The diameter of a large Ferris wheel is 48 meters and it takes 2.8 minutes for the wheel to complete one revolution. A rider gets onto the wheel at its lowest point which is 60 cm above ground at t = 0.
a) Find a sinusoidal function h(t) that gives the height h, in meters, of the rider above ground as a function of the time t in minutes.
b) Find the time intervals for which the rider is at a height less than 30 meters for the period of time from t = 0 to t = 2.8 minutes.
c) How many minutes, from t = 0, does it take the rider to reach the highest point for the second time?
Answers
Solution
a) The minimum height hmin above the ground is 0.6 meters. The maximum height hmax is equal to the minimum height plus the diameter of the weel.
hmax = 0.6 + 48 = 48.6
Since h(t) is minimum at t = 0, it would be easier to model it by a reflected cos(x) function. Hence
h(t) = a cos [b(t - d)] + c
|a| = (hmax - hmin) / 2 = -(48.6 - 0.6) / 2 = 24 , two solutions for a = ~+mn~24
We take a = -24 where the minus sign account for the reflection on the horizontal axis.
c = (hmax + hmin) / 2 = (48.6 + 0.6) / 2 = 24.6
The period = 2.8 = 2π/b , hence b = 2π/2.8
h(t) = - 24 cos [ (2π/2.8)t ] + 24.6
Check that a t = 0 , h(0) = 0.6 m the minimum height and a t = 1.4 (half a period later) , h(1.4) = - 24 cos ( (2π/2.8) 1.4 ) + 24.6 = - 24 cos ( π ) + 24.6 = 48.6 m is maximum.
b) We first need to solve the equation
- 24 cos [ (2π/2.8)t ] + 24.6 = 30
cos [ (2π/2.8)t ] = (30 - 24.6) / (-24)
t = 2.8 arccos [ (30 - 24.6) / (-24) ] / (2 π) = 0.8 mn
The solution found corresponds to t1 which is the intersection of the graph of h(t) and y = 30.
hence t1 = 0.8 mn and t2 = 2.8 - 0.8 = 2 mn.
h(t) is less than 30 from t =0 to t = 0.8 mn and from t = 2 to t = 2.8 mn a total of 1.6 mn.
Graph of y = h(t) and y = 30
c) The rider reaches the maximum at t = half a period for the first time and t = half a period + one period the second time. Hence it takes
(1/2)2.8 + 2.8 = 4.2 mn for the rider to reach the maximum for the second time.
Answer:
a) The minimum height hmin above the ground is 0.6 meters. The maximum height hmax is equal to the minimum height plus the diameter of the weel.
hmax = 0.6 + 48 = 48.6
Since h(t) is minimum at t = 0, it would be easier to model it by a reflected cos(x) function. Hence
h(t) = a cos [b(t - d)] + c
|a| = (hmax - hmin) / 2 = -(48.6 - 0.6) / 2 = 24 , two solutions for a = ~+mn~24
We take a = -24 where the minus sign account for the reflection on the horizontal axis.
c = (hmax + hmin) / 2 = (48.6 + 0.6) / 2 = 24.6
The period = 2.8 = 2π/b , hence b = 2π/2.8
h(t) = - 24 cos [ (2π/2.8)t ] + 24.6
Check that a t = 0 , h(0) = 0.6 m the minimum height and a t = 1.4 (half a period later) , h(1.4) = - 24 cos ( (2π/2.8) 1.4 ) + 24.6 = - 24 cos ( π ) + 24.6 = 48.6 m is maximum.
b) We first need to solve the equation
- 24 cos [ (2π/2.8)t ] + 24.6 = 30
cos [ (2π/2.8)t ] = (30 - 24.6) / (-24)
t = 2.8 arccos [ (30 - 24.6) / (-24) ] / (2 π) = 0.8 mn
The solution found corresponds to t1 which is the intersection of the graph of h(t) and y = 30.
hence t1 = 0.8 mn and t2 = 2.8 - 0.8 = 2 mn.
h(t) is less than 30 from t =0 to t = 0.8 mn and from t = 2 to t = 2.8 mn a total of 1.6 mn.
Graph of y = h(t) and y = 30
c) The rider reaches the maximum at t = half a period for the first time and t = half a period + one period the second time. Hence it takes
(1/2)2.8 + 2.8 = 4.2 mn for the rider to reach the maximum for the second time.
Step-by-step explanation: