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.
--------------------
but wouldn't you have to rearange it and end up subtracting Xo
This is what i did
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
i do not see were i went wrong
Please check your previous posts. Here is my answer from several hours ago:
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.
Of course you have to rearrange it
X -Xo = (V-Vo)(V+Vo)/2a
(X-Xo)2a = (V-Vo)(V+Vo)
2 a (X-Xo) = V^2-Vo^2
Yes, you are correct. To demonstrate the derivation properly, let's rearrange the equation step by step.
We start with the equation given:
V^2 - Vo^2 = 2a(X - Xo)
To rewrite it as:
(V-Vo)(V+Vo) = 2a(X - Xo)
Now, let's express average velocity during the interval as (V+Vo)/2.
We know that displacement (X) can be calculated by multiplying average velocity with time (t). So, we have:
X = Xo + t(V+Vo)/2
Next, we substitute the value of velocity (V) into the equation using the equation of motion:
V = Vo + at
By rearranging the equation above, we can solve for time (t):
t = (V - Vo)/a
Substituting this expression for t back into the equation for displacement (X), we get:
X = Xo + [(V - Vo)/a](V + Vo)/2
Finally, simplifying the expression further, we have:
X = Xo + (V^2 - Vo^2)/2a
As you can see, we have subtracted Xo from the equation during the simplification process.