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March 31, 2015

March 31, 2015

Posted by **TO ]BOBPURSLEY** on Thursday, June 18, 2009 at 9:40pm.

(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

OK so far

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

(2a)x = (v^2-v0^2) + x0 the last term should be xo*2a ng

--------------------------------------

why i thought you had to subtract Xo from both sides???

please help me understand

- Physics -
**Physics**, Friday, June 19, 2009 at 8:23amdef 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

OK so far

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

(2a)x = (v^2-v0^2) + x0 the last term should be xo*2a ng

--------------------------------------

why i thought you had to subtract Xo from both sides???

please help me understand

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