Question

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

Answer 1

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