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Formula for calculating distance traveled with uniform acceleration 
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#1
Dec2111, 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+ (VfVo)/2.t Where Vf= a.t +Vo Be correct? 


#2
Dec2111, 08:18 PM

PF Gold
P: 7,120

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



#3
Dec2211, 05:28 AM

P: 8

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? 


#4
Dec2211, 10:59 AM

P: 1,135

Formula for calculating distance traveled with uniform acceleration



#5
Dec2211, 11:14 AM

P: 1,195




#6
Dec2211, 11:23 AM

P: 1,135




#7
Dec2211, 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).



#8
Dec2211, 09:16 PM

PF Gold
P: 7,120

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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