I need help with a ferris wheel sin function problem

[kaebun]

ok so heres the problem..... A ferris wheel 50 ft in diameter makes one revolution every 40 sec. if the center of the wheel is 30 ft above the ground, how long after reaching the low point is a rider 50 ft above the ground? ... ok i started doing my hmk in class because thats what i try to do if i can understand it so i figured out how to find the amplitude it 25 and the movement up 30, then my teacher went over that stuff but didn't bother to tell us how to find the period from 40sec the only part i couldn't figure out myself grrrrrr. So anyway thats what i need help with what i have is f(x)=25sin(bx)+30 is that right and how do you find b or the period i can find b from that ... thanx

 

[Gene]

Close. It goes 2*pi radians in 40 seconds so that is 2pi/40 radians per second.
A better function is f(t)=-25cos(t*pi/20) + 30.
The height at t=0 seconds is 5 feet. The height at 20 seconds is 55 feet.

 

[galactus]

We can try it another way:

Using the triangle in the drawing, we can use \(\displaystyle 180-tan^{-1}(3/4)=143.13\)degrees is the arc swept out from the low point to the point 50 feet above the ground.

Since the ferris wheel rotates at \(\displaystyle 2{\pi}\) radians per 40 seconds, we have:

\(\displaystyle \frac{143.13}{360}(40)=15.9\) seconds.

So, a quick equation would be: \(\displaystyle \frac{180-tan^{-1}(\frac{3}{4})}{9}=15.9\)