Proving the Velocity Formula: vf^2 = vi^2 + 2aΔd

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In summary, the conversation is discussing how to prove the equation vf2 = vi2 + 2aΔd, which relates velocity, acceleration, and displacement. The participants suggest using energy conservation and kinematic equations to derive the equation, and the conversation ends with doubts about whether this can be considered a valid proof.
  • #1
killaI9BI
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Homework Statement



Can you prove that vf2 = vi2 + 2aΔd?

Homework Equations


The Attempt at a Solution



I don't know where to start. I've not been given any values to use so I'm not sure how to go about answering the question.
 
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  • #2
What did your course cover so far?
This is just an application of energy conservation, but I don't know what you can use in such a proof.
 
  • #3
This lesson is about acceleration. It's covered vectors and vector components, relative velocity and displacement so far.
 
  • #4
killaI9BI said:
This lesson is about acceleration. It's covered vectors and vector components, relative velocity and displacement so far.

If you know any of the other kinematic equations then write them down and think about how you can eliminate time as a variable. are there any combinations or substitutions you can make given any of the other kinematic equations?
 
  • #5
Maybe it'll help if I post the entire problem:

Derive an equation that relates vi, vf, Δd, and a. (Hint: Notice that Δt is not invloved.) Solve for Δt in the first equation. Substitute that value into Δt in the third equation. Solve for vf2. Can you prove that vf2 = vi2 + 2aΔd?

I think that this equation relates all of the variables that the problem is asking for:

vf = [itex]\sqrt[]{}[/itex]{vi2 + 2a X d}

There is a table on the page that the problem doesn't reference specifically but I'm now realizing that it must be relevant:

DSC_0343.jpg


solve for Δt for the first equation:

Δt = (vf - vi) X a

Substitute that value into Δt for the third equation:
Δd = ((vi - vf)/2) X ((vf - vi) X a)

Solve for vf2:
vf2 = vi2 + 2a X d

I'm not sure if I did all of that correctly but it leaves me with the initial problem of proving that that vf2 = vi2 + 2aΔd

I'm not sure I understand where to start with that.
 
  • #6
killaI9BI said:
solve for Δt for the first equation:

Δt = (vf - vi) X a

Substitute that value into Δt for the third equation:
Δd = ((vi - vf)/2) X ((vf - vi) X a)

Solve for vf2:
vf2 = vi2 + 2a X d

I'm not sure if I did all of that correctly but it leaves me with the initial problem of proving that that vf2 = vi2 + 2aΔd

I'm not sure I understand where to start with that.
What do you mean you still have to prove it? you started with true relations and ended up with the required equation; that's how this equation comes about. That's all you need to do.
 
  • #7
My answer should then be yes I can prove that equation because... ?

Like I said, I'm not sure I understand where to start with that.
 
  • #8
what you did is proof enough. In this sense derivation would be proof, because like I said, you are taking relations that are already known to be true, and ending up with the time independent kinematic equation.
 
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  • #9
I really do appreciate your input. I am pretty stunned when it comes to this stuff...

I just don't understand how what I've done above is enough because as you can see here:
DSC_0361.jpg


Proving that equation is the last problem. I have no idea how to answer it.
 
  • #10
Rewrite the equation as ##v_f^2-v_i^2=(v_f+v_i)(v_f-v_i)=2aΔd##

How is ##(v_f-v_i)## related to a and Δt?

How is ##(v_f+v_i)## related to the average velocity?

How is Δd related to the average velocity and Δt?

Chet
 
  • #11
Chestermiller said:
Rewrite the equation as ##v_f^2-v_i^2=(v_f+v_i)(v_f-v_i)=2aΔd##...

Could this really be considered a proof? One could easily derive this relation through this method as well.
 
  • #12
matineesuxxx said:
Could this really be considered a proof? One could easily derive this relation through this method as well.
Who knows what they were thinking when they asked for a proof?

Chet
 
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  • #13
lol thanks for your help guys!
 

FAQ: Proving the Velocity Formula: vf^2 = vi^2 + 2aΔd

1. What is the velocity formula and how is it derived?

The velocity formula, vf^2 = vi^2 + 2aΔd, represents the relationship between an object's final velocity (vf), initial velocity (vi), acceleration (a), and displacement (Δd). It can be derived from the equations of motion, specifically the equation vf = vi + at, by solving for displacement and substituting it into the equation Δd = vit + 1/2at^2.

2. How do we know that this formula is accurate?

This formula has been extensively tested and validated through experiments and observations in the field of classical mechanics. It has consistently shown to accurately predict the final velocity of an object based on its initial velocity, acceleration, and displacement.

3. What is the significance of the term "2aΔd" in the formula?

The term "2aΔd" represents the change in velocity due to acceleration over a given displacement. It is derived from the equation Δd = vit + 1/2at^2, where 1/2at^2 represents the displacement due to acceleration. By rearranging the equation, we can see that the change in velocity (Δv) is equal to 2aΔd.

4. Can this formula be used for any type of motion?

This formula is specifically applicable to objects moving with constant acceleration, such as in the case of free-fall or projectile motion. It may not accurately predict the velocity of an object in cases of non-uniform or changing acceleration.

5. How does this formula relate to other equations of motion?

The velocity formula vf^2 = vi^2 + 2aΔd is a combination of the equations of motion vf = vi + at and Δd = vit + 1/2at^2. It can also be derived from the kinematic equations by substituting the equations for velocity and displacement into the formula Δd = vit + 1/2at^2 and solving for vf.

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