# Formula for calculating distance traveled with uniform acceleration

by An Indiot
Tags: acceleration, distance, formula, traveled, uniform
 P: 8 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?
 PF Gold P: 7,120 Sure, that works, although the standard formula is typically just given as $x = x_0 + v_0 t + {{1}\over{2}}at^2$ which you could solve using what you have
P: 8
 Quote by Pengwuino Sure, that works, although the standard formula is typically just given as $x = x_0 + v_0 t + {{1}\over{2}}at^2$ which you could solve using what you have
Much appreciated.

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

like
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?

P: 1,135
Formula for calculating distance traveled with uniform acceleration

 Quote by An Indiot m/s^2 by s would be m/s and m/s^2 divided by s would be m/s^3
I really can't understand what are you trying to ask ...
P: 1,195
 Quote by An Indiot Much appreciated. I am having some problems figuring how opperators affect different types of numbers. like 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?
Units are handled algebraically just like variables.
P: 1,135
 Quote by LawrenceC Units are handled algebraically just like variables.
Oh so by "by" he means multiplication ?
 P: 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).
PF Gold
P: 7,120
 Quote by An Indiot Much appreciated. I am having some problems figuring how opperators affect different types of numbers. like 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?
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, $x = vt$, but made the units work out like $[position] = [meters/second]^{[second]}$, 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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