Proving a known function of position via Chain Rule

[kylera]

Homework Statement

Use the Chain Rule to prove that for rectilinear motion, when the acceleration is a known function of position, you can find the velocity as a function of position via the integral

$$\frac{v^{2}-v_{0}^{2}}{2} = \int^{s}_{s_{0}}a(s)ds$$

Homework Equations

$$v^{2}=v_{0}^{2}\times2as$$

The Attempt at a Solution

I took the left fraction, substituted v^2, simplified and got $$as$$. I let A be as, then took dA to get a da. Now I'm stuck.

 

[ace123]

I think you simplified wrong. It's a multipication sign not addition. See what I mean?

 

[kylera]

Hold up, I wrote the question on the board wrong -- it is supposed to be a plus for the relevant equation part.

 

[calef]

I'm not exactly sure what they're asking here. For constant accelerations, you "relevant equation" is basically the answer, assuming you swap out the "x" for a "+" and take a square root. What's throwing me is the request for proof by chain rule.

 

[tiny-tim]

Chain Rule

Homework Statement

Use the Chain Rule to prove that for rectilinear motion, when the acceleration is a known function of position, you can find the velocity as a function of position via the integral

$$\frac{v^{2}-v_{0}^{2}}{2} = \int^{s}_{s_{0}}a(s)ds$$

Homework Equations

$$v^{2}=v_{0}^{2}\times2as$$

The Attempt at a Solution

I took the left fraction, substituted v^2, simplified and got $$as$$. I let A be as, then took dA to get a da. Now I'm stuck.

Hi kylera!

You were asked to use the Chain Rule. So …

Hint: the LHS is ∫vdv. So use the Chain Rule on dv.