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babunello [35]
3 years ago
15

In a normal respiratory cycle the volume of air that moves into and out of the lungs is about 508 mL. The reserve and residual v

olumes of air that remain in the lungs occupy about 2020 mL and a single respiratory cycle for an average human takes about 4 seconds. Find a model for the total volume of air V(t) in the lungs as a function of time.
Physics
2 answers:
Amanda [17]3 years ago
8 0

Answer:

V(t)=2020+508sin((pi/2)t))

Explanation:

breathing can be seen as an oscillatory function, so it would be perfect to use a sine or cosine function.

for this reason this is an equation of the form

V(t)=C+Asin(Wt)

as we know that there is always a fixed amount of air in the lungs (2020ml) which indicates that it is a constant amount.

on the other hand, the amount of air that moves in the lungs varies between 0l and 508ml for this reason this will be the amplitude

W=2pi/T

the period (T) is defined as the amount of time it takes for a function to repeat itself, in our case it is 4

W=2pi/4=pi/2

considering all of the above

V(t)=2020+508sin((pi/2)t))

Lynna [10]3 years ago
6 0

Answer:

v(t) = 2274 + 254 sin(\frac{\pi}{2}t)

Explanation:

V_{min} = 2020 ml

V_{max} = 2020 + 508 =2528 mL

Assuming the given function is cyclic function

v_{avg} = \frac{v_{max} - v_{min}}{2} = \frac{2528 + 2020}{2} = 2274 ml

Amplitude = A = \frac{V_{max} - V_{min}}{2}= \frac{2528 - 2020}{2}= 254

time period is 4 sec

we know that \omega  =  \frac{2\pi}{t} = \frac{2\pi}{4} = \frac{pi}{2} rad/s

so function is

V(t) = V_{avg} + Asin (\omega t})

v(t) = 2274 + 254 sin(\frac{\pi}{2}t)

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