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Homework Help: Formula for calculating distance traveled with uniform acceleration

  1. Dec 21, 2011 #1
    I had to calculate the distance a car traveled in a given time given its starting speed and acceleration.

    Would this formula
    x=Vo.t+ (Vf-Vo)/2.t
    Where Vf= a.t +Vo

    Be correct?
  2. jcsd
  3. Dec 21, 2011 #2


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    Gold Member

    Sure, that works, although the standard formula is typically just given as [itex]x = x_0 + v_0 t + {{1}\over{2}}at^2[/itex] which you could solve using what you have
  4. Dec 22, 2011 #3
    Much appreciated.

    I am having some problems figuring how opperators affect different types of numbers.

    m/s^2 by s would be m/s and m/s^2 divided by s would be m/s^3
    what should I search for explanations on how and why this works?
  5. Dec 22, 2011 #4
    I really can't understand what are you trying to ask ...
  6. Dec 22, 2011 #5
    Units are handled algebraically just like variables.
  7. Dec 22, 2011 #6
    Oh so by "by" he means multiplication ?
  8. Dec 22, 2011 #7
    Yes, the way i think of it is as almost an equation in itself as this means you can handle the units algebraically and include any factors of magnitude (such as kilo).
  9. Dec 22, 2011 #8


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    It works because we created it such that it did work consistently. The units represent a sense of "physicalness" to the math. If I have a distance moved, say 100m, and the amount of time it took, 5 seconds, you know the velocity must be in some unit that isn't meters and isn't seconds because you know, physically, a velocity is not a distance nor a time. It works algebraically, as another poster mentioned, because we made it work algebraically.

    If for example, you created a rule such that when 2 values multiplied each other, the units would exponentiate each other, you wouldn't be able to create a consistent way of doing physics. So say you wanted a position = velocity * time, [itex] x = vt [/itex], but made the units work out like [itex] [position] = [meters/second]^{[second]}[/itex], you wouldn't be able to create a consistent system where you could move between times, energies, positions, accelerations, etc. in a consistent manner (or I really really really doubt you could). Plus, with this example in mind, you'd clearly realize somethings wrong because our physical idea of "positions" are measured as a length in meters or feet or whatever.
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