Question

A Ferris Wheel is built such that the height in feet above ground of a seat on the wheel at time t seconds can be modeled by h(t) = 53 + 50sin((pi/16)t - (pi/2)) The wheel makes one revolution every 32 seconds. The ride begins when t = 0. During the first 32 seconds of the ride, when will a person on the Ferris Wheel be 53 feet above the ground?

Answer 1

Answer:

at time, t = 8 seconds and t = 24 seconds Ferris Wheel be 53 feet above the ground

Step-by-step explanation:

Data provided in the question:

height in feet above ground of a seat on the wheel at time t seconds is

modeled as

h(t) = $53 + 50\sin((\frac{\pi t}{16} - \frac{\pi}{2})$

now,

at height 53 above the ground, we get the equation as:

53 = $53 + 50\sin(\frac{\pi t}{16} - \frac{\pi}{2})$

or

$50\sin(\frac{\pi t}{16} - \frac{\pi}{2})$ = 53 - 53

or

$\sin(\frac{\pi t}{16} - \frac{\pi}{2})$ = 0

also,

sin(0) = 0

and,

sin(π) = 0

therefore,

$(\frac{\pi t}{16} - \frac{\pi}{2})$ = 0

or

$\frac{\pi t}{16} = \frac{\pi}{2}$

or

t = 8 seconds

and,

$(\frac{\pi t}{16} - \frac{\pi}{2})$ = π

or

$\frac{\pi t}{16}= \pi + \frac{\pi}{2}$

or

$\frac{\pi t}{16}= \frac{3\pi}{2}$

or

t = 24 seconds

Hence,

the at time, t = 8 seconds and t = 24 seconds Ferris Wheel be 53 feet above the ground