Formula for calculating distance traveled with uniform acceleration


by An Indiot
Tags: acceleration, distance, formula, traveled, uniform
An Indiot
An Indiot is offline
#1
Dec21-11, 08:03 PM
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?
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Pengwuino
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#2
Dec21-11, 08:18 PM
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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
An Indiot
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#3
Dec22-11, 05:28 AM
P: 8
Quote Quote by Pengwuino View Post
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
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?

cupid.callin
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#4
Dec22-11, 10:59 AM
P: 1,135

Formula for calculating distance traveled with uniform acceleration


Quote Quote by An Indiot View Post
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 ...
LawrenceC
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#5
Dec22-11, 11:14 AM
P: 1,195
Quote Quote by An Indiot View Post
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.
cupid.callin
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#6
Dec22-11, 11:23 AM
P: 1,135
Quote Quote by LawrenceC View Post
Units are handled algebraically just like variables.
Oh so by "by" he means multiplication ?
Dovahkiin
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#7
Dec22-11, 08:42 PM
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).
Pengwuino
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#8
Dec22-11, 09:16 PM
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P: 7,125
Quote Quote by An Indiot View Post
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, [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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