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What is the radius of a bobsled turn banked at 75.0° and taken at 30.0 m/s, assuming it is ideally banked?

a) 42.4 m
b) 53.1 m
c) 65.8 m
d) 78.6 m

User Superfell
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1 Answer

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Final answer:

The radius of the bobsled turn banked at 75.0° and taken at 30.0 m/s, assuming it is ideally banked, is approximately 42.4 m.

Step-by-step explanation:

To find the radius of a bobsled turn banked at 75.0° and taken at 30.0 m/s, we can use the formula for centripetal force:

F = m * v^2 / r

where F is the centripetal force, m is the mass of the bobsled, v is the velocity, and r is the radius of the turn.

Since the bobsled is ideally banked, we can also use the formula:

r = v^2 / (g * tan(θ))

where g is the acceleration due to gravity and θ is the angle of banking.

Plugging in the values, we get:

r = (30.0 m/s)^2 / (9.8 m/s^2 * tan(75.0°))

Simplifying the equation, the radius of the bobsled turn is approximately 42.4 m.

User Butterscotch
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