Posted by AP Physics on Thursday, June 18, 2009 at 6:44pm.
I'm trying to derive the formula
v^2 = v0^2 + 2a(xx0)
were zeros are subscripts
my book tells me to derive it this way
use the definition of average velocity to derive a formula for x
use the formula for average velocity when constant acceleration is assumed to derive a formula for time
rearange the defintion of aceleration for a formula for t
then combine equations to get the derived formula for v^2
so here's my work please show me were I won't wrong
def of average velocity = t^1 (x  x0)
(average velocity = t^1(xx0))t=(avearge velocity)t + x0= x  x0 + x0 = x = (average velocity)t + x0
x = (average velocity)t + x0
def of average velocity were costant acceleration is assumed = 2^1(v0 + v)
plug into
x = (average velocity)t + x0
x = 2^1(v0 + v)t + x0
def of acceleration = t^1(vv0)
(a=t^1(vv0))t=(at=(vv0))a^1 = t = a^1(vv0)
t = a^1(vv0)
plug into x = 2^1(v0 + v)t + x0
x = 2^1(v0 + v)a^1(vv0) + x0
solve for v^2
x = 2^1(v0 + v)a^1(vv0) + x0
simplfy
x = (a2)^1(v^2 v0^2)+ x0
(x = (a2)^1(v^2 v0^2)+ x0)2a
(2a)x = (v^2v0^2) + x0
(2a)x  x0 = (v^2v0^2) + x0  x0
(2a)x  x0 + v0^2= (v^2  v0^2) + v0^2
(2a)x  x0 + v0^2 = v^2
so here's what I got for my equation
v^2 = v0^2 +(2a)x  x0
here's what I was suppose to get
v^2 = v0^2 + 2a(xx0)
please show me were I went wrong
thank you!

AP Physics  bobpursley, Thursday, June 18, 2009 at 8:37pm
def of average velocity = t^1 (x  x0)
(average velocity = t^1(xx0))t=(avearge velocity)t + x0= x  x0 + x0 = x = (average velocity)t + x0
x = (average velocity)t + x0
def of average velocity were costant acceleration is assumed = 2^1(v0 + v)
plug into
x = (average velocity)t + x0
x = 2^1(v0 + v)t + x0
def of acceleration = t^1(vv0)
(a=t^1(vv0))t=(at=(vv0))a^1 = t = a^1(vv0)
t = a^1(vv0)
plug into x = 2^1(v0 + v)t + x0
x = 2^1(v0 + v)a^1(vv0) + x0
solve for v^2
x = 2^1(v0 + v)a^1(vv0) + x0
simplfy
x = (a2)^1(v^2 v0^2)+ x0
OK so far
(x = (a2)^1(v^2 v0^2)+ x0)2a
(2a)x = (v^2v0^2) + x0 the last term should be xo*2a ng