There is an object which has an initial velocity of $`V_1`$ and has an acceleration of $`V^2 k+d+b`$ in the opposite direction, where k,d,b are constants. What will be the distance traveled until reaching the velocity of $`V_2`$?
I have tried some integration and derivation and all the time I found something useful but I stuck after some point. Are these equations right and can you help me finding the result as $V_1,k,d,b$?
We have the differential equation
$$ \frac{dV}{dt} = -(kV^2 +d+b) $$
$$ \frac{V'}{kV^2 + d+b} = -1 $$ $$ \frac{V'}{\frac{k}{d+b} V^2 + 1} = -d-b $$
Let $\alpha = \sqrt{\frac{k}{d+b}}$ and $\beta = \alpha(d+b) = \sqrt{k(d+b)}$.
$$ \frac{\alpha V'}{\alpha^2 V^2+1} = -\beta $$
$$ \int \frac{\alpha V'}{\alpha^2 V^2+1} dt = -\int \beta\, dt $$
$$ \arctan(\alpha V) = -\beta t + C $$
$$ V = \frac{1}{\alpha} \tan(C - \beta t) $$
At $t=0$ we must have $V=V_1$, so $\tan C = \alpha V_1$. These equations are valid for $0 \leq t \leq \frac{\arctan(\alpha V_1)}{\beta}$.
If $t_f$ is the time when $V=V_2$, then $C-\beta t_f = \arctan(\alpha V_2)$.
Then the total distance from time $t=0$ to time $t=t_f$ is
$$ X = \int_0^{t_f} V\, dt $$ $$ X = \frac{1}{\alpha} \int_0^{t_f} \tan(C-\beta t)\, dt $$ $$ X = \frac{1}{\alpha \beta} \left[ \ln \cos(C-\beta t) \right]_{t=0}^{t=t_f} $$ $$ X = \frac{1}{k} \big[\ln(\cos \arctan(\alpha V_2)) - \ln(\cos \arctan(\alpha V_1))\big] $$
If $\theta = \arctan y$, then $y=\tan \theta$, and $1+y^2 = 1+\tan^2 \theta = \sec^2 \theta = (\cos \theta)^{-2}$, so $\cos \arctan y = \frac{1}{\sqrt{1+y^2}}$.
$$ X = \frac{1}{k} \left[ \ln \frac{1}{\sqrt{1+\alpha^2 V_2^2}} -\ln \frac{1}{\sqrt{1+\alpha^2 V_1^2}} \right] $$
$$ X = \frac{1}{2k} \ln \frac{1+\frac{k}{d+b} V_1^2}{1+\frac{k}{d+b} V_2^2} $$
$$ X = \frac{1}{2k} \ln \frac{kV_1^2 + d + b}{kV_2^2 + d + b} $$