Question

Please HELP!!
A ferris wheel of diameter 18.5m rotates at a rate of 0.2rad/s. if passengers board the lowest car at a height of 3m above the ground, determine a sin function that models the height \, h, in meters, of the car relative to the ground as a function of the time, t, in seconds.

Answer 1

radius is \(9.25\) and starts at 3 up, lets see
you want \(12+9.25\sin(t)\) almost but that won't quite do.
we need to change the period and also arrange it so at \(t=0\) we get \(-1\) for sine

Answer 2

wait i spoke too soon, that is all wrong, we need \(12.25+9.25\sin(t)\) i think and then we change the period and the phase shift

Answer 3

how do we change the period and phase shift

Answer 4

i think i am messing this up

Answer 5

its ok i cant wrap my head around it either

Answer 6

ok lets try this
if it rotates at a rate of \(.2\) radians per second, how long does it take to make a complete rotation? a complete rotation is \(2\pi\) so since distance is rate times time, we get \(.2t=2\pi\) and so \(t=10\pi\) in other words it takes \(10\pi\) seconds to make a complete rotation
this sounds a little odd to me because it only takes about 31 seconds to go all the way around, but so be it

Answer 7

this tells you that the period is \(10\pi\) i think, so we can make sure that our period is \(10\pi\) because we know the period of \(\sin(bt\)) is \(\frac{2\pi}{b}\)
set \(\frac{2\pi}{b}=10\pi\) solve for \(b\) get \(b=.2\) i must be an idiot today, because i guess we have written the obvious

Answer 8

so it looks like we should be almost done. if we put \(12.5+9.25\sin(.2t)\) we almost have it, but not quite because we want \(-1\) for sine when \(t=0\) not \(0\)

Answer 9

sorry, i meant \(12.25+9.25\sin(.2t)\)

Answer 10

now lets arrange it so at time \(t=0\) we get \(-1\) instead of \(0\) by putting \(.2(t-\frac{5\pi}{2})\) as our input i think this might work

Answer 11

or maybe simpler to write \(.2t-\frac{\pi}{2}\)
that way, when \(t=0\) we get \(\sin(-\frac{\pi}{2})=-1\) and so
\[12.25+9.25\sin(-\frac{\pi}{2})=3\] as needed

Answer 12

so to get a "final answer" i think it is \(21.25+9.25\sin(.2t-\frac{\pi}{2})\) but my brain does not seem to be working too well.
maybe we can check this and see if it is right

Answer 13

neither are my fingers, i meant
\[12.25+9.25\sin(.2t-\frac{\pi}{2})\]

Answer 14

actually it looks like it might be ok
here is a picture of the height vs time
http://www.wolframalpha.com/input/?i=12.25%2B9.25sin%28.2t-pi%2F2%29

Answer 15

starts at a height of 3 when \(t=0\) goes up to a height of \(21.5\) which is good because it is 3 feet above the ground and the diameter is \(18.5\)
and take \(10\pi\) seconds to make a complete rotation, which is the main thing that is bothering me, but that is what is said in the problem

Answer 16

Thank you