1. The following graph shows the height of a chair on a Ferris wheel as the wheel rotates.
How long does the Ferris wheel take to make 5 revolutions?
a. 100 s
b. 10 s
c. 20 s
d. 4 s
2. A ferris wheel takes 32 seconds to complete one revolution of the ride. The maximum height of the wheel is 23 metres and the minimum height is 1 metres.
How high is the hub or center of the ferris wheel off the ground?
3. A ferris wheel takes 33 seconds to complete one revolution of the ride. The maximum height of the wheel is 17 metres and the minimum height is 3 metres. If a trigonometric equation were created (using radians) to represent the height of the ferris wheel as a function of time, what would be the value of 'b' for the equation, to the nearest hundredth?
4. The distance above the ground of a passenger on a circular ferris wheel is given by the equation, h(t)= 5sin[0.22(t - 6)] + 2
where h, is the distance above the ground, in metres and t, is the time, in seconds, after the passenger passes the lowest point of the ride for the first time.
The distance of the passenger above the ground 6 seconds after passing the lowest point of the ride, to the nearest tenth of a metre, is
5. On a typical day at an oceanport, the water has a maximum depth of 22 m at 6:00am. The minimum depth of 14 m occurs 6.1 hours later. Assume that the relationship between the depth of the water and time is a sinusoidal function. The equation to represent this function can be written in the form y = a cos[b(t - c)] + d.
a) Determine the positive values of a, b, c and d.
b. What is the depth of the water, to the nearest tenth of metre, at 11:18 am?
c. At what time, to the nearest tenth of hour, does the water reach a depth of 15 m for the third time on the same day? Record your answer in 24 hour time system.
6. The pedals of a bicycle are mounted on a bracket whose centre is 28 cm above the ground. Each pedal is 15 cm from the centre of the bracket. Assuming that the bicycle is pedalled at 20 cycles per minute and that the pedal starts at time t = 0 s at the topmost position.
The equation to represent this function can be written in the form y = a cos[b(t - c)] + d, where y is the height of the pedal from the ground in cm and t is the time in seconds.
a. Determine the positive values of a, b, c and d.
b. Use your values from the previous bullet to find the height, to the nearest tenth of cm, of the pedal above the ground at time t = 5 seconds.
At what time, to the nearest tenth of second, does the pedal reaches its lowest position for the third time?
I think 1 is c. and the rest do not know how to do it.
That's a lot of questions in one post.
You are correct - the answer to number 1 is (c). You can see from the graph that it does one revolution in 4 seconds, so it will do 5 revolutions in 20 seconds.
All of the rest of your answers have to do with your equation y = a cos[b(t - c)] + d: finding or interpreting values for a, b, c, and/or d. So what do these parameters represent?
a is the amplitude - one half the difference between the maximum and minimum.
d is the average value - (max + min)/2
b has to do with how fast y goes up and down - it is divided by the length of a whole cycle
c is used to adjust the graph to the right or left - it is one of the values of t where y is at its maximum
I hope this helps you on your way. If you are having problems with one of the questions in particular, please post again.
I got answers for 2,3,4. Please check if this is right.
Originally Posted by hollywood
Please help me with 5 and 6 I do not know how to do it.
The answer to number 2 is 12 meters. You have the distance from the end of the Farris Wheel to the center (which is 11 meters), but they are asking from the distance between the center and the ground (which would be 11 meters + the 1 meter the Farris wheel is raised off the ground).
In 5 and 6, you are still using the equation y = a cos[b(t-c)] + d. In the first part of the problem, you identify a, b, c, and d, and in the subsequent parts you are given either y or t and need to find the other.
In problem 5, you have the max, min, and time of max, but instead of the whole cycle time, you are given the time from a max to a min, which is half the cycle time. In problem 6, you are given a and d - the height of the bracket and the distance from the bracket to the pedal. The cycle time is 1/20, since there are 20 cycles per minute.
Once you have the equation, it should be easy to get 5(b) and 6(b) - just plug in the values. In 6(c), it reaches a minimum at 1/2 cycle (and 3/2, 5/2, 7/2, etc.).
Problem 5(c) might be a little tricky. It's at a minimum 6.1 hours before 6:00am, so that's 0.1 hours before midnight. If it gets from the minimum of 14m to 15m before midnight, the first time at 15m that day will be just before the minimum at about noon, and the third time at 15m will be late that night. I think you'll find that it is still below 15m at midnight, so the first time at 15m will be a little after midnight, and the third time will be a little after the minimum at about noon.