You 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.
I'm trying to derive the formula
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???
1 answer