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

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Discussion Overview

The discussion centers around the equation vf^2 = vi^2 + 2ad, focusing on understanding its derivation and the underlying principles of motion under constant acceleration. Participants explore different methods of arriving at this equation, including proofs and substitutions from fundamental kinematic equations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant requests an explanation of the equation vf^2 = vi^2 + 2ad, expressing a desire to understand its derivation rather than just memorizing it.
  • Another participant provides a proof involving the average velocity and acceleration, leading to the equation through a series of steps that include integration and manipulation of variables.
  • A different approach is suggested by another participant, who starts with the general equations for position and velocity under constant acceleration, proposing to eliminate time to derive vf^2.
  • A later reply expresses appreciation for the explanations and indicates a newfound understanding of the concepts presented.

Areas of Agreement / Disagreement

Participants appear to agree on the importance of understanding the derivation of the equation, but multiple methods of deriving it are presented without consensus on which is preferable or more intuitive.

Contextual Notes

The discussion includes various mathematical steps and assumptions related to constant acceleration, but does not resolve any potential limitations or dependencies on specific definitions.

Who May Find This Useful

Individuals interested in kinematics, physics students seeking deeper understanding of motion equations, and those looking for different methods of deriving fundamental physics equations.

zedjay
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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) [tex]X = \frac{V_{f} + V_{i}} {2} \Delta t [/tex]
you can prove this with an integral. (constant acceleration)
2) [tex]a = \frac{V_{f} - V_{i}}{ \Delta t }[/tex]
3)[tex]\Delta t = \frac{V_{f} - V_{i}}{ a }[/tex]
4) [tex]X = \frac{V_{f} + V_{i}} {2} \cdot \frac{V_{f} - V_{i}}{ a }[/tex]

5) [tex]X = \frac{V_{f}^2 - V_{i}^2} {2 a}[/tex]

6) [tex]V_{f}^2 = V_{i}^2 + 2 a X[/tex]
:smile:
 
Last edited:
Here's another way. Start with the general equations for position and velocity under constant acceleration:
[tex]x = x_0 + v_0 t + \frac{1}{2} a t^2[/tex]
[tex]v = v_0 + a t[/tex]
Make the following substitutions: [itex]x_0 = 0[/itex], [itex]x = d[/itex], [itex]v_0 = v_i[/itex] and [itex]v = v_f[/itex].
Solve the two equations together to eliminate [itex]t[/itex], then rearrange to solve for [itex]v_f^2[/itex].
 
Thx a lot

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

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