system
Given the following post, respond as if you're commenting on Reddit:
human
If time slows the closer to the speed of light someone travels, what is the effect on an airline pilot over the course of their career?

Of course, we'll have to do some assumptions her, and I'll start.

1. The pilot has a 50 year career.
1. The average speed of the pilot over their career will be 300MPH.

Any more assumptions and I'll edit them in with an asterisk.

*And please show the reasoning behind it, so that I can further understand the concepts.
gpt
Since you asked for a step-by-step approach with actual numbers, I'll try to break this down a bit.

The easiest thought experiment here is to imagine a laser-clock that is defined by two mirrors `A` and `B`, separated by a distance L:

    mirror A   mirror B
       |         |
       |--------»|
       |«--------|
       |         |
       |(---L---)|  # mirrors separated by distance L

The clock ticks once every time a laser pulse leaves A, hits B, and returns to A. From the point of view of someone watching the clock when it is at rest relative to them, how long does it take the clock to tick once?

Before we answer that, let's answer a slightly simpler question. You roll a ball towards a wall that is 10 meters away at 4 m/s. The ball bounces off the wall and rolls straight back towards you. How long does it take the ball to return?

Well, if the wall is 10 meters away, then the ball traveled a total distance of 20 meters. And if the ball travels at 4 m/s, then that means it took **5 seconds**, since `20 m / 4 m/s = 5 s`. In other words, we can use the simple kinematics expression of 

    distance / speed = time

to decide the answer. Likewise, with our mirror clock, the light pulse travels at `c`, and the distance is `L`, so the stationary observer measures the clock's period as

    t_0 = 2 L/c

A moving observer, however, sees something different. This time, the observer moves at a speed `v` parallel to the observer's frame of reference.

        mirror A   mirror B
           |         |
         X |--       |
           |  --     |
           |    --   |
           |      --»|
         W |(---L---)| Y
           |       --|
           |     --  |
           |   --    |
         Z |«--      |
           |         |

This time, the light traced a longer path, from `X` to `Y` to `Z`. Clearly, this is a longer route than `2L`. What is the total distance it traveled?

Well, we see that `X`, `Y`, and `Z` form a triangle. Let's call the distance `XY` (which is the same as the distance `YZ`) `h`. Notice that `h` is the hypotenuse of the right triangle `XWY`. Now `h` is easy to calculate. It's just:

    h^2 = XY^2 =      WY^2 + XW^2
    h^2 = XY^2 =      L^2  + XW^2
    h          = sqrt[L^2  + XW^2]

So, similar to before, the light takes `t' = 2 h/c` for a moving observer. But what's XW^2 ? Well, if we moved with speed `v`, then this is `1/2 * v * t'` -- half the distance covered by our movement at speed `v`. So we can express `h` as

    h = sqrt[L^2 + (1/2 * v * t')^2]

If we substitute this into `t' = 2 h/c` and solve for `t'`, then we get the time dilation equation, which is 

                 t_0
    t' = --------------------
          sqrt(1 - v^2 / c^2)

Since the denominator here is a number that will be less than 1 for `v > 0`, a moving observer will observe that the clock ticks more slowly relative to the stationary observer's measurements.

Now, onto actual numbers. Let's pick some nice round numbers so the math will be easier. Also, by picking extremely generous numbers we'll wind up with an convenient upper bound.

So, to be extremely generous, let's say you're a spaceship pilot who travels at a constant 250 m/s for 50 years. Relative to a stationary observer, how much time has passed? Let's crunch the numbers:

    t' = t_0 / sqrt[1 - v^2 / c^2]
    t' = (50 years) / sqrt[1 - (250 m/s)^2/(300,000,000 m/s)^2]

Right away, we can see that 250 m/s divided by the speed of light is not a very big fraction, and so will not have a very big impact on the difference in time the two references measure. Let's keep going:

    t' = (50 years) / sqrt[1 - 6.944 × 10^-13]
    t' = (50 years) / 0.999999999999652777...
    t' = 50 years + ~5 × 10^-4 seconds

In other words, a stationary observer would measure about **an extra 500 microseconds** of time relative to your own clock if all you did was travel at airplane speeds for five decades. To put that in perspective, 3,000 microseconds is roughly the time it takes for a housefly to flap its wings once. Sorry! Relativity's a bummer sometimes.

-----

Perhaps a more satisfying answer to the question would come if we reworded it a bit. What if we wanted to know how fast you need to go before there was an appreciably large difference in times measured? Let's say that we want to have a time difference of at least one year. How fast would our pilot need to travel?

     51 years = 50 years / sqrt[1 - v^2/c^2]

Rearranging this equation and solving for v gives us

     v = ~59,000,000 m/s, or about 0.0002 c

The fastest rocketships we have can go about 15,000 m/s, so if you had one that was about 4,000 times faster, you would be able to gain an extra year after traveling for 50 years.