attached is the graph of your position function ... period is not 15 seconds and the position is always negative. ???

I recommend you take some time to review the basics of trig graphs ...

Graphing Trigonometric Functions

Printable View

- Aug 17th 2011, 01:33 PMskeeterRe: Amusement park problem.
attached is the graph of your position function ... period is not 15 seconds and the position is always negative. ???

I recommend you take some time to review the basics of trig graphs ...

Graphing Trigonometric Functions - Aug 17th 2011, 04:49 PMexplodingtoenailsRe: Amusement park problem.
Why isn't the period 15 seconds?

- Aug 17th 2011, 05:25 PMskeeterRe: Amusement park problem.
- Aug 18th 2011, 01:12 AMexplodingtoenailsRe: Amusement park problem.
That's the problem I had from the beginning...I didn't know how to relate the motion to the height in an equation. I need help.

- Aug 18th 2011, 03:35 AMskeeterRe: Amusement park problem.
graph this equation and compare it to the first graph I posted ...

... then see if you can determine why it models the motion of a seat on the wheel. - Aug 18th 2011, 11:09 AMexplodingtoenailsRe: Amusement park problem.
I'll give my best go at it...

Uh 19 is half of 38, so it would go up 19 and down 19. Would it matter if it was -19 or positive 19?

Also, why is it cosine instead of sine? Would it matter since they are just the same graph with different translations?

2pi is one revolution, times t number of seconds, and it's divided by 15 because a revolution happens every 15 seconds. - Aug 18th 2011, 12:04 PMexplodingtoenailsRe: Amusement park problem.
19sin((2pi/15)t - pi/2) produces the same graph as the one you gave to me. I think I'm finally getting this! :D

- Aug 18th 2011, 01:52 PMexplodingtoenailsRe: Amusement park problem.
a) 19sin((2pi/15)t - (pi/2))

b) 18.6 feet

c) I used the equation 27 = 19sin((2pi/15)t - (pi/2)) + 3 but apparently you can't arcsine anything greater than 1 so I'm not sure how to do this one. - Aug 18th 2011, 02:16 PMskeeterRe: Amusement park problem.
- Aug 18th 2011, 02:19 PMexplodingtoenailsRe: Amusement park problem.
Can you help me out with part c, please? I don't know what I'm doing wrong.

- Aug 18th 2011, 02:43 PMskeeterRe: Amusement park problem.
as mentioned in the problem statement, the t-axis represents the middle of the wheel.

where would the ground be relative to the t-axis?

where would 27 ft above the ground be? - Aug 18th 2011, 04:15 PMexplodingtoenailsRe: Amusement park problem.
The ground would be 3 feet below the 19 foot radius of the wheel.

There's no way to specify where 27 feet above the ground would be. - Aug 18th 2011, 04:39 PMskeeterRe: Amusement park problem.
- Aug 18th 2011, 05:10 PMexplodingtoenailsRe: Amusement park problem.
Uhhh it would be somewhere in the upper part of the curve. The maximum height is 41, so 27 is somewhere above the t-axis.

- Aug 18th 2011, 05:12 PMskeeterRe: Amusement park problem.