Question

Advanced Quadratic drag - How to solve?

So I have gotten this far...

`a=((F_a-(c_d*rho_h*A*v^2)/2)-m_t*g_r)`

This is based upon `F_a-(F_d+F_g)/m =a`

But `a` is `(dv)/(dt)`

So this becomes a differential equation because the faster you go the harder it is to go faster.

How can one solve this?

Please note I'm currently a Calc AB student.

Answer 1

Supposing that
`c_0 = F_a-m_t g_r = C^{te}`
`c_2=(c_d rho_h A)/2 = C^{te}`
we have #v(t) = sqrt[c_0/c_2] Tanh(sqrt[c_0c_2] ( t+t_0))# with `t_0` integration constant

Explanation:

This equation

`dot v(t) = c_0(t) + c_1(t)v(t) + c_2(t)v(t)^2`

with `V = c_2(t)v`, `S(t) = c_0(t)c_2(t)` and `R(t) = c_1(t)+(dot c_2(t))/(c_2(t))`

can be reduced to

`dot V = V^2+R(t)V+S(t)`

which is the known Riccati equation.
https://en.wikipedia.org/wiki/Riccati_equation

Now making `V = -(dot u)/u` the Riccati equation can be reduced to
a second order ODE which is

`ddot u -R(t) dot u + S(t) u = 0`

A solution `u` to this equation can be used to find the solution to the original equation

`v = -(dot u)/(c_2(t)u)`

As an example, supposing that

`c_0 = F_a-m_t g_r = C^{te}`
`c_2=(c_d rho_h A)/2 = C^{te}`

we obtain

#v(t) = sqrt[c_0/c_2] Tanh(sqrt[c_0c_2] (t + t_0))#

after solving

`ddot u-c_0c_2 u = 0`

with general solution

`u(t) = C_1e^{sqrt(c_0c_2)t}+C_2e^{-sqrt(c_0c_2)t}`

Explanation . substituting in

`v(t) = -(dot u(t))/(c_2(t)u(t))`

we get

#v(t) = (sqrt[c_0 c_2] C_2 e^(-sqrt[c_0 c_2] t) - C_1 sqrt[c_0 c_2] e^( sqrt[c_0 c_2] t))/(c_2 (C_2 e^(-sqrt[c_0 c_2] t) + C_1 e^(sqrt[c_0 c_2] t)))#

grouping

#v(t) = sqrt(c_0c_2)/c_2((C_2 e^(-sqrt[c_0 c_2] t) - C_1e^(sqrt[c_0 c_2] t))/ (C_2 e^(-sqrt[c_0 c_2] t) + C_1 e^(sqrt[c_0 c_2] t)))#

or

#v(t) = sqrt(c_0c_2)/c_2((C_2 e^(-sqrt[c_0 c_2] t) - C_1e^(sqrt[c_0 c_2] t))/ (C_2 e^(-sqrt[c_0 c_2] t) + C_1 e^(sqrt[c_0 c_2] t)))#

but

#(C_2 e^(-sqrt[c_0 c_2] t) - C_1e^(sqrt[c_0 c_2] t))/ (C_2 e^(-sqrt[c_0 c_2] t) + C_1 e^(sqrt[c_0 c_2] t))=(e^(-sqrt[c_0 c_2] t) - C_1/C_2e^(sqrt[c_0 c_2] t))/ (e^(-sqrt[c_0 c_2] t) + C_1/C_2 e^(sqrt[c_0 c_2] t))#

and `EE a | abs(C_1/C_2) = e^{2a}` then

#(C_2 e^(-sqrt[c_0 c_2] t) - C_1e^(sqrt[c_0 c_2] t))/ (C_2 e^(-sqrt[c_0 c_2] t) + C_1 e^(sqrt[c_0 c_2] t))=(e^(-sqrt[c_0 c_2] t-a) - e^(sqrt[c_0 c_2] t+a))/ (e^(-sqrt[c_0 c_2] t-a) + e^(sqrt[c_0 c_2] t+a)) = sinh(sqrt(c_0c_2t+a))/cosh(sqrt(c_0c_2t+a)) = tanh(sqrt(c_0c_2t+a))#

Finally putting all together

`v(t) = sqrt[c_0/c_2]Tanh(sqrt[c_0c_2] t + a)` or

#v(t) = sqrt[c_0/c_2] Tanh(sqrt[c_0c_2] (t + t_0))#

with `a = t_0 sqrt(c_0c_2)`