Chapter 5
Modeling with Differential Equations
5.1 Raindrops
5.1.3 The Velocity-Squared Model
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For large raindrops, specifically, those with diameter inches, the component of acceleration due to air resistance is found to be proportional to the square of velocity. That is, Stokes' Law is replaced by
In this case, experimental determination of the proportionality constant shows that \(c\), in units of feet and seconds, is approximately
This model is called the Velocity-Squared Model.
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The Velocity-Squared Model, like Stokes' Law, implies a terminal velocity . Use the Velocity-Squared differential equation to calculate as a function of the diameter \(D\). (Be careful with the units.)
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What is the terminal velocity of a raindrop of diameter \(0.05\) inches? How does that compare with the drizzle drop? If the raindrop falls \(3000\) feet at terminal velocity, how long does it fall?
Answer the questions in part (b) for drops with twice the diameter.
Answer the questions in part (b) for drops with four times the diameter.
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The fastest known times for raindrops to fall \(3000\) feet are between one and two minutes. Are your answers consistent with that fact?
We won't attempt to find a symbolic solution of the Velocity-Squared equation until later. In the meantime, we can state what you will find later as the solution: The function defined by
satisfies the Velocity-Squared equation and has the right initial value.
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Does the formula just stated as the solution confirm your calculation of the terminal velocity, , in Activity 4?
Use the solution formula to write a formula for the ratio
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How long does it take each of the raindrops in parts (b), (c), and (d) of Activity 4 to reach \(99\)% of terminal velocity?
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Was it reasonable to assume in parts (b), (c), and (d) of Activity 4 that these drops travel at terminal velocity for their entire fall? Why or why not?