A Ferris wheel is 25 meters in diameter and boarded from a platform that is 3 meters above the ground. The six o'clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 4 minutes. How many minutes of the ride are spent higher than 18 meters above the ground?
so what fraction of the circle circumference is 18 - 3 = 15 meters above the bottom of the circle?
R = 25/2 = 12.5 meters
15 - 12.5 = 2.5 meters above circle center
A = angle above circle center
sin A = 2.5 / 12.5
A = 11.54 degrees
2 A = 23.07 degrees
fraction above = [360 - 23.07-180 ] / 360= 0.436
0.436 * 4 minutes = 1.743minutes
a = 25/2 = 12.5
period = 2π/k
k = 2π/4 = π/2
so let's start with height = 12.5sin (π/2 t)
this would give us a minimum of -12.5 and a max of 12.5, but we want the
minimum to be 3, so we have to raise our curve 15.5
so next adjustment:
height = 12.5sin(πt/2) + 15.5
but, a rough sketch shows that our min of 3 is at t = -1 , to we have to move
our curve 1 unit to the right:
Final equation: height = 12.5 sin(π/2 (t-1)) + 15.5
( I graphed this using DESMOS, and it is correct)
we want 12.5 sin(π/2 (t-1)) + 15.5 ≥ 18
let's consider 12.5 sin(π/2 (t-1)) + 15.5 = 18
12.5sin(π/2 (t-1)) = 2.5
sin(π/2 (t-1)) = .2
Calculator time , set to radians :
π/2 (t-1) = .201358 or π/2 (t-1) = π - .201358
t-1 = .128188 or t-1 = 1.8718
t = 1.128 or t = 2.8718
between those two times, the rider is ≥ 18 m
(confirmed by DESMOS graph)
So the rider spends 2.8718-1.128 or 1.744 minutes above 18 m.
good one
To find out how many minutes of the ride are spent higher than 18 meters above the ground, we need to analyze the position of the Ferris wheel at different times during its rotation.
First, let's find the radius of the Ferris wheel. The diameter is given as 25 meters, so the radius is half of that, which is 12.5 meters.
Next, let's figure out the highest point of the Ferris wheel. Since it starts at the six o'clock position and completes one full revolution in 4 minutes, we can determine that it reaches the highest point halfway through the ride, at the three o'clock position. At this point, the Ferris wheel is at its maximum height above the ground.
To calculate the height above the ground at the highest point, we need to add the height of the platform (3 meters) to the radius of the Ferris wheel (12.5 meters). This gives us a total height of 15.5 meters above the ground.
Now, we can determine the minutes spent higher than 18 meters above the ground. Since the highest point is at 15.5 meters, we need to figure out how many minutes the Ferris wheel stays above 18 meters.
To do this, we need to find the angle of the Ferris wheel when it is at 18 meters above the ground. Using trigonometry, we can calculate that the angle is equal to the inverse sine of the height above the ground (18 meters) divided by the radius (12.5 meters). This gives us an angle of approximately 63.69 degrees.
Since a full circle is 360 degrees, we can calculate the fraction of the ride spent higher than 18 meters by dividing the angle above 18 meters (63.69 degrees) by the total angle for a full revolution (360 degrees). This comes out to be approximately 0.1775.
Finally, to find the minutes of the ride spent higher than 18 meters above the ground, we multiply the fraction of the ride (0.1775) by the total time for one full revolution (4 minutes). This gives us an answer of approximately 0.71 minutes, which is equal to 42.6 seconds.
Therefore, approximately 0.71 minutes (or 42.6 seconds) of the ride are spent higher than 18 meters above the ground.