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Music Freak's question at Yahoo! Answers regarding Newtonian mechanics (linear drag)

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MarkFL

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Feb 24, 2012
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Here is the question:

Differential Equation Help Please?


According to a newspaper account, a paratrooper survived a training jump from 1200 ft when his parachute failed to open but provided some air resistance by flapping unopened in the wind. Allegedly he hit the ground at 100 mph after falling for 8 seconds. Test the accuracy of this account. (Suggestion: Find p(rho) in Equation 4 by assuming a terminal velocity of 100mph. Then calculate the time required to fall 1200 ft.)

And Equation 4: dv/dt= -pv-g where p= the greek symbol rho.

I'm trying to solve this question as a review for my upcoming exam, but I'm not sure how to do this or approach this question. Please show step by step and explain. Thank you.
I have posted a link to this topic so the OP can see my work.
 
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MarkFL

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Feb 24, 2012
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Hello Music Freak,

I believe there to be a typo in your statement of "equation 4."

I would begin with Newton's second law of motion:

\(\displaystyle ma=F\)

Since acceleration $a$ is defined to be the time rate of change of velocity $v$, we may write:

\(\displaystyle m\frac{dv}{dt}=F\)

We have two forces acting on the paratrooper...the force of gravity and drag, which we are told is proportional to his velocity. The force of gravity is in the same direction as his velocity (both are in the downward direction), which the force of drag, opposing the motion, points upwards, hence:

\(\displaystyle m\frac{dv}{dt}=mg-kv\)

where $0<k$ is called the coefficient of drag. Now, if we divide through by the mass of the paratrooper, we obtain:

\(\displaystyle \frac{dv}{dt}=g-\frac{k}{m}v\)

If we define \(\displaystyle \rho=\frac{k}{m}\) then we may write:

\(\displaystyle \frac{dv}{dt}=g-\rho v\) where \(\displaystyle v(0)=v_0\)

Writing the ODE in linear form, we have:

\(\displaystyle \frac{dv}{dt}+\rho v=g\)

Computing the integrating factor, there results:

\(\displaystyle \mu(t)=e^{\rho\int\,dt}=e^{\rho t}\)

and so the ODE becomes, after multiplying through by this factor:

\(\displaystyle e^{\rho t}\frac{dv}{dt}+\rho e^{\rho t}v=ge^{\rho t}\)

Now, the left side may be expressed as the derivative of a product:

\(\displaystyle \frac{d}{dt}\left(e^{\rho t}v \right)=ge^{\rho t}\)

Integrating with respect to $t$, we have:

\(\displaystyle \int\,d\left(e^{\rho t}v \right)=g\int e^{\rho t}\,dt\)

\(\displaystyle e^{\rho t}v=\frac{g}{\rho}e^{\rho t}+C\)

Dividing through by $e^{\rho t}$, we obtain the general solution:

\(\displaystyle v(t)=\frac{g}{\rho}+Ce^{-\rho t}\)

Using the initial condition, we may determine the value of the parameter $C$:

\(\displaystyle v(0)=\frac{g}{\rho}+C=v_0\,\therefore\,C=v_0-\frac{g}{\rho}\)

and so the solution to the IVP is:

\(\displaystyle v(t)=\frac{g}{\rho}+\left(v_0-\frac{g}{\rho} \right)e^{-\rho t}\)

Now, it is easy to see that the terminal velocity is:

\(\displaystyle v_{\max}=\lim_{t\to\infty}v(t)=\frac{g}{\rho}\)

Thus, we may state the solution as:

\(\displaystyle v(t)=v_{\max}+\left(v_0-v_{\max} \right)e^{-\frac{g}{v_{\max}} t}\)

Next, using the fact that the time rate of change of position (or distance fallen at time $t$ which we'll call $x(t)$) is velocity, we have the IVP:

\(\displaystyle \frac{dx}{dt}=v(t)=v_{\max}+\left(v_0-v_{\max} \right)e^{-\frac{g}{v_{\max}} t}\) where \(\displaystyle x(0)=x_0\)

Treating $dx$ and $dt$ as differentials, we may separate variables to obtain:

\(\displaystyle dx=\left(v_{\max}+\left(v_0-v_{\max} \right)e^{-\frac{g}{v_{\max}} t} \right)\,dt\)

\(\displaystyle \int\,dx=\int v_{\max}+\left(v_0-v_{\max} \right)e^{-\frac{g}{v_{\max}} t}\,dt\)

\(\displaystyle x(t)=v_{\max}t+\frac{v_{\max}\left(v_{\max}-v_0 \right)}{g}e^{-\frac{g}{v_{\max}} t}+C\)

Using the initial condition, we may determine the value of the parameter $C$:

