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## Homework Statement

A race boat is traveling at a constant speed v

_{0}= 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(d

^{2}y/dx

^{2}))

*edit* a

_{n}= v

^{2}/ρ

**a**= a

_{t}

**u**

_{t}+ a

_{n}

**u**

_{n}

**v**= v

**u**

_{t}

at t = 0;

**u**

_{t}= -sin90

**i**+ cos90

**j**

## 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

v

^{2}= v

_{0}

^{2}+ 2 a

_{c}(s - s

_{0})

[(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.

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