A woman is riding a jet ski at a speed of 23.7 m/s and notices a seawall straight ahead. the farthest she can lean the craft in order to make a turn is 26.5°. this situation is like that of a car on a curve that is banked at an angle of 26.5°. if she tries to make the turn without slowing down, what is the minimum distance from the seawall that she can begin making her turn and still avoid a crash?

Question

03-07-2018
Mathematics

contestada

A woman is riding a jet ski at a speed of 23.7 m/s and notices a seawall straight ahead. the farthest she can lean the craft in order to make a turn is 26.5°. this situation is like that of a car on a curve that is banked at an angle of 26.5°. if she tries to make the turn without slowing down, what is the minimum distance from the seawall that she can begin making her turn and still avoid a crash?

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sqdancefan sqdancefan · Answer 1 · 2018-07-03T07:31:11+02:00

If you solve the equations in physics for the radius of curvature, you find it is
r = v²/(g·tan(θ)) . . . . where g is the acceleration due to gravity, 9.8 m/s²

The required turn radius is
r = (23.7 m/s)²/(9.8 m/s²·tan(26.5°)) ≈ 114.96 m

The minimum distance she can begin her turn is 115 m.

_____
This question is in the math section, so we presume the required formula and its constants are known. If this were in the physics section, we might show the derivation of the formula based on the free body diagram of the jet ski, the centripetal acceleration, and the horizontal force vector resulting from the bank angle.