Yes quite right 212 total agree tried it before with not traction
rearrange and forget about U^2 (zero)
s= v^2 / 2a
then for 4000fpm (roughly 20m/s) stopping at acceleration (a) of 1g ((10m.s^-20) additional to usual g) then the distance required is
v^2 / 2 a = 400/20 = 20m
20m is not that dramatic at 1 g
at ROD 3000fpm stopping in 20m gives
a=v^2/2d = 225/40 = 5.6m^s^-2 is about 1/2 'g' ... no big deal