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How do you guys deal with problems having only 2 variable? Good fomula sheet?

  1. Oct 16, 2006 #1
    Hey guys, sorry this is kinda vauge (Can't sleep) How do you guys deal with problems listing only 2 variables, so you cant straight use the 5 main velocity motion formulas? I know some like t=v-vnaught/a
    But, how many all are there that are common? I've ran into 2*vnaught/a
    But I'm not used to them, and can't quite see always when to use em..
    Any tips or stuff I can read on the net on these? And what formulas would you guys put down on a formula sheet? I wanna commit those two to memory (I have, but I might forget em in a weeK)
    Anyways..I'll try to check on this tomorrow and maybe expand..thx.
  2. jcsd
  3. Oct 16, 2006 #2
    Well, I'm not sure. If you knew more math, then once you know the relation between s, v, and a, then you can derive anything. However, I'm assuming you are in highschool and you probably don't have the neat math tricks that you will learn later on in your education.

    Anyhoo, one golden equation that really helped me in highschool physics was:
    [tex] s = v_{o}t + \frac{1}{2}at^{2} [/tex]
    Most kinematic problems you do in highschool will use them and also in your first year university physics.

    I don't know what else to say, but that's the start!

  4. Oct 16, 2006 #3
    Oh, I was good at HS Physics, I even help a friend out with his a bit..but, the uni ones are def more complicated..
    Well this 1 isn't that complicated..but, worried like freeze up in the test totally forget the velocity is 0
    (a) with what speed mst a ball be thrown vertically from ground level to rise to a maximum height of 50m? (b) how long will it be in the air?
    31m and 6.4 s

    I guess it's kinda like the 1 I helped friend with..ball stays in air for 8 seconds being shot straight up, what was its intial velocity? I just took half and basically saw how fast it hit the ground from it's highest point..

    But, like, your eqn...I dont see how some gets t = 2(v initial)/a from it..
    the t's kinda cancel so its not square rooted?

    I'm not sure how to integrals..so I'm sure that doesnt help..
  5. Oct 16, 2006 #4


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    Staff: Mentor

    All constant-acceleration problems can be solved using these two equations:

    [tex]x = x_0 + v_0 t + \frac{1}{2} a t^2[/tex]

    [tex]v = v_0 + a t[/tex]

    All those other equations can be derived from these two. So if you practice how to apply these two equations to a wide variety of problems, you shouldn't have to worry about the other equations. You'll effectively be re-deriving them as necessary.

    The one "catch" in using this method is that you sometimes have to solve two equations in two unknowns, so you may need to brush up on that.
  6. Oct 16, 2006 #5
    On the side note, the equation (2) is just a derivative of equation (1). I.e.

    [tex] \frac{dx}{dt} = \frac{d}{dt}\left(x_0 + v_0 t + \frac{1}{2} a t^2\right)[/tex]

    [tex]\frac{dx}{dt} = v = v_0 + a t[/tex]

    So if you know how to do derivative. You only need one equation :smile:
  7. Oct 16, 2006 #6


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    Homework Helper

    I would assume that jtbell is aware of that fact. :smile:
  8. Oct 16, 2006 #7


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    Staff: Mentor

    If you want to go the calculus route, the only equation you ever really need is a = constant! Just integrate "a" twice with respect to t and you get the equations for v and x. :wink:

    But take Torec_Scrail's problem as an example:

    I drew a diagram illustrating all the variables, showing which ones are known and unknown, then substituted into the x and v equations and solved them. See the attachment...

    Attached Files:

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