TO BOBPURSLEY
posted by TO ]BOBPURSLEY .
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

why i thought you had to subtract Xo from both sides???
please help me understand

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

why i thought you had to subtract Xo from both sides???
please help me understand
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