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 = 2π,
therefore the we have T = 2π/k = 15
Therefore k = 2π/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 sin⁻¹(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 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, π, 2π
Therefore [tex]-sin(\frac{2\pi t}{15} )[/tex] = 0 whereby t = 0 or
[tex]\frac{2\pi t}{15}[/tex] = π and t = 7.5 s, at 2·π 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] - 2/3π 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 2/3π m/s or 2.094 m/s.
