Engineering Dynamics: Normal-Tangential Components

[ben.tien]

Homework Statement

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 v_f = 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.

Homework Equations

ρ(x) = ([1 + (dy/dx)²]^3/2)/(absvalue(d²y/dx²))

*edit* a_n = v²/ρ

a = a_tu_t + a_nu_n

v = vu_t

at t = 0;
u_t = -sin90i + cos90j

The Attempt at a Solution

So the distance traveled by the boat is 1/4 of a circle, s = rθ = ρ($\pi$/2)

converting mph to ft / s
v² = v₀² + 2 a_c(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.

 

[PhanthomJay]

The equation you have written assumes tangential acceleration only at -1g. You want the total resultant acceleration (centripetal plus tangential, vectorially added) not to exceed 2g's.