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

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The equation vf^2 = vi^2 + 2ad describes the relationship between final velocity, initial velocity, acceleration, and distance under constant acceleration. It can be derived using kinematic equations and calculus, starting with the average velocity and integrating to find displacement. By substituting values and eliminating time, the equation can be rearranged to show how final velocity relates to initial velocity and acceleration over a distance. Understanding this derivation enhances comprehension of motion dynamics. The discussion concludes with appreciation for the clarity gained from the explanation.
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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 a lot :biggrin:
 
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Here is the proof:
1) X = \frac{V_{f} + V_{i}} {2} \Delta t <br />
you can prove this with an integral. (constant acceleration)
2) a = \frac{V_{f} - V_{i}}{ \Delta t }<br />
3)\Delta t = \frac{V_{f} - V_{i}}{ a }<br />
4) X = \frac{V_{f} + V_{i}} {2} \cdot \frac{V_{f} - V_{i}}{ a }<br />

5) X = \frac{V_{f}^2 - V_{i}^2} {2 a}<br />

6) V_{f}^2 = V_{i}^2 + 2 a X<br />
:smile:
 
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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.
 
Thx alot

ohhhh now i c,
why didnt i see that?
thanks a lot :biggrin:
 
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