Michelson and Morley experiment time troubles When discussing the Michelson and Morley experiment (as though the aether existed) we say that the different beams of light would take different times to travel the two distances due to the aether wind and so would arrive out of phase.
I'm having a lot of trouble intuitively seeing how they take different times. I can't seem to grasp that the beam moving against/with the wind will be this take a longer time. I think this is as no equation made from it directly shows this conclusion. When I think about it I always get confused by the fact that the beam would be sped up and then a lower down, so in my mind it would still arrive at the same time. 
If anybody could try and possibly explain why they arrive at different times in a simple way that would be lovely! 
 A: Suppose the aether is flowing past the Earth at a speed $v$, then when the light is travelling with the flow its net speed is $c+v$ and when it's travelling against the flow the net speed is $c-v$.
We'll call the length of the arm $\ell$, so for the trip with the flow the time taken is:
$$ t_1 = \frac{\ell}{c+v} $$
and for the trip against the flow the time taken is:
$$ t_2 = \frac{\ell}{c-v} $$
The average speed is then just the distance travelled, $2\ell$, divided by the total time taken, $t_1 + t_2$:
$$\begin{align}
v_{av} &= \frac{2\ell}{t_1 + t_2} \\
       &= \frac{2\ell}{\frac{\ell}{c+v} + \frac{\ell}{c-v}} \\
       &= \frac{c^2 - v^2}{c} \\
       &= c\,\left(1 - \frac{v^2}{c^2}\right)
\end{align}$$
So even though you're quite correct that one leg of the trip is speeded up and the other leg is slowed down, the average velocity doesn't stay constant.
A: I seem to have misunderstood your question. Your reasoning is correct. If you take time needed to cover a distance whilst going against the aether vs the time needed to cover a distance whilst going with the aether, then the time will be the same. 
Given $Velocity=\frac{Displacement}{Time}$ where $d$ is the distance between mirror and light source

Case 1: Light source pointed with movement of aether

Time needed to complete first leg of journey: 
$t_1=d\div(V_{light}-V_{aether})$
Time needed to complete second leg of journey, after reflection: 
$t_2=d\div(V_{aether}+V_{aether})$

Case 2: Light source pointed against movement of aether

Time needed to complete first leg of journey: 
$t_3=d\div(V_{light}+V_{aether})$
Time needed to complete second leg of journey (after reflection): 
$t_4=d\div(V_{light}-V_{aether})$
Clearly the total time taken in Case 1 is the same as Case 2; $(t_1+t_2)=(t_3+t_4)$ but that's not what the Michelson-Morley experiment is about.
The key here is realising that the experiment's setup made light travel in directions perpendicular to each other through different times. Not one with the aether and one against aether, but perpendicular to each other and then measuring the interference pattern when they recombine. If one of the light beams had "assistance", the timing would be off. They did this by suspending the entire setup in liquid mercury and then rotating it. In other words, one of the light rays would have a different arrival time because it would be influenced by the aether movement, if there was any.
Here is a gif from Wikipedia that illustrates the difference perpendicular light rays would have if there was aether movement.

If there was aether movement in the horizontal direction (like in the gif), then the time taken for the horizontal light beam to reach the beam-splitter/mirror would be 
$T_{horizontal}=t_{right}+t_{left}$ 
where $t_{right}=d\div(V_{light}-V_{aether})$ 
and $t_{left}=d\div(V_{light}+V_{aether})$
giving you $T_{horizontal}=\frac{d(2V_{light})}{V_{light}^2-V_{aether}^2}$
Think of the aether shifting the "endpoint" of the beam, that's why you see the blue dot bounce after the red dot bounces. Whereas the time taken for the vertical light beam to reach the beam-splitter/mirror would be 
$T_{vertical}=t_{up}+t_{down}$ 
where $t_{up}=d\div V_{light}$ 
and $t_{down}=d\div V_{light}$ 
giving you $T_{vertical}=\frac{2d}{V_{light}}$ 
This is clearly different from the horizontal case, making the light beams out of phase and causing some interference. In this case, the horizontal motion of the aether doesn't affect vertical velocity. 
