Assuming that a sub-orbital flight doesn't greatly increase the distance needed to travel to reach another spot on Earth and half the distance and time to accelerate and the other half to decelerate:
circumference of Earth is 40 x 10^6 m, therefore, greatest distance to travel is half of that or 20 x 10^6 m and half of that is 10 x 10^6 m
s = 10 x 10^6 m
t = 30 minutes = 1800 seconds
s = 1/2 a t^2
a = 2 * s / (t^2)
a = 6 m/s^2 = 0.6 g
So, unless I've screwed up the calculations, it looks like the acceleration/deceleration wouldn't be a problem.
g is about 9.81 m/s^2
therefore a is about 0.63g
So I think your calculations are accurate enough for this discussion - except that the take off acceleration will be 1.6 g (assumes rocket is initially climbing vertically), as the calculated acceleration needs to be added to the Earth's gravity!
Though in practice, the take off acceleration will be a lot higher, as most of the journey will probably be coasting, and then there will be deacceleration at the other end (assuming human a safe landing is required!).
If the entire flight is powered, then the maximum acceleration will be at least 4 times greater (before you allow for the Earth's gravity!).
Note that we are doing rough calculations here, and that we are ignoring many details such:
(1) as air resistance
(2) gravity falls off with height
(3) the rocket is not moving in a straight line
(4) acceleration instantly changes from one value to another
(0, +a, -a)
(5) accelerating slowly burns up more fuel
(6) reduced mass of rocket due to fuel consumption