Respuesta :
Answer:
The answers to the question is
(a) Jamie is gaining altitude at 1.676 m/s
(b) Jamie rising most rapidly at t = 15 s
At a rate of 2.094 m/s.
Step-by-step explanation:
(a) The time to make one complete revolution = period T = 15 seconds
Here will be required to develop the periodic motion equation thus
One complete revolution = 2Ļ,
therefore the Ā we have T = 2Ļ/k = 15
Therefore k = 2Ļ/15
The diameter = radius of the wheel = (diameter of wheel)/2 = 5
also we note that the center of the wheel is 6 m above ground
We write our equation in the form
y = [tex]5*sin(\frac{2*\pi*t}{15} )+6[/tex]
When Jamie is 9 meters above the ground and rising we have
9 = [tex]5*sin(\frac{2*\pi*t}{15} )+6[/tex] or 3/5 = [tex]sin(\frac{2*\pi*t}{15} )[/tex] = 0.6
which gives sinā»Ā¹(0.6) = 0.643 =[tex]\frac{2*\pi*t}{15}[/tex]
from where t = 1.536 s
Therefore Jamie is gaining altitude at
[tex]\frac{dy}{dt} = 5*\frac{\pi *2}{15} *cos(\frac{2\pi t}{15}) =[/tex] 1.676 m/s.
(b) Jamie is rising most rapidly when Ā the velocity curve is at the highest point, that is where the slope is zero
Therefore we differentiate the equation for the velocity again to get
[tex]\frac{d^2y}{dx^2} = -5*(\frac{\pi *2}{15} )^2*sin(\frac{2\pi t}{15})[/tex] =0, Ļ, 2Ļ
Therefore [tex]-sin(\frac{2\pi t}{15} )[/tex] = 0 whereby t = 0 or
[tex]\frac{2\pi t}{15}[/tex] = Ļ and t = Ā 7.5 s, at 2Ā·Ļ t = 15 s
Plugging the value of t into the velocity equation we have
[tex]\frac{dy}{dt} = 5*\frac{\pi *2}{15} *cos(\frac{2\pi t}{15}) =[/tex] - 2/3Ļ m/s which is decreasing
so we try at t = 15 s and we have [tex]\frac{dy}{dt} = 5*\frac{\pi *2}{15} *cos(\frac{2\pi *15}{15}) = \frac{2}{3} \pi[/tex]m/s
Hence Jamie is rising most rapidly at t = 15 s
The maximum rate of Jamie's rise is 2/3Ļ m/s or 2.094 m/s.
