Physics Asked by QuantumGravity on June 28, 2021
I am modeling an object under a force $F$ that diminishes with distance but is considered constant at small time intervals. I am also incorporating a drag force $D$ that is opposite the direction of $F$ and dependent on velocity squared. The net acceleration can be written as:
$$F-D=ma$$
As $D$ is velocity dependent, we can write a differential equation for velocity. $C$ is the coeffecient of drag, $A$ is the area, and $rho$ is the air pressure.
$$dfrac{Delta v}{Delta t} + dfrac{CArho}{2m}v^2 = dfrac{F}{m}$$
Setting acceleration $dfrac{Delta v}{Delta t}$ to zero, we find can find terminal velocity.
$$v^2_{term} = dfrac{2F}{CArho}$$
Another solution that got me this far indicates that we should rewrite the differential equation. I myself am unsure of how to make this transformation.
$$dfrac{1}{1-v^2/v^2_{term}}dv = dfrac{F}{m}dt$$
Taking the integrals of both sides is indicated next.
$$tanh^{-1}(v/v_{term}) = dfrac{F}{mv_{term}}t + C$$
Setting $t$ to zero gives $C$ equal to $tanh^{-1}(v_{0}/v_{term})$, and we can solve for $v_{t}$.
$$v_{t} = v_{term} tanh(frac{F}{mv_{term}}t + tanh^{-1}(v_{0}/v_{term}))$$
Each term of this equation makes sense to me, as the equation of $tanh$ approximates the change in velocity towards terminal velocity, velocity increases toward terminal velocity faster under higher force, and initial velocity independently shifts the curve to the left. My model shows a smooth reduction in acceleration as velocity approaches terminal velocity asymptotically for each time step.
My problem arises as such: because the force $F$ is distance dependent, there will come a time step where $v_{0}$ is greater than $v_{term}$, which is outside the bounds of the $tanh^{-1}$. I am looking for an equation of motion for the approach of terminal velocity from a higher speed.
I deeply suspect it will take the form of $v_{t} = v_{term} coth(something + coth^{-1}(v_{0}/v_{term}))$, but I’ve been unable to find $coth$ on any integral lists, so I’m not certain how to get there.
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