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How many equations for kinematics?

  1. Jun 3, 2015 #1
    This is a small question about the actual number of kinematic equations. I've seen many websites derive 4 different equations, but some only use 3.

    The 4th equation that I have yet to see be put to use is xfinal=xinitial+1/2(vinitial+vfinal)(time)

    Could someone explain why some do not mention this 4th equation? It is it necessary for working one dimensional motion problems?
     
  2. jcsd
  3. Jun 3, 2015 #2

    SteamKing

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    This equation is just calculating the average velocity and using that to find distance traveled over an interval.

    It's accurate only if the velocity is constant (v initial = v final) or if the change in velocity w.r.t. time is linear (i.e., acceleration is constant).

    IOW, this equation can be derived from the other kinematic equations very easily.
     
  4. Jun 3, 2015 #3

    haruspex

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    Most formulations involve five variables (omitting initial displacement). Since any three values are sufficient to deduce the remaining two, there are in principle five equations, each omitting one variable. Read more in section 7 of
    https://www.physicsforums.com/insights/frequently-made-errors-mechanics-kinematics/
     
  5. Jun 3, 2015 #4

    NascentOxygen

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    Let's simplify your equation just a little, and write it as distance moved, x = ½ (vi + vf)t

    How to get that using equations you already know?

    Start with x = vit + ½at2

    Remember, for motion under constant acceleration, the acceleration is given as
    a = (vf - vi)/t

    Now, your exercise is to substitute this expression for 'a' into the preceding equation for x, and simplify to get the equation at the top.
     
  6. Jun 3, 2015 #5

    SammyS

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    Actually, average velocity is defined as ##\displaystyle \ v_\text{average} = \frac{\text{displacement}}{\text{elapsed time}}=\frac{x_\text{final}-x_\text{initial}}{t} \ .##

    Solving for xfinal gives :

    ##\displaystyle x_\text{final}=x_\text{initial} + (v_\text{average})\cdot t\ ##

    As pointed out above, if acceleration is constant, then the following is also true.

    ##\displaystyle v_\text{average} =\frac{v_\text{initial}+v_\text{final}}{2}\ ##
     
    Last edited: Jun 3, 2015
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