Distance between events at relativistic speeds The problem: A spaceship is travelling between Mountain $A$ and Mountain $B$, both of which erupt simultaneously as recorded by a scientist on the ground. The mountains are $300 \; \text{km}$ apart on Earth, and the spaceship is travelling at $v = 0.8c$. 
When you do the transformations to solve for $\Delta t'$, it works out that the event of Mountain $B$ erupts before Mountain A, by $1.334 \; \text{ms}$. The weird thing is, when you solve for how far apart the two events occurred in the $S'$ frame, $\Delta x'$ = $500 \; \text{km}$, because $\Delta x' = \gamma (\Delta x -v\Delta t) = \frac{5}{3}(300 - 0)$. 
I don't understand this, because my teacher says that the ship's odometer doesn't read $\mathbf{500 \; \text{km}}$, even though from its reference frame, the events occurred $\mathbf{500 \; \text{km}}$ apart. Even weirder, length contraction would make the distance look like the mountains are closer than $300 \; \text{km}$, not farther. So what is it, exactly, that is $\mathbf{500 \; \text{km}}$ apart? If the answer is: "the events," then what does that mean if the actual distance is $300 \; \text{km}$ (or less when the ship travels it)?
(Also, if the ship is to travel between A and B, does it really read less miles on the odometer, or does it simply appear to take less time (but still the same distance)?)
This is all quite confusing. Thanks for any insights!
 A: Let's have a look at the situation in the rest frame of the mountains, $S$:

In this frame the two eruptions happen simultaneously. We'll take this to be time zero, so the events for the two eruptions are $(0,0)$ and $(0,300000)$ where I'm writing the event as $(t,x)$.
The rocket passes the first eruption at time zero, so the origins of the two frames $S$ and $S'$ coincide and the position of the eruption $A$ is $(0,0)$ in both frames. To find the position of the eruption in the second frame we use the Lorentz transformations:
$$ \begin{align}
 t' &= \gamma \left( t - \frac{xv}{c^2} \right ) \\
 x' &= \gamma \left( x - vt \right) 
\end{align}$$
And putting $t=0$ and $x=300000$ into this gives us the position of the mountain $B$ eruption:
$$\begin{align}
 t' &= -0.00133 ~ \mathrm{seconds}\\
 x' &= 500000 ~ \mathrm{m}
\end{align}$$
As you say, the time is coming out as $-1.33$ ms so let's draw the situation in frame $S'$ at $t = -0.00133$:

We know the rocket was at mountain $A$ at $t=0$, so if we go back $1.33$ ms the rocket must have been to the left of mountain $A$ by a distance of its speed, $0.8c$, times $1.33$ ms, and this comes to $320$ km.
So what we find is that at $t=-1.33$ ms when the mountain $B$ erupts the distance from the rocket to the mountain, measured in the rocket's rest frame, is $500$km. That's where your figure of $500$ km comes from. At this instance the distance from the rocket to mountain $A$ is $320$ km, and the distance from mountain $A$ to mountain $B$ is $180$ km.
And for completeness we'll draw the situation in $S'$ at time zero:

At time zero the rocket passes mountain $A$ as it's erupting. The distance to mountain $B$ (which erupted $1.33$ ms ago) is now $180$ km.
A: This is what we know as the "rear clock ahead" effect, which is better known as loss of simultaneity. If two events are a proper length $L$ apart, and an observer is traveling at a speed $v$ parallel to the same line, the "rear" event will be ahead of the "front" event by a time $$\frac{Lv}{c^2}$$ So in your case, when your ship is traveling from $A$ to $B$, observer on the ship will see $B$ as the "rear" event, and thus $B$ occurs first.
Lengths are measured by taking two readings at the same time. As you have found out, $A$ and $B$ do not occur at the same time in $S'$ and thus, we need to correct for this "extra" time that results in some "extra" measured distance. Making this correction will then produce the desired length, which would be the measured length had $\mathbf A$ and $\mathbf B$ happened simultaneously in $\mathbf{S'}$. Thus, the actual length between the two mountains, as seen from $S'$, is then $$L' = \gamma (L-v \Delta t) - v\frac{Lv}{c^2} = \gamma(L -0) - \frac{Lv^2}{c^2} = \gamma L(1-\frac{v^2}{c^2}) = \frac{L}{\gamma}$$ which is the standard result of length contraction.
