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Help On Equations Of Motion

  1. Oct 8, 2004 #1
    DEAR MEMBERS,

    I AM LEARNING THE EQUATIONS FOR MOTION, AND WE ARE TAUGHT 'SUVAT'

    BUT I DON'T QUITE UNDERSTAND WHY I NEED ALL 4 OF THEM AND WHAT EACH IS USED FOR ?

    FOR EXAMPLE : V^2 = U^2 + 2AS

    WHERE DOES THE SQUARE COME FROM ?

    PLEASE HELP ME.

    FROM ROGER :cry:
     
  2. jcsd
  3. Oct 8, 2004 #2

    arildno

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    Dearly Missed

    Now, Newton's laws of motion are basically the only laws you NEED (at least for a long time!);
    The equation you mentioned is, for example a RESULT from Newton's 2.law of motion.
    Just asking:
    Is the 4 equations you're talking about Newton's 3 laws, plus the one you wrote down?
     
  4. Oct 8, 2004 #3
    Dear Arildno,

    The Equations Were :

    V=u+at
    S= (u+v)t/2
    S=ut + 1/2at^2

    Please Can You Explain Them For Me...
    I Don't Understand Why I Need 4 Of Them.

    And Also It Says To Substitute One Term From One Equation Into The Other But Again I Dont Understand ?

    Thankyou.

    Roger
     
  5. Oct 8, 2004 #4

    Diane_

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    Roger - this is not an easy thing to explain in a forum like this. Part of the problem is that you're taking an algebra-based physics course. It may not seem it right now, but calculus-based physics is so much easier. There, to get the equations you mention, all you need are the definitions of the terms and a little integration.

    However: Let's start with definitions. You know that acceleration is the rate of change of velocity. If we assume a constant acceleration, then, we can get the acceleration by dividing the change in velocity by the time:

    a = (v - u)/t

    Solving for v, we get the first of your equations:

    v = u + at

    For your second, remember the definition of average velocity: average velocity is displacement over time. If the acceleration is constant, then you can also get the average velocity by averaging the initial and the final velocities. Let V_ be the average velocity (and one day I'll figure out who to write it properly here), then we have:

    V_ = s/t

    and, under constant acceleration,

    V_ = (u + v)/2

    So,

    (u + v)/2 = s/t

    which, when solved for s, gives

    s = (u + v)t/2

    your second equation.

    To get your third equation, take your second and substitute into it for v:

    s = (u + v)t/2 = (u + u + at)t/2 = (2ut + at^2)/2

    So,

    s = ut + (1/2)at^2

    You'll note, then, that the first two come straight from definitions (plus, in the second case, an assumption). The third comes from the first two - which means that, technically, you don't need it. You can get the proper results by using the first two. You're given the third one because sometimes it's easier to see your way clear to an answer if you don't have to go through too many steps.

    Please note that there are other, better ways to derive all of these equations. In particular, there is a geometric approach to the third one that I like a lot. However, it's hard to do that one without being able to draw a lot of diagrams. For right now, just concentrate on the differences between position, velocity, and acceleration, and the relationships between them.

    Hope this helps.
     
  6. Oct 8, 2004 #5
    There are some ways to obtain such a formula, the easiest one is to go from substituting one after the other.
    These kinds of equation describe how an object is in motion. For example, you need to know and see how people are walking, how a ball bounces, how Mickey Mouse performs his funny plays to suck veryone especially old ladies in, etc. :tongue2:

    -------You can also prove the above formula from energy conservation
     
  7. Oct 8, 2004 #6
    Hello, you yourself have helped me myself learn alot, but how can derive the third one using the approach you said ? I would really love it 2 if you could tell me :smile: thnx
     
  8. Oct 8, 2004 #7
    Well if you accept v = u + at and either of v^2 = u^2 + 2aS or S = ut + (1/2)at^2 the other follows naturally...

    [tex]v = u + at[/tex]
    [tex]v^2 = u^2 + 2as
    \Rightarrow u^2 + 2uat + a^2t^2 = u^2 + 2as
    \Rightarrow s = ut + \frac{1}{2}at^2[/tex]
    ---------------------------------------------------
    [tex]v = u + at[/tex]
    [tex]s = ut + \frac{1}{2}at^2[/tex]
    [tex]\Rightarrow s = u\frac{v-u}{a} + \frac{1}{2}a(\frac{v-u}{a})^2[/tex]
    [tex]\Rightarrow 2as = 2uv - 2u^2 + (v-u)^2[/tex]
    [tex]\Rightarrow 2as = 2uv - 2u^2 + (v^2 - 2uv + u^2)[/tex]
    [tex]\Rightarrow 2as = 2uv -2u^2 + v^2 -2uv + u^2[/tex]
    [tex]\Rightarrow 2as = v^2 - u^2[/tex]
     
  9. Oct 9, 2004 #8

    Diane_

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    Well, we can try.

    I want you to draw a graph - velocity verses time for an object that starts at rest and accelerates at a constant rate for a time t. You should end up with a straight line, slope a, starting at the origin. Draw a vertical line from the end of your velocity/time line to the time axis - you will end up with a right triangle, one leg of length t and the other of length at. (Note: that last part is important. Make sure you understand why the coordinates of the end of your line are (t, at)).

    OK - is it clear that the area under the curve is the displacement of the object? You might be able to convince yourself of that if you back up and make a few graphs of velocity vs. time for constant velocity. You'll always end up with a rectangle of area vt, which is clearly the displacement. Without using early calculus arguments, I can't make it any clearer right now. (Obviously this is something I need to work on. :) ) Anyway, if we accept that, then we're home free. The area of a triangle is (1/2)*base*height. Here we have a triangle with base t and height at, so the area is

    A = (1/2)*t*at = (1/2)at^2

    which is the displacement for an object at constant acceleration a for time t starting at rest.

    Now - redraw your graph of velocity vs. time for an object with a positive starting velocity. Do you see that what you have is the previous right triangle on top of a rectangle? And the rectangle has length t and height u? The area of that rectangle is ut, so the area of the entire figure is

    s = ut + (1/2)at^2

    Hope that helps. Once again, I want to emphasize that most of the gyrations I just went through are unnecessary if you have calculus available. As an added bonus, calculus let's you derive the equations of motion for non-constant acceleration very easily. You can also do so using the approach I just used, but I shudder to think of explaining it. :)
     
  10. Oct 10, 2004 #9
    Yes, calculus is more than just mathematics. It is everything (skeptics who think space is discrete may ignore this statement please).

    Roger, perhaps you should spend a few minutes trying to understand how calculus works with kinematic relationships. Once you become comfortable with d/dt, you'll easily understand things...besides, the graphical approach that Diane has suggested has its advantages and will be even better supplemented by calculus.

    Cheers
    Vivek
     
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