A race boat is traveling at a constant speed v0 = 130 mph when it performs a turn with constant radius ρ to change its course by 90°. The turn is performed while losing speed uniformly in time so that the boat's speed at the end of the turn is vf = 116 mph. If the magnitude of the acceleration is not allowed to exceed 2g, where g is the acceleration due to gravity, determine the tightest radius of curvature possible and the time needed to complete the turn.
ρ(x) = ([1 + (dy/dx)2]3/2)/(absvalue(d2y/dx2))
*edit* an = v2/ρ
a = atut + anun
v = vut
at t = 0;
ut = -sin90i + cos90j
The Attempt at a Solution
So the distance traveled by the boat is 1/4 of a circle, s = rθ = ρ([itex]\pi[/itex]/2)
converting mph to ft / s
v2 = v02 + 2 ac(s - s0)
[(170.13)2 - (190.67)2] / 2(32.2 ft/ s ^2) = s
s = 57.5 ft = 17.5 meters
ρ = 57.5 ft * 2 / pi
I feel like I'm missing something BIG because it can't be this simple.