Please note that I have only a superficial understanding of complex physics; this is simple physics (and in truth, only some equations associated with some simple physics) from someone with a simple understanding. I'd love to hear any advice or corrections from the experts!

The equations for position, velocity, and acceleration are some of the first and most basic taught to any physics student. For me - a dabbler without the time or talent for the complex stuff - they're the most beautiful bridge between thought and reality I've ever seen. Three equations: all related mathematically, expressions of each other; all related conceptually, expressions of the physical reality in which we exist. You can derive any one from any of the others, or from the motion you see around you; you can predict what you'll see by their workings. They're all one and the same - the concepts, the equations, the tiny facet of Being they represent - all wrapped up in a packet of understanding. It's Motion in form of Reason.

Plus, if you ever need to figure out how long it'll take an egg to hit the ground or when a car that's braking slower than the one in front of it is going to crash, nothing serves better!

Here are a few concepts you need to understand in orders to use these equations:

**position:** position is, simply, where an object is along the given axis. This is in units of distance, usually meters (m).

**velocity:** velocity is the rate at which an object's position is changing, i.e. how fast it's moving. Velocity is given in units of distance per unit of time, which is usually one second (s). Thus, the position of an object with a velocity of 5 m/s is changing by 5m every second. You should use the same unit of distance for your velocity values as you use for your position values, or you'll have to do a lot of extra conversion.
**acceleration:** acceleration is the rate at which an object's velocity is changing - basically, how quickly it's speeding up or slowing down. Acceleration is given in units of velocity per unit of time; that is, units of distance per unit of time, per unit of time. So, an object that is speeding up by 5m/s every second has an acceleration of 5m/s^{2}. Again, you should use the same units as you do for position and velocity, or face painful complication.

This, also, is key: remember that all of these equations are scalar, which means we only deal with one axis (or dimension) at a time; however, velocity and acceleration are vectors. That means that they have a magnitude *and* a direction, both which you have to take into account. Imagine it this way: you're thinking about where a car is on a certain road, but it turns left before it starts to drive. It has an acceleration and a velocity, but with respect to your road both are zero. Objects can have positions measured along more than one axis as well, so if you're dealing with things occurring in more than one dimension, you need to use some trigonometry to break all of your vectors down into x, y, and z components, and solve for each component separately. SOH CAH TOA!

Here are the equations:

**Position: **

**S = S**_{0} + v_{0}t + 1/2at^{2}

**S**_{0} is the initial position of the object in question (that is, the object's position when time = 0.)

**v**_{0} is the initial velocity of said object.

**t** is time.

**a** is the object's acceleration.

Now, as you'll recall, velocity is the rate of change of position. If you're mathematically inclined (or just had your calculus drilled into you really, really thoroughly,) you know that means derivative. So, it's no surprise that one equation for velocity is the derivative of the one for position (and thus, that the position equation is the integral of the velocity equation!) Too cool.

**Velocity:**

**v = v**_{0} + at

**v**_{0} is the object's velocity at time 0.

**a** is the object's acceleration.

**t** is, once again, time.

(There's also another, which you can use if you don't have (or don't care) about time: **v**^{2} = v_{0}^{2} + 2aS. If only it were so easy to remove the tyranny of time from the rest of life!)

Let's repeat our logic from above: if you can derive velocity from position because velocity is the rate of change of position, you can do just the same with acceleration. It's the rate of change of velocity, so its equation is the derivative of the velocity equation:

**Acceleration:**

**a = a**

**a** is, of course, acceleration.

Now, that's kind of a let-down! But if you think about it, this tautology is the simple truth that the other equations rest upon, as you get them by integrating it. So, it's not so useless after all, at least conceptually.

What do you do with these equations? Gain a greater understanding of your world - and solve story problems, of course! So, imagine you're 100m off the ground, and you drop an egg - how long will it take to land? Let's find out.

First, you need to pick an equation. Since we're dealing with an issue that involves position, we'd better pick an equation that does too, which leaves only one possibility (S = S_{0} + v_{0}t + 1/2at^{2}.)

Now, let's figure out what we know. In this case, we already know the final position of the egg - ground level. Let's call that 0. So:

**s** = 0.

We also know how high up we are, which is the egg's initial position. So:

**s**_{0} = 100m.

We're just dropping the egg, so:

**v**_{0} = 0.

It'd be easy to put in a value that'd represent throwing the egg, but that would just be unnecessarily destructive, wouldn't it?

Anyway, we also know the acceleration due to gravity, since that's a constant: 9.8 m/s. Now, we're saying that ground level is 0 and counting up as we go up, but the acceleration due to gravity is clearly in the opposite direction (take a look at the excellent Sign Convention article for some good advice on how to decide this sort of thing.) So:

**a** = -9.8m/s^{2} (Note: if we want to be *really* pointy-headed, we could adjust this for air resistance!)

Now, plug everything in and do some solving:

- 0m = 100m + 0m/s
**t**s + 1/2(-9.8m/s^{2})**t**s^{2}

- 0m = 100m + 0m + -4.9m/s
^{2}**t**^{2}s^{2}

- 4.9m/s
^{2}**t**^{2}s^{2} = 100m

**t**^{2}s^{2} = 20.408s^{2} (we divided both sides by 4.9m/s^{2} here)

**t**s = 4.517s.

Notice how the units worked out just right? That's a sign that you did everything correctly. If you get **t** in units of meters, for example, you can be sure something is wrong. Putting the units in the equation complicates an easy math problem, though, and you can always check your answer by plugging the value you get back into the equation and solving it. So, let's do that instead.

- 0 = 100 + 0
**t** + 1/2(-9.8)**t**^{2}

- 0 = 100 + 0 + -4.9
**t**^{2}

- 4.9
**t**^{2} = 100

**t**^{2} = 20.408 (we divided both sides by 4.9 here)

**t** = 4.517.

Now, if we solve 0 = 100 + 0(4.517) + 1/2(-9.8)(4.517)^{2}, we get 0 = .023: we're a little off due to rounding errors, but correct. So now you know that if you drop an egg from 100m up, it'll take almost exactly four and a half seconds to hit. And do you know what? That's always what happens. Now you know why, and understand how it all works. That's what makes this stuff worth knowing, and what makes it beautiful.

Thanks to Kidas for suggesting that I add the second velocity equation, and to stupot, who called me on my inadequate treatment of vectors. Thanks also to Anark who points out that the mathematical relationships between the equations don't have any profound physical significance (like E=MC^{2}, which does say something profound about the universe we live in.) The derivative of the position equation describes velocity because velocity is defined as the rate of change of position, not for any deeper reason. For someone (like me) who likes the whole thing because of its pure "make-senseness," though, that's enough to make it seem cool!

I searched as thoroughly as I could and couldn't find these anywhere else. If they do exist elsewhere in a writeup I couldn't find, please let me know so I can either move this to the right place, or remove it if it's totally redundant. Thanks!