# Can any1 explain vf^2 = vi^2 + 2ad?

1. Oct 30, 2005

### zedjay

Hi

can anyone explain the logistics of vf^2 = vi^2 + 2ad to me? i mean its one thing to simply remember the equation but, to understand how they came about it is way more helpful.

thx alot

2. Oct 30, 2005

### daniel_i_l

Here is the proof:
1) $$X = \frac{V_{f} + V_{i}} {2} \Delta t$$
you can prove this with an integral. (constant acceleration)
2) $$a = \frac{V_{f} - V_{i}}{ \Delta t }$$
3)$$\Delta t = \frac{V_{f} - V_{i}}{ a }$$
4) $$X = \frac{V_{f} + V_{i}} {2} \cdot \frac{V_{f} - V_{i}}{ a }$$

5) $$X = \frac{V_{f}^2 - V_{i}^2} {2 a}$$

6) $$V_{f}^2 = V_{i}^2 + 2 a X$$

Last edited: Oct 30, 2005
3. Oct 30, 2005

### Staff: Mentor

Here's another way. Start with the general equations for position and velocity under constant acceleration:
$$x = x_0 + v_0 t + \frac{1}{2} a t^2$$
$$v = v_0 + a t$$
Make the following substitutions: $x_0 = 0$, $x = d$, $v_0 = v_i$ and $v = v_f$.
Solve the two equations together to eliminate $t$, then rearrange to solve for $v_f^2$.

4. Oct 30, 2005

### zedjay

Thx alot

ohhhh now i c,
why didnt i see that?
thanks alot