Asked by syd
A Ferris wheel is boarding platform is 1 meters above the ground, has a diameter of 70 meters, and rotates once every 5 minutes.
How many minutes of the ride are spent higher than 44 meters above the ground?
How many minutes of the ride are spent higher than 44 meters above the ground?
Answers
Answered by
R_scott
center of wheel is 36m high ... (70 / 2) + 1
how much time >8m above center?
draw a sketch ... diameter is horizontal at 36m
... 8m perpendicular from diameter to edge of wheel
... sine of central angle is ... 8 / 35
180º plus twice central angle is portion of circumference BELOW 44m
how much time >8m above center?
draw a sketch ... diameter is horizontal at 36m
... 8m perpendicular from diameter to edge of wheel
... sine of central angle is ... 8 / 35
180º plus twice central angle is portion of circumference BELOW 44m
Answered by
Reiny
Or , using strictly trig and a sine curve:
period : 2π/k = 5
k = 2π/5
possible equation:
y = 35sin (2π/5 t) + 36
we want our curve to be such that, when t = 0, y = 1, so we need a phase shift.
y = 35sin 2π/5(t + d) + 36
when t=0,y=1
35sin 2π/5(0 + d) + 36 = 1
35sin 2π/5(d) = -35
sin 2π/5d = -1
we know sin 3π/2 = -1
2π/5d = 3π/2
d = 15/4
<b>y = 35sin 2π/5(t + 15/4) + 36</b>
35sin 2π/5(t + 15/4) + 36 > 44
sin 2π/5(t + 15/4) > 8/35
let's look when sin 2π/5(t + 15/4) = 8/35
2π/5(t+15/4) = .23061.. or 2π/5(t+15/4) = π - .23061 = 2.91098
t+15/4 = .183514.... or t+15/4 = 2.316485
t = -3.5665 or t = -1.4335
but the period of our curve is 5 , let's add 5 to our answers to get more
so t = 1.4335 or t = 3.5665
then we are above 44 m from time 1.4335 min to 3.5665 min or for
a period of 2.133 minutes
confirmation:
https://www.wolframalpha.com/input/?i=plot+sin+(2%CF%80%2F5(t+%2B+15%2F4))++%3D+8%2F35+for+t+%3D+0+to+8
period : 2π/k = 5
k = 2π/5
possible equation:
y = 35sin (2π/5 t) + 36
we want our curve to be such that, when t = 0, y = 1, so we need a phase shift.
y = 35sin 2π/5(t + d) + 36
when t=0,y=1
35sin 2π/5(0 + d) + 36 = 1
35sin 2π/5(d) = -35
sin 2π/5d = -1
we know sin 3π/2 = -1
2π/5d = 3π/2
d = 15/4
<b>y = 35sin 2π/5(t + 15/4) + 36</b>
35sin 2π/5(t + 15/4) + 36 > 44
sin 2π/5(t + 15/4) > 8/35
let's look when sin 2π/5(t + 15/4) = 8/35
2π/5(t+15/4) = .23061.. or 2π/5(t+15/4) = π - .23061 = 2.91098
t+15/4 = .183514.... or t+15/4 = 2.316485
t = -3.5665 or t = -1.4335
but the period of our curve is 5 , let's add 5 to our answers to get more
so t = 1.4335 or t = 3.5665
then we are above 44 m from time 1.4335 min to 3.5665 min or for
a period of 2.133 minutes
confirmation:
https://www.wolframalpha.com/input/?i=plot+sin+(2%CF%80%2F5(t+%2B+15%2F4))++%3D+8%2F35+for+t+%3D+0+to+8