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New here, need some understanding about acceleration.

  1. Aug 27, 2009 #1
    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 :confused:? thanks and have a good day!
  2. jcsd
  3. Aug 27, 2009 #2
    It's a matter of learning to translate from math to English (and back).

    x is position, t is time.

    [tex]v={{dx}\over{dt}}[/tex] or "velocity is the rate of change of the position"

    [tex]a={{dv}\over{dt}}[/tex] 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,


    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
  4. Aug 27, 2009 #3
    hi, staetualex, take a little time on chain rules.
    It's not dx^2/dv; it's vdv/dx.
  5. Aug 27, 2009 #4


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    I can't help you, but I thank you for adding a cool new word to my vocabulary. :rofl:
  6. Aug 27, 2009 #5
    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.


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


    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.

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