A woman is riding a Jet Ski at a speed of 30 m/s and notices a seawall straight

Question

A woman is riding a Jet Ski at a speed of 30 m/s and notices a seawall straight ahead. The farthest she can lean the craft in order to make a turn is 23

A woman is riding a Jet Ski at a speed of 30 m/s and notices a seawall straight ahead. The farthest she can lean the craft in order to make a turn is 23 degree . This situation is like that of a car on a curve that is banked at an angle of 23 degree . 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?

Explanation / Answer

This problem is analogous to the classic problem of a car rounding a banked curve. The equation for centripetal force in 2 dimensions is

(1) Fc = m(v^2)/r

The woman leaning on the jet ski is like the car sitting on a banked road of angle (b). The horizontal force pushing the jet ski in towards the center of the curve acts as the centripetal force and is given by

(2) Fc = Fn sin(b), where Fn is the normal force

Furthermore, the normal force on the incline is related to the gravitational force by

(3) Fn = mg / cos(b)


Substitute equation 3 into equation 2 to get

(4) Fc = mg tan(b)

You can now set equations 1 and 4 equal to each other

m(v^2)/r = mg tan(b)

30^2/r = 9.8*tan23

r = 216.35

(v) is the woman's initial velocity, (b) is the angle at which she leans on the jet ski, and you are solving for the turning radius (r)

The final answer should be about 216.35 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.