Final answer:
The elevation speed of a rider when they are 24 meters above the ground on a ferris wheel with a diameter of 30 meters, making 1 revolution every 2 minutes, is approximately 1.45 m/s.
Step-by-step explanation:
To find the speed at which the elevation of a rider is rising, we need to calculate the linear velocity of the rider at a height of 24 meters above the ground.
First, we need to calculate the linear velocity of the ferris wheel. The circumference of the ferris wheel can be found using the formula C = πd, where d is the diameter. In this case, the diameter is 30 meters, so the circumference is 30π meters.
Next, we need to find the time it takes for the ferris wheel to complete one revolution. In this case, it takes 2 minutes, which is equal to 120 seconds.
Now we can calculate the linear velocity of the ferris wheel using the formula v = C/t. Substituting the values, we have v = (30π)/120 m/s.
Finally, we can find the speed at which the elevation of a rider is rising when they are 24 meters above the ground. To do this, we need to find the linear velocity of a rider at that height.
Since the linear velocity of the ferris wheel is constant, we can use the equation v = ωr, where ω is the angular velocity and r is the radius. Rearranging the equation, we have ω = v/r. Substituting the known values, we have ω = ((30π)/120)/15 rad/s.
Now we can use the linear velocity of the rider to find their elevation speed. At 24 meters above the ground, the radius of the ferris wheel is 24 + 5 = 29 meters. Multiplying the angular velocity by the radius, we have ((30π)/120)/15 * 29 m/s.
Simplifying the expression, the elevation speed of the rider when they are 24 meters above the ground is approximately 1.45 m/s.