= | dv |
· | dt |
= a |
1 |

dx |
dx |
dt |
dt |
dx |
v |

Next step, separation of variables. Get things that are similar together and integrate them. Here's what we get when acceleration is constant…

dv |
= | a |
1 | ||||||

dx |
v |
||||||||

v dv |
= | a dx |
|||||||

v |
x |
||||||||

⌠ ⌡ |
v dv |
= | ⌠ ⌡ |
a dx |
|||||

v_{0} |
x_{0} |
||||||||

½(v^{2} − v_{0}^{2}) |
= | a(x − x_{0}) |
|||||||

v^{2} |
= | v_{0}^{2} + 2a(x − x_{0}) |
[3] |

Certainly a clever solution, and it wasn't all that more difficult than the first two derivations. However…. It really only worked because acceleration was constant — constant in time and constant in space. If acceleration varied in any way, this method would be uncomfortably difficult. We'd be back to using algebra just to save our sanity. Not that there's anything wrong with that. Algebra works (and sanity is worth saving).

v = |
v_{0} + aΔt |
[1] | |

+ | |||

x = |
x_{0} + v_{0}Δt + ½aΔt^{2} |
[2] | |

= | |||

v^{2} = |
v_{0}^{2} + 2a(x − x_{0}) |
[3] |