But an average speed of 142 mph is cutting it very close: 95% of top speed. It may be achievable if one can keep the line's curvature and grades down, but it may be difficult if the line is to follow I-15.
I measured the radii of curvature of some of the twistier parts of that freeway's route, and I found numbers like 0.6, 0.7, and 1 mi -- I'll take 0.6 mi or 1 km as the route's minimum radius of curvature. At 150 mph, that curvature will produce about 0.4 g of centrifugal acceleration. It is not as noticeable for highway traffic, which has 0.07 g's for 60 mph / 100 kmh, because the centrifugal acceleration goes as the square of the velocity:
a = v^2/r
where r is the radius of curvature. Using k = 1/r for the curvature, one gets a = k*v^2 The curvature for a curve given by (x(t),y(t)) is
k = (x'*y'' - x''*y')/(x'^2 + y'^2)^(3/2)
where x' = dx/dt, y' = dy/dt, x'' = d^2x/dt^2, and y'' = d^2y/dt^2
If I had the coordinates for positions along a route, like for Vic-LV I-15, I could easily calculate the curvature at each point, and find the most curved parts from that.
According to Design & Build a High Speed Railway: the French Experience
, the maximum tilt height or cant used in French high-speed lines is 180 mm (200 mm for passengers only), and the maximum "cant deficiency" (extra amount needed to make the train's acceleration exactly perpendicular to the tracks) is 100 mm. For 300 kmh, that paper states a minimum radius of curvature of 3793 m / 2.4 mi, which is close to my calculations.
A cant deficiency of 100 mm yields a sideways acceleration of 0.07 g while one of 180 mm yields 0.125 g, with a total of nearly 0.2 g. So the existing I-15 route is too curved for full-speed DXP duty.