How far does an object travel under resisted motion?

Question

A submarine of mass $m$ is travelling at max power and has engines that deliver a force $F$ at max power to the submarine. The water exerts a resistance force proportional to the square of the submarine's speed $v$.

The submarine increases its speed from $v_1$ to $v_2$, show that the distance travelled in this period is $\frac{m}{2k}\ln\frac{F-kv_1^2}{F-kv_2^2}$ where $k$ is a constant.

I've found $\frac{dv}{dt}=\frac{1}{m}(F-kv^2)$ using $\sum{F}=ma$ but I am struggling to progress further using integration or the fact that $\frac{dv}{dt}=v\frac{dv}{dx}=\frac{d}{dx}(\frac12v^2)=\frac{d^2x}{dt^2}$ which is how I've been taught to solve resisted motion questions.

Answer 1

You are close:

$$\frac{mv}{F-kv^2}\frac{dv}{dx}=\frac{m}{-2k}\frac{d\left(F-kv^2\right)}{F-kv^2}\frac{1}{dx}=1$$

And therefore the desired result by integration.

Answer 2

Using chain rule, $\displaystyle \frac{dx}{dv} = \frac{dx}{dt} \cdot \frac{dt}{dv} = \frac{v}{a}$

So, $dx = \cfrac{v}{a} \cdot dv = \cfrac{m v}{F- k v^2} \cdot dv$

Integrating both sides and given $v$ increases from $v_1$ to $v_2$,

$ \displaystyle x = - \frac{m}{2k} \ \ln \left(\frac{F-k v_2^2}{F-k v_1^2}\right) = \frac{m}{2k} \ \ln \left(\frac{F-k v_1^2}{F-k v_2^2}\right)$