Real boat situation.

On simplification, $t= \frac{bv_{br}}{{v_{br}}^2 - v_r^2} = 200$

$\theta$ is the angel between direction of boats velocity and line connecting initial and final position.

$u$ is the speed of the water

$V$ is the speed of the boat relative to the water

$r$ is the distance between boat and the final position

$d$ is the river width

$\frac{dr}{dt}=V-u\sin(\theta)$

$\int_{0}^{d}{dr}=V\int_{0}^{T}{dt}-u\int_{0}^{T}{sin(\theta)dt}$

$d=VT-u\int_{0}^{T}{sin(\theta)dt}$

Horizontal distance boat travels is equal to $0$ , so we can write:

$0=\int_{0}^{T}{udt}-\int_{0}^{T}{Vsin(\theta)dt}$

$0=uT-V\int_{0}^{T}{sin(\theta)dt}$

$\int_{0}^{T}{sin(\theta)dt}=\frac{uT}{V}$

By putting value of this integral in first equation we get:

$d=(V-\frac{u}{V})T$

So the time is equal to:

$T=\frac{Vd}{V^2-u^2}=200s$

I did not use the water's frame of reference, I do not see how that helps.

(x,y) is the position of the boat measured such that the boat starts at (0,0), y is the downstream-direction, and x is the across-stream direction

$\theta$ is the direction of travel, measured such that it is zero at the starting position

$V$ is the speed of the boat relative to the water

$W$ is the speed of the water (relative to the shore)

$\tan (\theta)$ = $\frac{y}{L-x}$

$\frac{dx}{dt}=\dot x=V\cos(\theta)=\frac{V}{\sqrt{1+(\frac{y}{L-x})^2}}$

$\frac{dy}{dt}=\dot y=V\sin(\theta)=\frac{Vy}{(L-x)\sqrt{1+(\frac{y}{L-x})^2}}$

$\frac{dy/dt}{dx/dt}=\frac{dy}{dx}=\frac{W}{V}\sqrt{1+(\frac{y}{L-x})^2}-\frac{y}{L-x}$

I will define $z=\frac{y}{L-x}$ and thus $y=z(L-x)$

$\frac{dy}{dx}=\frac{d}{dx}[z(L-x)]=\frac{dz}{dx}(L-x)-z=\frac{W}{V}\sqrt{1+z^2}-z$

$\frac{dz}{dx}=\frac{W\sqrt{1+z^2}}{V(L-x)}$

$\int \frac{dz}{\sqrt{1+z^2}}=\int \frac{W}{V} \frac{dx}{L-x}$

$arcsinh(z)=\frac{-W}{V}ln(L-x)=arcsinh(\frac{y}{L-x})$

Now we know the path the boat takes

(It is the (x,y) solutions to the above relationship)

And we also know the x-speed as a function of the (x,y) position, $\dot x(x,y)$ , it was the first equation I wrote. Since we know y(x) we can reduce it to $\dot x(x,y(x))=\dot x(x)$

$\dot x=\frac{v}{\sqrt{1+(sinh(\frac{W}{V}ln(\frac{L}{L-x})))^2}}=\frac{v}{cosh(\frac{W}{V}ln(\frac{L}{L-x}))}=\frac{2v}{(\frac{L}{L-x})^{W/V}+(\frac{L}{L-x})^{-w/v}}=\frac{dx}{dt}$

$\int \left((\frac{L}{L-x})^{W/V}+(\frac{L}{L-x})^{-W/V}\right)dx=2v\int dt$

$-L\left( \frac{(1-\frac{x}{L})^{\frac{W}{V}+1}}{\frac{W}{V}+1}+\frac{(1-\frac{x}{L})^{-\frac{W}{V}+1}}{-\frac{W}{V}+1}\right)=2vt+C$

Use the initial condition (x=0,t=0) to find C, and then solve for t when x=L and you get 200 seconds. (I will not do it, this post is long enough.)

I did not solve this alone! A friend helped me, and I would not have done it without them. I just wanted to share the solution