Deriving the Kinematic equation without time? $x = (v^2 - v_0^2)/2a + x_0$

Question

Ok so this is a short question (maybe a little bit more about my inability to do math).

I need to derive $x = \frac{v^2 - v_0^2}{2a} + x_0$ from the position function $ = v_0t + \frac{1}{2}at^2 + x_0$. I solved for time by using $v = v_0 + at$, which is $\frac{v - v_0}{a} = t$. I plugged that into $x = v_0t + \frac{1}{2}at^2 + x_0$ for $t$ and get:

$$x = v_0\biggl(\frac{v - v_0}{a}\biggr) + \frac{1}{2}a\biggl(\frac{v - v_0}{a}\biggr)^2 + x_0$$

When I foiled I got:

$$x = -\frac{v_0^2}{a} + \frac{v^2}{2a} + \frac{v_0^2}{2a} + x_0$$

I checked my work 3+ times (redoing the problem each time). Either my math is wrong, or the substitution is rubbish. Can you give me any tips? Thanks.

Answer 1

Looks good so far. Add your terms with $V_o^2$ in the numerators. After this, factor out your $1/(2a)$. You're well on your way.

Answer 2

look at my answer to the question immediately above this one, where all five equations for constant acceleration are given

well I see that question was on the physics page. This is so confusing.

but:-

v - u = a.t

v + u = 2s/t

v^2 - u^2 = 2a.s

s = u.t + at^2 / 2

s = v.t - at^2 / 2

The five equations can be characterized by which of the five variable is absent.

Of course these are only valid for constant acceleration problems.