\(\displaystyle x(0)=\frac{v_{\max}\left(v_{\max}-v_0 \right)}{g}+C=x_0\,\therefore\,C=x_0-\frac{v_{\max}\left(v_{\max}-v_0 \right)}{g}\)

and so the solution to the IVP is:

\(\displaystyle x(t)=v_{\max}t+\frac{v_{\max}\left(v_{\max}-v_0 \right)}{g}e^{-\frac{g}{v_{\max}} t}+x_0-\frac{v_{\max}\left(v_{\max}-v_0 \right)}{g}\)

\(\displaystyle x(t)=v_{\max}t+\frac{v_{\max}\left(v_{\max}-v_0 \right)}{g}\left(e^{-\frac{g}{v_{\max}} t}-1 \right)+x_0\)

Now, if we use:

\(\displaystyle v_0=0\frac{\text{ft}}{\text{s}}\)

\(\displaystyle x_0=0\text{ ft}\)

\(\displaystyle v_{\max}=100\frac{\text{mi}}{\text{hr}}\cdot\frac{5280\text{ft}}{1\text{ mi}}\cdot\frac{1\text{ hr}}{3600\text{ s}}=\frac{440}{3}\,\frac{\text{ft}}{\text{s}}\)

\(\displaystyle \frac{g}{v_{\max}}=\frac{32.17405\dfrac{\text{ft}}{\text{s}^2}}{\dfrac{440}{3}\, \dfrac{\text{ft}}{\text{s}}}=\frac{1930443}{8800000}\,\frac{1}{\text{s}}\)

\(\displaystyle t=8\text{ s}\)

We then have:

\(\displaystyle x\left(8\text{ s} \right)=\left(\frac{440}{3}\,\frac{\text{ft}}{ \text{s}} \right)\left(8\text{ s} \right)+\frac{\left(\dfrac{440}{3}\,\dfrac{\text{ft}}{\text{s}} \right)^2}{\dfrac{643481}{20000}\,\dfrac{\text{ft}}{\text{s}^2}}\left(e^{-\left(\dfrac{1930443}{8800000}\,\dfrac{1}{\text{s}} \right)\left(8\text{ s} \right)}-1 \right)\)

\(\displaystyle x\left(8\text{ s} \right)=\left(\frac{3520}{3}+\frac{38720000000}{57911329}\left(e^{-\dfrac{1930443}{1100000}}-1 \right) \right)\text{ ft}\approx620.338173295040\text{ ft}\)

So, after 8 seconds, the paratrooper would have fallen only about 620 ft. To find the time required to fall 1200 ft, we may set:

\(\displaystyle x(t)=1200\text{ft}\)

\(\displaystyle 1200\text{ ft}=\left(\frac{440}{3}\,\frac{\text{ft}}{ \text{s}} \right)\left(t\text{ s} \right)+\frac{\left(\dfrac{440}{3}\,\dfrac{\text{ft}}{\text{s}} \right)^2}{\dfrac{643481}{20000}\,\dfrac{\text{ft}}{\text{s}^2}}\left(e^{-\left(\dfrac{1930443}{8800000}\,\dfrac{1}{\text{s}} \right)\left(t\text{ s} \right)}-1 \right)\)

Using a numeric root finding technique, since we cannot solve for $t$ explicitly, we find:

\(\displaystyle t\approx12.442929881626764089\text{ s}\)
 

topsquark

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Aug 30, 2012
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Nicely done! Any day now you will find yourself saying "Math isn't fun any more. I wish I did Physics."

-Dan
 
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MarkFL

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Feb 24, 2012
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Nicely done! Any day now you will find yourself saying "Math isn't fun any more. I wish I did Physics."

-Dan
I have always enjoyed applied math...especially classical physics. (Sun)
 

topsquark

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Aug 30, 2012
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I have always enjoyed applied math...especially classical physics. (Sun)
As do I. Over the past few years I've been drifting closer and closer to Mathematical Physics rather than strictly QM. I've had to learn a heck of a lot of Math outside the usual Physics menu to do QFT. Or maybe it's trying to get a date with Flo. I can't tell the difference any more.

-Dan
 

Ackbach

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Jan 26, 2012
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As do I. Over the past few years I've been drifting closer and closer to Mathematical Physics rather than strictly QM. I've had to learn a heck of a lot of Math outside the usual Physics menu to do QFT. Or maybe it's trying to get a date with Flo. I can't tell the difference any more.

-Dan
Mathematical Physics is where the action is. I would argue that's where the action always has been. Think about it. What are the great unsolved problems in physics? One of the biggies is a firm theoretical foundation for the SM. No mathematician yet has been able to provide a rigorous definition of the Feynman path integral. Or think about the Navier-Stokes equation.

The biggest unsolved problem, I'll admit, is not directly mathematical in nature: reconcile the SM with gravity to achieve a ToE. But no doubt there'll be plenty of mathematics involved there as well.