# New here, need some understanding about acceleration.

1. Aug 27, 2009

### staetualex

Hy folks! I am new in here.Anyone can give a basic intuition about why the acceleration formula actually works? I mean, my brain is not capable to "nerd-ize" formulas, i either forgot them , i either on a test write the coulomb formula something like (q1 q2 r^2)/2pi eps0. I am shure if i remember now the (d^2 x)/(dt^2) in a couple of months it will be something like dx^2/dv or whatsoever . Is anyone who can describe this formula, how it holds in a real example, some associations with other processes, etc. what every number/letter in that formula represent? I mean change in time square, what ? thanks and have a good day!

2. Aug 27, 2009

### PhaseShifter

It's a matter of learning to translate from math to English (and back).

x is position, t is time.

$$v={{dx}\over{dt}}$$ or "velocity is the rate of change of the position"

$$a={{dv}\over{dt}}$$ or "acceleration is the rate of change of the velocity"

Now that we're done defining terms, let's look at what that means:

Since v is the first derivative of x, and a is the first derivative of v, then a can be written as a second derivative of x.
Since both first derivatives are taken with respect to t,

$$a={{dv}\over{dt}}={{d^{2}x}\over{dt^{2}}}$$

I don't think I can stress enough just how much it helps to understand the math--I took a course once where I had to memorize 37 equations for the first test alone. Needless to say, it didn't go well. But on a second attempt at the course, with a better understanding of the math involved, I only had to memorize four or five equations and could derive all of the others in about a minute total.

Last edited: Aug 27, 2009
3. Aug 27, 2009

### kyiydnlm

hi, staetualex, take a little time on chain rules.
It's not dx^2/dv; it's vdv/dx.

4. Aug 27, 2009

### Danger

I can't help you, but I thank you for adding a cool new word to my vocabulary. :rofl:

5. Aug 27, 2009

### PhaseShifter

First, look at the simplest possible case (constant acceleration).

If a is constant, then the solution for velocity looks like v=at+k.

When t=0, we know that at=0. This is pretty obvious so far, but it does tell us that k is the velocity at that time. Since it's the initial velocity of the object, I'll change the name of that variable to something that lets me remember what it represents.

$$v=at+v_{i}$$

Now if we go back to the definition of velocity, we have $${{dx}\over{dt}}=at+v_{i}$$. We integrate again to solve for position, and get the following:

$$x={{{at^{2}}\over{2}}+v_{i}t+k$$

Again we note the first two terms are zero, when t=0, so this time k must be equal to the initial position. So once again, we rename that variable to something easier to remember.

$$x={{{at^{2}}\over{2}}+v_{i}t+x_{i}$$

Last edited: Aug 28, 2009
6. Aug 27, 2009

### staetualex

well, i know, i was just predicting my future, in case i mechanically learn the formulas, not knowing the principle it works. I may forgot them, or scramble the terms involved. Others who gave a reply on post, thank you, you helped me to understand better the schrodinger equation proof.

Phase shifter, you are great, now i'm not dependent of the formula anymore, i can derive it by myself.

to danger, i am glad i can help , sometimes i can invent new words .