This question is an extension of the following trivial problem:
on a stationary body of water, a motorized device is 100m S and 10m E of a buoy. Given the device can move 5m/s in stationary water, what is the quickest path it can take to the buoy, and how long does it take to traverse?
In this case the line directly between the points - the shortest path - is the quickest and the time can easily be computed using trigonometry.
If a further condition is added though, namely that there is a current drifting N of speed 5m/s the problem becomes more difficult.
Attempting to solve the problem, I break down the speed of the device into horizontal and vertical speeds $v_h(t)$ and $v_v(t)$, and letting $T$ be the whole time of the trip I then obtain the equations
$$100=\int_0^T (v_v(t) + 5)dt = 5T + \int_0^T \sqrt{25-v_h^2(t)}dt$$
$$10=\int_0^T v_h(t)dt$$
With the bounds $0\leq v_h \leq 5$. With these conditions, is it possible to minimize $v_h(t)$ to get a minimal T?
Is there also an easier way to solve this problem?
If the path is a straight line, then you would aim at a point that is $5t$ south of the buoy. And the path you take to get to that point (without a current) would also be $5t$, coincidentally since boat speed = current speed.
Then you solve for $t$ using pythagorean theorem to get $t=10.1$, $v_v=4.9$, and $v_h=0.99$
But, like I commented, no idea if a constant bearing is faster than a curving route.