Related Rates, two ships 
There are two ships, A and B.
  Ship B is 32 miles south of Ship A.
  Ship A is travelling south at 16 mph. Ship B is travelling east at 12 mph. 
  After how long do the ships cease to approach one another and how far apart are they when this happens?

Basically, I drew a diagram putting Ship B southwest of Ship A. I found how long it would take Ship A to cross the 32 miles (2 hours). Then I found how long Ship B would have travelled in that time (24 miles).
Next, I designated point C as the point where they stop approaching one another (the vertex of the triangle with the right angle). I designated variable x to AC, y to BC, and z to AB. Then I solved for Z using the pythagorean theorem. Then I differentiated all the variables with respect to time. I substituted 32 miles for the x value, and the speeds 16 mph and 12 mph respectively for dx/dt and dy/dt. Finally, I recognized that when they "cease to approach one another", the length z, which was previously decreasing, goes to zero (and then begins to increase). Thus I substituted in zero for dz/dt. Finally I solved for y. I get approximately 42.67 miles. This is how far ship B started from point C! Since I already found that it travels 24 miles within the 2 hour timeframe, the final answer I gave for the distance between the ships when they cease to approach each other was 18.67 miles.

I just need to know if this is right LOL. If not, what is wrong with my approach?
My approach was in fact wrong simply due to the way I set up the question. B is DIRECTLY SOUTH of A! When it says 'cease to approach' all that means is that the distance between the ships is at a minimum! There's a discussion in the comments if you want to check it out. Thanks guys :)
 A: Ship A is only moving along $y$ axis. The position is given by $x_A=0$ and $$y_A=32-16t$$
Ship B is only moving along $x$ axis. The position is given by $y_B=0$ and $$x_B=12t$$
The distance squared is $$d^2(t)=(x_A-x_B)^2+(y_A-y_B)^2=(12t)^2+(32-16t)^2$$
The ship stop approaching when the absolute value of the distance (or distance square) have a minimum.
Just set the derivative to zero: $$\frac{d}{dt}d^2(t)=0$$
Now calculate $t$. Take this value, and plug it back into the equation for the distance.
A: Here is perhaps a simpler approach. Put $B$ initially at the origin, traveling along the positive $x$-axis at $12$ mph, and $A$ at $(0,32)$ initially, travelling at $-16$ mph. 
Then, at any time $t \in [0,2]$, you have $A(t) = (0, 32-16t)$ and $B(t) = (12t,0)$, so the square of the distance between them is given by
$$
d(t) = (32-16t)^2 + (12t)^2.
$$
It's clear $d(0) = 32^2$ and $d(2) = 24^2$. Can you find where does $d$ attain its minimum?

UPDATE
After you posted your answer, I will complete this to show the solution without Calculus techniques. We have
$$
d(t) = (12^2+ 16^2) t^2 - 2 \cdot 32 \cdot 16 t + 32^2, 
$$
which is a parabola opening up, so it attains its minimum at its vertex, which occurs at
$$
t_\min
 = - \frac{-2\cdot 32 \cdot 16}{2\cdot \left(12^2+ 16^2\right)}
 = \frac{32}{9 + 16}
 = 1\frac{7}{25}
 = 1.28 \text{ hours}.
$$
You can now find the actual distance as $d(t_\min)$.
