Speed of object sliding to a stop w/ air resistance with respect to position

[Bill M]

Okay, first a brief intro. I studied math/physics extensively as an undergrad...but that was over 10 years ago now. My day job keeps me doing more basic physics and math regularly, but I haven't, for example, solved a differential equation in over a decade! Anyway...

I'm trying to calculate the initial velocity of an object sliding to a stop over a single level surface with a known friction coefficient and a known distance (D). This is easy ignoring air resistance, but I want to factor that in. I'm setting up my forces so that the object is traveling in the positive x direction (therefore, friction and drag are negative).

The F = ma equation that I came up with is

F = -fMg - kv²(x) = ma = mV'(x)V(x)

where V(x) is the velocity as a function of x, f is the coeff. of friction, M is the object's mass, g is gravitational acceleration, k is a constant representing all the other constants in the drag equation before the v² (cross sectional area, density, drag coeff).

Solving for v(x) while breaking out my old diffy q book and looking for help on the internet (and having substitution/chain rule flashbacks), I came up with the following:

V(x)=Sqrt[ (-Mfg+Mfge^(-2kx+2kd)/M) / k]

I know I skipped a lot of steps, but it's a lot of typing! I can include more if needed, but I'm curious if I'm on the right track. I used the fact that when x=d, v=0 (end of deceleration) to solve it. The numbers vs. ignoring drag make sense when I do the calculations.

Any help/advice would be appreciated.

 

[Andrew Mason]

The problem is that the drag force is not always proportional to v^2. This is true for higher speeds with turbulent flow. For lower speeds where there is no turbulence, drag force is proportional to v. So I think you will have to know the speed at which turbulent flow ends.

AM

 

[Bill M]

Andrew, that's a good point. I suppose I'd like to ignore that for now and first figure out if I came up with the right solution:

$V(x) = \sqrt{-\frac{Mfg}{k}(1-e^{\frac{-2k(x-D)}{M}})}$

My calculations are showing that the drag produces very little difference in calculated speeds for a large object (say, a vehicle). This isn't surprising given the speeds involved and the strength of the frictional force (due to the object's weight) compared to drag in this instance, but I'd like to make sure my numbers are accurate. Then I can go through the more complicated process of factoring in the point where turbulence ends if I truly want to drive myself crazy.