Advanced Quadratic drag - How to solve?

Question

Advanced Quadratic drag - How to solve?

So I have gotten this far...

$a = ((F_{a} - \frac{c_{d} \cdot ρ_{h} \cdot A \cdot v^{2}}{2}) - m_{t} \cdot g_{r})$ $a=((F_a-(c_drho_hAv^2)/2)-m_tg_r)$

This is based upon $F_{a} - \frac{F_{d} + F_{g}}{m} = a$ $F_a-(F_d+F_g)/m =a$

But $a$ $a$ is $\frac{d v}{d t}$ $(dv)/(dt)$

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.

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Answer 1 · 2016-06-07T14:48:39

Supposing that
$c_{0} = F_{a} - m_{t} g_{r} = C^{t e}$ $c_0 = F_a-m_t g_r = C^{te}$
$c_{2} = \frac{c_{d} ρ_{h} A}{2} = C^{t e}$ $c_2=(c_d rho_h A)/2 = C^{te}$
we have $v (t) = \sqrt{\frac{c_{0}}{c_{2}}} tanh (\sqrt{c_{0} c_{2}} (t + t_{0}))$ $v(t) = sqrt[c_0/c_2] Tanh(sqrt[c_0c_2] ( t+t_0))$ with $t_{0}$ $t_0$ integration constant

Explanation:

This equation

$. v (t) = c_{0} (t) + c_{1} (t) v (t) + c_{2} (t) v {(t)}^{2}$ $dot v(t) = c_0(t) + c_1(t)v(t) + c_2(t)v(t)^2$

with $V = c_{2} (t) v$ $V = c_2(t)v$ , $S (t) = c_{0} (t) c_{2} (t)$ $S(t) = c_0(t)c_2(t)$ and $R (t) = c_{1} (t) + \frac{{. c}_{2} (t)}{c_{2} (t)}$ $R(t) = c_1(t)+(dot c_2(t))/(c_2(t))$

can be reduced to

$. V = V^{2} + R (t) V + S (t)$ $dot V = V^2+R(t)V+S(t)$

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

Now making $V = - \frac{. u}{u}$ $V = -(dot u)/u$ 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 $a = t_0 sqrt(c_0c_2)$

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Advanced Quadratic drag - How to solve?

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