Answer:
233.48s
3.84 min
Step-by-step explanation:
In order to solve this problem, we can start by drawing what the situation looks like. See attached picture.
We can model this situation by making use of a trigonometric function. Trigonometric functions have the following shape:
[tex]y=A cos(\omega t+\phi)+C[/tex]
where:
A= amplitude =-20m because the model starts at the lowest point of the trajectory.
f= the function to use, in this case we'll use cos, since it starts at the lowest point of the trajectory.
t= time
[tex]\omega=[/tex] angular speed.
in this case:
[tex]\omega=\frac{2\pi}{T}[/tex]
where T is the period, in this case 6 min or
[tex]6min(\frac{60s}{1min})=360s[/tex]
so:
[tex]\omega=\frac{2\pi}{360}[/tex]
[tex]\omega = \frac{\pi}{160}[/tex]
and
[tex]\phi[/tex]= phase angle
C= vertical shift
in this case our vertical shift will be:
2m+20m=22m
in this case the phase angle is 0 because we are starting at the lowest point of the trajectory. So the equation for the ferris wheel will be:
[tex]y=-20 cos(\frac{\pi}{180}t)+22[/tex]
Once we got this equation, we can figure out on what times the passenger will be higher than 13 m, so we build the following inequality:
[tex]-20 cos(\frac{\pi}{180}t)+22>13[/tex]
so we can solve this inequality, we can start by turning it into an equation we can solve for t:
[tex]-20 cos(\frac{\pi}{180}t)+22=13[/tex]
and solve it:
[tex]-20 cos(\frac{\pi}{180}t)=13-22[/tex]
[tex]-20 cos(\frac{\pi}{180}t)=-9[/tex]
[tex]cos(\frac{\pi}{180}t)=\frac{9}{20}[/tex]
and we can take the inverse of cos to get:
[tex]\frac{\pi}{180}t=cos^{-1}(\frac{9}{20})[/tex]
which yields two possible answers: (see attached picture)
so
[tex]\frac{\pi}{180}t=1.104[/tex] or [tex]\frac{\pi}{180}t=5.179[/tex]
so we can solve the two equations. Let's start with the first one:
[tex]\frac{\pi}{180}t=1.104[/tex]
[tex]t =1.104(\frac{180}{\pi})[/tex]
t=63.25s
and the second one:
[tex]\frac{\pi}{180}t=5.179[/tex]
[tex]t=5.179(\frac{180}{\pi})[/tex]
t=296.73s
so now we can build our possible intervals we can use to test the inequality:
[0, 63.25] for a test value of 1
[63.25,296.73] for a test value of 70
[296.73, 360] for a test value of 300
let's test the first interval:
[0, 63.25] for a test value of 1
[tex]-20 cos(\frac{\pi}{180}(1))+22>13[/tex]
2>13 this is false
let's now test the second interval:
[63.25,296.73] for a test value of 70
[tex]-20 cos(\frac{\pi}{180}(70))+22>13[/tex]
15.16>13 this is true
and finally the third interval:
[296.73, 360] for a test value of 300
[tex]-20 cos(\frac{\pi}{180}(300))+22>13[/tex]
12>13 this is false.
We only got one true outcome which belonged to the second interval:
[63.25,296.73]
so the total time spent above a height of 13m will be:
196.73-63.25=233.48s
which is the same as:
[tex]233.48(\frac{1min}{60s})=3.84 min[/tex]
see attached picture for the graph of the situation. The shaded region represents the region where the passenger will be higher than 13 m.
