# The use and meaning of the three beginning physics equations

• Jaq202
The concepts are the important thing. (This is another thing that @PhanthomJay was getting at).In summary, the given equations, Vf = V0 + at, Vf^2 = V0^2 + (1/2)at^2, and x = V0*t + (1/2)at^2, only apply for uniform acceleration where the acceleration remains constant. The variables, a, v_0, v_f, x, and t represent acceleration, initial velocity, final velocity, position/displacement, and time, respectively. The variable names can vary depending on the situation. These equations can be derived from the basic premise that acceleration isf

#### Jaq202

The first three equations i was given is as follows:
Vf = V0 + at
Vf^2 = V0^2 + (1/2)at^2
x = V0*t + (1/2)at^2 [why do they sometimes use y instead of x or V0,y?]

I don't understand in what situations i would use them in nor do i understand why you us V=gt or X=Vx*T. (i understand g can is 9.8 m/s.)When and why do you use these equations? What are they supposed to mean?

The first three equations i was given is as follows:
Vf = V0 + at
Vf^2 = V0^2 + (1/2)at^2
x = V0*t + (1/2)at^2 [why do they sometimes use y instead of x or V0,y?]

I don't understand in what situations i would use them in nor do i understand why you us V=gt or X=Vx*T. (i understand g can is 9.8 m/s.)When and why do you use these equations? What are they supposed to mean?
hi jaq and welcome to PF. One of your equations is wrong. But beyond that, it is most unfortunate that you were ‘given’ these equations without a knowledge of how they are derived and applied. Without that knowledge, the equations are meaningless.

• PeroK
hi jaq and welcome to PF. One of your equations is wrong. But beyond that, it is most unfortunate that you were ‘given’ these equations without a knowledge of how they are derived and applied. Without that knowledge, the equations are meaningless.
Thank you, but might i ask for assistance then? I need a clear understanding of them before i can really begin any real work relying on these equations. As @PhanthomJay mentioned, one of your equations is wrong. I'll leave it up to you and your notes to determine which one.

Here is some brief help to get you started.

These equations only apply for uniform acceleration. What does "uniform acceleration" mean? It means that once acceleration starts, the acceleration remains constant until the time that the acceleration ends. Those equations only apply for that interval where the body is accelerating (although the acceleration can be 0), and where the acceleration doesn't change. It means that the acceleration, represented by the variable $a$, is constant.

$a$: This represents the acceleration. In order for these, particular equations to hold true, $a$ must be uniform (i.e., constant).
$v_0$: Initial velocity.
$v_f$: Final velocity.
$x$: Position. Also, you can call this "displacement."
$t$: time.

why do they sometimes use y instead of x or V0,y

None of the variable names are set in stone. If the body is moving along the y-axis, then you might as well call the variable for displacement "y". Similarity, if you'd rather call the initial velocity $v_i$ instead of $v_0$, then that's fine too.

It doesn't really matter what the variable names are. The concepts are the important thing. (This is another thing that @PhanthomJay was getting at).

I'm not going to derive the equations for you here. But if you wanted to derive them yourself (or follow someone else's derivation), the following relationships are crucial:

Velocity is the rate of change of displacement.
Acceleration is the rate of change of velocity.

• Jaq202, PeroK and Delta2
Oh, and just to be clear, in the real-world, acceleration can and does change in many situations. (I'm not saying that non-uniform acceleration is impossible.)

My point is that if the acceleration is changing, then those particular equations don't apply; they only apply the special case where the acceleration is uniform.

You can solve systems undergoing non-uniform acceleration, but you'll need different equations, and [almost certainly] calculus.

• Jaq202
Here’s another helpful site, using algebra derivations, for motion in one dimension with constant acceleration, all starting from the basic premise that acceleration = change in velocity divided by change in time. So for example if a car starts from rest and accelerates uniformly to a velocity of 30 m/sec in 10 seconds, it’s acceleration is 30/10 or 3 m/sec^2. Watch units!

https://physics.info/motion-equations/

• Jaq202 As @PhanthomJay mentioned, one of your equations is wrong. I'll leave it up to you and your notes to determine which one.

Here is some brief help to get you started.

These equations only apply for uniform acceleration. What does "uniform acceleration" mean? It means that once acceleration starts, the acceleration remains constant until the time that the acceleration ends. Those equations only apply for that interval where the body is accelerating (although the acceleration can be 0), and where the acceleration doesn't change. It means that the acceleration, represented by the variable $a$, is constant.

$a$: This represents the acceleration. In order for these, particular equations to hold true, $a$ must be uniform (i.e., constant).
$v_0$: Initial velocity.
$v_f$: Final velocity.
$x$: Position. Also, you can call this "displacement."
$t$: time.

None of the variable names are set in stone. If the body is moving along the y-axis, then you might as well call the variable for displacement "y". Similarity, if you'd rather call the initial velocity $v_i$ instead of $v_0$, then that's fine too.

It doesn't really matter what the variable names are. The concepts are the important thing. (This is another thing that @PhanthomJay was getting at).

I'm not going to derive the equations for you here. But if you wanted to derive them yourself (or follow someone else's derivation), the following relationships are crucial:

Velocity is the rate of change of displacement.
Acceleration is the rate of change of velocity.
Thank you so much! This helped a lot!

• collinsmark