Use of d[1/2 v^2]/dx (1 Viewer)

[Mumma]

Im not sure on the correct use of d[¹/₂ v²]/dx

Now, say acceleration is a function of x, why cant I directly integrate to get velocity?

From my understanding
d²x/dt² = d[¹/₂ v²]/dx

So from what I can see, its used to get velocity in terms of x when acceleration is in terms of t.
Am I missing something fundamental ?

An example:
A particle moves in a straight line so that a = 3-2x. Find v in terms of x given that v = 2 when x = 1.

Cant I just go v = integral (3-2x) dx ?

Why v² = 2 integral (3-2x) dx ?

 

[SoulSearcher]

d²x/dt² is acceleration as a function of time.
d/dx(¹/₂ v²) is acceleration as a function of displacement.
d/dx(¹/₂ v²) is usually used to describe velocity in terms of the particle's displacement, while d²x/dt² is used to describe velocity in terms of time.
Even though they both describe acceleration, they deal with different variables, and as such must be treated differently.
In your example, acceleration is described as a function of x, and thus you integrate with respect to x, otherwise you will screw up the equation.

 

[Naylyn]

Did your teacher use a proof too show that a=d[1/2v^2]? because it all relate back to that.

 

[pLuvia]

There are different ways of expressing acceleration either:
dv/dt - when acceleration is a function of t
v(dv/dx) - when acceleration is a function of v
d(v²/2)/dx - when acceleration is a function of x

For your example since the acceleration is in terms of x you will need to express a as d(v²/2)/dx so:
d(v²/2)/dx=3-2x

 

[Riviet]

Did your teacher use a proof too show that a=d[1/2v^2]? because it all relate back to that.

Here's the proof, starting from scratch:
a=dv/dt
=dx/dt x dv/dx, by the chain rule
=v.dv/dx, since v=dx/dt

Now v.dv/dx = d/dv(1/2.v²) x dv/dx, by the chain rule
=d(1/2.v²)/dx as required.

 

[Mumma]

Thanks guys, makes sense. And yeah I knew of the proof.