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-   -   Getting around Mechanical Physics, Motion (https://www.askmehelpdesk.com/showthread.php?t=67711)

  • Mar 1, 2007, 12:46 PM
    Evil dead
    Getting around Mechanical Physics, Motion
    This is dedicated to some random dude called 'Capulchin':p You inspire me to become a better Scientist - I'm 16 lololol.

    Basically, this is thinking about motions logically rather than just punching in numbers into formulas. In the long run, this is more efficient and ensures you never get a mechanical question wrong again. The following also involves homegenity of equations which is essential if you need to proof your answers in a Physics test or if you are kind of stumped on a question.

    Legal stuff: This is the thinking and solely the thinking of Ade. My physics teacher checked over my work and approved it, nothing more nothing less. Some instances of these notes are still in construction, although I'm sure Calpulchin will iron them out for us - he's such a great guy:p

    The equations of motion

    v = u + at
    s = 1/2(u + v) x t
    s = ut + 1/2at2
    v2 = u2 + 2as

    where:
    v = velocity (ms-1)
    u = initial velocity (ms-1)
    s = displacement in a straight line (m)
    t = time taken (s)
    a = acceleration - the increase of speed over a certain amount of time (ms-2)

    Most people usually try and get the values that they've been given, punch it in, and usually get it wrong. I've devised a straight forward way to ensure that this doesn't happen.

    standard symbol equation
    unit equation
    calculated unit equation
    re-worded and easy to understand word equation

    Our first equation:

    v = u + at
    ms-1 = ms-1 + (ms-2 x s)
    ms-1 = ms-1 + ms-1
    final velocity = initial velocity + average velocity in the duration of time

    You see the homogeneity of the equation? The 'at' when multiplied equals ms-1 (using your indices rules and knowledge of powers). When you add that with the 'u' which also has a unit of ms-1, it will still equal 'v' which ultimately has ms-1 as it's unit.

    Our second equation:

    s = 1/2(u + v) x t
    ms = 1/2(ms-1 + ms-1) x s
    ms = 1/2(m + m)
    Total distance traveled = average velocity divided by 2

    For my work, my main flaw are my revised word equations. The above could be something like ' Total distance = average velocity x speed' or whatever. It is not written in stone, the main thing you should be focusing on is the homogeneity of the equation.

    Our third equation:

    s = ut + (1/2 x a x t2)
    m = (ms-1 x s) + 1/2(ms-2 x s2)
    m = m + m
    Total distance traveled = distance initially + average distance whilst accelerating

    The problem with this one, is what if the object moving started from rest? Then the distance initially would be 0. If that is the case, would the average distance traveled whilst accelerating be enough to compensate the whole distance?

    Our forth equation:

    v2 = u2 + 2as
    m2s-2 = m2s-2 + 2(m2 x s2)
    m2s-2 = m2s-2 + 2(m2s-2)
    erm..... final velocity squared = inital velocity squared + acceleration over distance given

    And that is it! Like I said before, what is above is not written in stone and I've already identified a few problems, like what if the object starts from rest, would the word equations agree? If me, Calpuchin and others bang heads to together we can easily rule out the flaws and improve my ideas on how to understand motion.

    Thank you for reading
  • Mar 1, 2007, 01:16 PM
    Capuchin
    I'm not entirely certain this is the right way to go about things, I think plugging numbers in is fine, but it must come after understanding of the problem.

    I like to not think of physics as numbers, but rather understand the question, and what needs to be done, and then after that use your knowledge of equations to come to a numerical answer.

    This prepares you to understand what should be the right answer, so you know if your answer looks right.

    You've obviously put a lot of work in here, and it will help you to understand before you bring out the equations. The way I look at constant acceleration questions is probably not as intelligent as you think I look at them.

    I understand the question, and that I need to use suvat equations. Then I will list the variable I know, and the variables I need to work out. And then I look through my suvat equations to find the equation(s) that have those variables in them.
  • Mar 1, 2007, 01:41 PM
    Evil dead
    That is easy enough to do Capulchin, but my method also teaches people homogenics. And also Physics is not just punching in numbers, make in my fathers day they were real physcist, not the kind of rabble we are pumping out of colleges and universes:mad:
  • Mar 1, 2007, 01:52 PM
    Capuchin
    Don't get me wrong, what you have written is very useful for high school physics, but beyond that you need a little better understanding, not of the equations, but of what is actually going on. That's where I feel the real understanding comes from.
  • Mar 1, 2007, 01:58 PM
    Nosnosna
    In the end, physics (or any science) isn't about using equations at all. All six physics classes I had in college (mechanics, magnetics/electricity, optics/relativity, thermodynamics, electrodynamics, and transistor theory) stressed that pretty thoroughly... understand the system, describe it with a series of equations, and then generalize it so that you can use it later. And if you forget any equation, you can simply derive it later.

    It's never about punching in numbers, but if you understand why the equation is what it is and how it got there, then there's nothing wrong with doing just that. Science is about building on what came before, and there's no reason to redo the same work over and over again every time you need to use it.

    Edit: Dealing with the system directly will make much more sense once you get into calc-based physics... since, at that point, all of the units used are demonstrably related, those equations will more intuitively match up with how you expect things to work.
  • Mar 2, 2007, 09:41 AM
    Evil dead
    So to conclude, are you saying what I put up there is all jumble?
  • Mar 2, 2007, 09:46 AM
    Capuchin
    Not at all.
  • Mar 4, 2007, 05:48 PM
    bob1
    Thanks I've been looking for that too

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