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April 20, 2014

April 20, 2014

Posted by **Physics PLEASE** on Friday, June 19, 2009 at 10:39am.

v^2 = v0^2 + 2a(x-x0)

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(x-x0))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(v-v0)

(a=t^-1(v-v0))t=(at=(v-v0))a^-1 = t = a^-1(v-v0)

t = a^-1(v-v0)

plug into x = 2^-1(v0 + v)t + x0

x = 2^-1(v0 + v)a^-1(v-v0) + x0

solve for v^2

x = 2^-1(v0 + v)a^-1(v-v0) + x0

simplfy

x = (a2)^-1(v^2 -v0^2)+ x0

(x = (a2)^-1(v^2 -v0^2)+ x0)2a

(2a)x = (v^2-v0^2) + x0

(2a)x - x0 = (v^2-v0^2) + 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(x-x0)

please show me were I went wrong

thank you!

Show me step by step and as to why because I thing your suppose to subtract x0 from both sides why do you multiply???

- Physics PLEASE -
**Damon**, Friday, June 19, 2009 at 5:49pmYou want to show that

V^2 - Vo^2 = 2 a (X - Xo)

first rewrite that as:

(V-Vo)(V+Vo) = 2a (X-Xo)

now

your average velocity during the interval = (V+Vo)/2

so

X = Xo + t(V+Vo)/2

but

V = Vo + a t

so

t = (V-Vo)/a

so

X = Xo +(V-Vo)(V+Vo)/2a

we are there.

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