Motorboat starts at rest and its motor produces constant acceleration at $4 ft/s^2$.... Motorboat starts at rest and its motor produces constant acceleration at $4 ft/s^2$, while the water resistance provides deceleration of $v^2/400$. What is v at 10 seconds and what is the limiting velocity?
$dv/dt = 4 - v^2/400$
$dv/(1600-v^2) =dt/400$
$(1/80)(ln|(v+40)/(v-40)|) = t/400 + K$
we get $K=0$ upon applying $v=0$ when $t=0$
$ln|(v+40)/(v-40)|=t/5$
$(v+40)/(v-40) = e^{t/5}$
$(v-40+80)/(v-40) = e^{t/5}$
$1+ 80/(v-40) = e^{t/5}$
$80/(v-40) = e^{t/5} - 1$
$80/(e^{t/5}) + 40 = v$
$v(10) = 80/(e^2 -1)$
$v(10)=52.5214$
for limiting velocity taking $t \to \infty$
$v_{limiting} = 40$
is this correct ?
 A: Well, in general: when there is an object that gets and constant acceleration of $
\text{n}_1\space\left[\text{m}/\text{s}^2\right]$ and feels a opposite accelaration that is proportional to the speed: $\frac{\text{v}\left(t\right)^2}{\text{n}_2}\space\left[\text{m}/\text{s}^2\right]$:
$$\text{a}\left(t\right):=\text{v}\space'\left(t\right)=\text{n}_1-\frac{\text{v}\left(t\right)^2}{\text{n}_2}\space\Longleftrightarrow\space\int\frac{\text{v}\space'\left(t\right)}{\text{n}_1-\frac{\text{v}\left(t\right)^2}{\text{n}_2}}\space\text{d}t=\int1\space\text{d}t\tag1$$
Where $\text{a}\left(t\right)$ is the acceleration function.
Now, for the integrals:


*

*Substitute $\text{u}:=\text{v}\left(t\right)$:
$$\int\frac{\text{v}\space'\left(t\right)}{\text{n}_1-\frac{\text{v}\left(t\right)^2}{\text{n}_2}}\space\text{d}t=\frac{1}{\text{n}_1}\int\frac{1}{1-\frac{\text{u}^2}{\text{n}_1\cdot\text{n}_2}}\space\text{d}\text{u}\tag2$$

*Substitute $\text{p}:=\frac{\text{u}}{\sqrt{\text{n}_1}\cdot\sqrt{\text{n}_2}}$
$$\frac{1}{\text{n}_1}\int\frac{1}{1-\frac{\text{u}^2}{\text{n}_1\cdot\text{n}_2}}\space\text{d}\text{u}=\frac{\sqrt{\text{n}_2}}{\sqrt{\text{n}_1}}\int\frac{1}{1-\text{p}^2}\space\text{d}\text{p}=\frac{1}{2}\cdot\frac{\sqrt{\text{n}_2}}{\sqrt{\text{n}_1}}\cdot\ln\left|\frac{\text{p}+1}{\text{p}-1}\right|+\text{C}_1\tag3$$

*$$\int1\space\text{d}t=t+\text{C}_2\tag4$$


So, we get:
$$\frac{1}{2}\cdot\frac{\sqrt{\text{n}_2}}{\sqrt{\text{n}_1}}\cdot\ln\left|\frac{\frac{\text{v}\left(t\right)}{\sqrt{\text{n}_1}\cdot\sqrt{\text{n}_2}}+1}{\frac{\text{v}\left(t\right)}{\sqrt{\text{n}_1}\cdot\sqrt{\text{n}_2}}-1}\right|+\text{C}_1=t+\text{C}_2\space\Longleftrightarrow$$
$$\left|1+\frac{2}{\frac{\text{v}\left(t\right)}{\sqrt{\text{n}_1}\cdot\sqrt{\text{n}_2}}-1}\right|=\text{C}\cdot\exp\left[2\cdot\frac{\sqrt{\text{n}_1}}{\sqrt{\text{n}_2}}\cdot t\right]\tag5$$
And, when we use $\text{v}\left(0\right)=0$:
$$\left|1+\frac{2}{\frac{\text{v}\left(t\right)}{\sqrt{\text{n}_1}\cdot\sqrt{\text{n}_2}}-1}\right|=\exp\left[2\cdot\frac{\sqrt{\text{n}_1}}{\sqrt{\text{n}_2}}\cdot t\right]\tag6$$
For the limit, try to prove:
$$\lim_{t\to\infty}\space\text{v}\left(t\right)=\sqrt{\text{n}_1}\cdot\sqrt{\text{n}_2}\tag7$$
