equations for uniformly accelerated motion

These equations are found in any physics text. Practice your algebra and isolate the different variables in these equations.

Remember, if an object is starting from rest (zero m/s) then Vo=0 and if an object comes to rest, then Vf=0
DX=((Vo+Vf)t)/2

Acceleration Defined: a=Vf-Vo/t ==>Vf=Vo+at and Vo=Vf-at and t = Vf-Vo/a

DX=Vo(t)+at²/2

Vf²=Vo²+2aDX

HOW DO WE GET THESE EQUATIONS ???

Xf-Xo=Vt, where v=average velocity, or Vavg

Vavg=(Vf+Vo)/2 because to average is to sum and divide by # total

if we insert second equation into first one (where V is),

we get Xf-Xo=((Vf+Vo)t)/2

introduce 2nd equation (from line 2 above)Vf=Vo+at insert this into third equation above,
(1 box above this one)

get Xf-Xo=((Vo+at+Vo)t)/2

clean it up a bit

get, Xf-Xo=(2Vot+at²)/2

which is the same as 2Vot/2 + at²/2

so we cancel out a couple of 2's,

and get Xf-Xo=Vot+at²/2

quite often we are asked for some final position, so physics books usually write the equation as

Xf=Xo+Vot+at²/2

Xf-Xo=Vt, where v=average velocity, or Vavg
Vavg=(Vf+Vo)/2 because to average is to sum and divide by # total
if we insert second equation into first one (where V is),
we get Xf-Xo=((Vf+Vo)t)/2
introduce 2nd equation (from line 2 above)Vf=Vo+at insert this into third equation above, (1 box above this one)
	get Xf-Xo=((Vo+at+Vo)t)/2
	clean it up a bit
	get, Xf-Xo=(2Vot+at²)/2
	which is the same as 2Vot/2 + at²/2
	so we cancel out a couple of 2's,
	and get Xf-Xo=Vot+at²/2
	quite often we are asked for some final position, so physics books usually write the equation as
	Xf=Xo+Vot+at²/2