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Derive Formula for Final Velocity (v2) Without Time

  1. Feb 28, 2013 #1
    1. The problem statement, all variables and given/known data
    Derive the formula for final velocity, [itex]v_2[/itex], without time included in said formula.

    [itex]\overrightarrow{a}[/itex]: Known

    [itex]\overrightarrow{s}[/itex]: Known

    [itex]v_1[/itex]: Known

    [itex]\Delta t[/itex]: Unknown (can be calculated, but defeats the purpose)

    [itex]\overrightarrow{v_2}[/itex]: Unknown and to be calculated


    2. Relevant equations
    [tex]
    \begin{align*}
    \Delta t &= \frac{\overrightarrow{s}}{\overrightarrow{\Delta v}} \\ \\
    \overrightarrow{a} &= \frac{\overrightarrow{\Delta v}}{\Delta t}
    \end{align*}
    [/tex]

    3. The attempt at a solution
    Since I've always disliked memorizing formulas, I just figure out how they're derived and then if I forget the formula, I can get it regardless; so I went about trying to get the formula as follows:
    [tex]
    \begin{align*}
    \overrightarrow{a} &= \frac{\overrightarrow{\Delta v}}{\Delta t} \\
    &= \frac{\overrightarrow{\Delta v}}{\frac{\Delta \overrightarrow{s}}{\Delta \overrightarrow{v}}} \\
    &= \frac{\Delta \overrightarrow{v}^2}{\Delta \overrightarrow{s}} \\
    &= \frac{(v_2 - v_1)^2}{\Delta \overrightarrow{s}} \\
    \overrightarrow{a} \Delta \overrightarrow{s} &= (v_2 - v_1)^2 \\
    &= v_2^2 - 2v_2v_1 - v_1^2 \\ \\
    v_2^2 &= -2v_2v_1 - v_1^2 + \overrightarrow{a} \Delta \overrightarrow{s}

    \end{align*}
    [/tex]

    I ended up with a result that different than the formula taught in my class. I can't remember the formula taught in class, as I wasn't actually there for the lesson, but I vaguely remember what the formula was like, when I saw it later. I foolishly forgot to write it down, so now I'm not too sure, but I think it was similar to [itex]\v_2^2 = v_1^2 + \overrightarrow{a}\Delta \overrightarrow{s}[\itex].

    4. What I really want to know
    I basically need to know 4(ish) things: Is my derived formula okay? If not, where'd I go wrong? What's the formula similar to what I listed above (or is it correct; this question also isn't too big of a deal since I can get it tomorrow)? And if the second formula is correct (or the correct, similar one has been provided), how is it derived?

    EDIT: I remember the formula in class, which is [itex]v_2^2 = v_1^2 + 2 \overrightarrow{a} \Delta \overrightarrow{s}[/itex]. How is this derived?
     
    Last edited: Feb 28, 2013
  2. jcsd
  3. Feb 28, 2013 #2

    tms

    User Avatar

     
  4. Feb 28, 2013 #3
     
  5. Feb 28, 2013 #4

    tms

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    That's just the first step.
    But your initial result is wrong.

    BTW, you can use \vec for vectors, which is easier to type than \overrightarrow.
     
  6. Feb 28, 2013 #5
    I found out where the equation is derived from when I noticed that one of the formulas looked a lot like the area of a trapezoid formula, which lead me to a velocity-time graph and I was able to derive from there.

    The reason I asked was to check if it worked, and if not, check why it didn't work. Although it doesn't really matter much anymore, since I found out how the correct one is derived.

    Ah, much better. I was recently typing LaTeX in an online LaTeX equation editor, and \overrightarrow looked better than \vec, but on Physics Forums, \vec appears nicer, and easier to write too.
     
  7. Feb 28, 2013 #6

    SammyS

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    It seems that [itex]\ v_2^2 = v_1^2 + 2 \vec{a}\cdot \Delta \vec{s}\ [/itex] should be derived in your textbook !

    This (erroneous) equation is causing at least some of the problem.
    This is wrong → [itex] \displaystyle \Delta t = \frac{\vec{s}}{\vec{\Delta v}}\ \ [/itex]
    (Besides the fact that division of a vector by a vector is not defined:)

    The correct equation comes from
    ## \displaystyle \vec{v}_\text{average}=\frac{\Delta \vec{s}}{\Delta t}##​
    So that ## \displaystyle \ \ \vec{v}_\text{average}\,\Delta t= \Delta \vec{s}\ .##

    For uniform acceleration, ## \displaystyle \ \ \vec{v}_\text{average} = (\vec{v}_1 + \vec{v}_2)/2\ .##

    Also, ## \displaystyle \ \ \vec{a}_\text{average}=\frac{\Delta \vec{v}}{\Delta t}\ .\ \ ## Furthermore, for uniform acceleration, ## \displaystyle \ \ \vec{a} = \vec{a}_\text{average}\ . ##

    Now, consider ## \displaystyle \ \ \vec{v}_\text{average}\cdot(\Delta \vec{v})\ . ##
     
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