About air resistance in horizontal direction

[amiras]

Hello, I was playing with formulas on air resistance and derived formula for velocity in horizontal direction after object was thrown. It looks reasonable but I started have confusion about it, i'd like for you great experts check on its derivation, so here it goes every v is v in x direction.

So I assumed that then object is thrown the only force that acts on it in x direction is air resistance opposite to its velocity, f = Dv^2, where D some constant.

so it should be: f = ma (since the force is in the same direction as acceleration (opposite to object velocity)), but I skipped this part and wrote:

-f = m dv/dx * dx/dt = mv dv/dx

-Dv^2 = mv dv/dx

-D/m dx = dv/v

and after integrating x from 0 to x, and from v0 to v, I've got: v = v0 * e^(-Dx/m)

It looks reasonably to me since velocity always decrease from its maximum value (v0), how its suppose to be.


Now after a while I got confused about this part: -f = mv dv/dx
if f = ma, according to me f [itex]\neq[/itex] mv dv/dx

Is it right because change in velocity over distance is in opposite direction as the force (and acceleration)? But velocity is also in different direction so why it is not like f = m(-v)(-dv/dx)

 

[Chestermiller]

The differential equation was solved incorrectly. The correct form should have been:
$$-\frac{dv}{v^2}=\frac{Ddx}{m}$$This integrates to $$\frac{1}{v}-\frac{1}{v_0}=\frac{Dx}{m}$$or$$v=\frac{v_0}{\left(1+\frac{Dv_0x}{m}\right)}$$