Fastest hyperloop-like vehicle which a human could endure I'm currently writing on a science-fiction novel where a transportation system like the hyperloop has been built.
Now, I'm wondering how fast I can make this. Since the plot takes place on many different places around the earth it is crucial that the transportation is fast.
It should be possible to drive through Europe (Lissabon->Moscow, 4.500km) in about one hour.
My question is if there could be a Hyperloop that manages to do that (from a technical perspective) and would it be possible to accelerate that Hyperloop fast enough but such that a human would survive the drive?
 A: This answer assumes the following:


*

*There is no practical limit on the maximum speed of the hyperloop-style device; this is at least somewhat realistic, as with enough expense, the main dissipative forces can be almost completely eliminated (for example, rolling friction can be eliminated with magnetic levitation, and air resistance can be eliminated by traveling in a giant vacuum chamber).

*Not much is known about the tolerance of the human body to extremely intense, short accelerations, as this is (perhaps understandably) not a particularly well-tested area of human experience. The designers of this hyperloop-like device would rather not risk fatally squishing their passengers' internal organs into each other, so the acceleration is carried out over a prolonged period.

*The centrifugal acceleration from traveling along a curved path is negligible compared to the translational acceleration.
The maximum acceleration that an average untrained human can survive for more than a short time seems to be around $4g$ (see https://space.stackexchange.com/questions/6154/maximum-survivable-long-term-g-forces), which is $49$ m/s$^2$.
With that in mind, the fastest possible safe transit would be to uniformly accelerate at $4g$ up to some maximum instantaneous speed, and then decelerate at $4g$ until stopped. The total distance covered must be 4500 km. With these two pieces of information, our problem is already totally constrained, and the time that this journey must take is fixed by the limits of the human body.
Suppose we accelerate from rest for a time $t_{accel}$, reaching a maximum speed $4gt_{accel}$. Since we are accelerating and decelerating at the same rate, we must also decelerate for the same amount of time in order to stop, which means that the total time this journey takes is $2t_{accel}$. The total distance $d$ traveled is the sum of the distance traveled when acclerating from rest to $4gt_{accel}$ and the distance traveled when decelerating from $4gt_{accel}$ to rest:
$$d=\frac{1}{2}(4g)t_{accel}^2+(4gt_{accel})t_{accel}-\frac{1}{2}(4g)t_{accel}^2=4gt_{accel}^2$$
Therefore, we have that the total time $t$ that it takes is
$$t=2t_{accel}=2\sqrt{\frac{d}{4g}}\approx680\;\mathrm{s}=11\;\mathrm{minutes}$$
So you can definitely do it in one hour.
A: If the hyperloop track follows the curvature of the earth then the centripetal acceleration required to keep the passengers on the circular path is $a=v^2/r$, so $v=\sqrt{a_{max} R_{earth}}$. The Wikipedia page on g-force quotes the estimate that most people without training and special suits can handle a maximum of about $5$ g's of accelertation, or $5\times 9.8 \simeq 49 \text{ m/s}^2$. That gives a maximum speed of about $17 \text{ km/s}$, not including the forward acceleration required to get up to that speed.
If track is dug through the earth so that the track is straight the only concern is the linear acceleration required to speed up and slow down the vehicle. If we assume that the vehicle speeds up at a constant rate and then halfway through slows down at a constant rate the required acceleration is $a = 4d/t^2$. For your example this gives $a \simeq 1.39 \text{ m/s}^2$, or  $0.14 \text{ g}$, so covering 4,500 km in one hour is manageable.
If you combine the two effects for your example above you get $a_{max} = a_{linear} \times \sqrt{1+(d/R_{Earth})^2}\simeq 0.14 \text{ g}\times 1.22 = 0.17\text{ g}$. Still manageable.
If you put that all together and set $a = a_{max} \simeq 5 \text{ g}$ you can find the shortest feasible trip time for a given distance. For 4,500 km it is about 11 minutes.
