TO BOBPURSLEY

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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
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 -

    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
    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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