Re: Farah's question at Yahoo! Answers ragarding motion of a falling body where drag varies as veloc
Hello Farah,
I would begin with Newton's second law of motion:
$$ma=F$$
Since acceleration $a$ is defined to be the time rate of change of velocity $v$, we may write:
$$m\frac{dv}{dt}=F$$
We have two forces acting on Luke...the force of gravity and drag, which we are told is proportional to his velocity, hence:
$$m\frac{dv}{dt}=mg-kv$$
We subtract the force due to drag since it opposes the motion. Treating $dv$ and $dt$ as differentials, we may separate variables to obtain:
$$\frac{dv}{mg-kv}=\frac{dt}{m}$$
Integrating we find:
$$\int\frac{1}{mg-kv}\,dv=\frac{1}{m}\int\,dt$$
$$\ln(mg-kv)=C-\frac{kt}{m}$$
Converting from logarithmic to exponential form, we find:
$$mg-kv=Ce^{-\frac{kt}{m}}$$ where $$0<C$$
Solving for $v$, we obtain:
$$v(t)=\frac{mg}{k}-Ce^{-\frac{kt}{m}}$$
Now, we are told Luke begins at rest, which means $$v(0)=0$$, and so from this initial value, we may determine the parameter $C$:
$$v(0)=\frac{mg}{k}-C=0\,\therefore\,C=\frac{mg}{k}$$
And so the solution satisfying the IVP is:
$$v(t)=\frac{mg}{k}-\frac{mg}{k}e^{-\frac{kt}{m}}=\frac{mg}{k}\left(1-e^{-\frac{kt}{m}} \right)$$
With $k=4\,\dfrac{\text{kg}}{\text{s}},\,m=100\text{ kg},\,g=10\,\dfrac{\text{m}}{\text{s}^2}$, we then have:
$$v(t)=250\left(1-e^{-\frac{t}{25}} \right)$$
Next, we may determine Luke's position $x(t)$ above the ground by using the definition:
$$\frac{dx}{dt}\equiv v(t)$$
If we orient our vertical $x$-axis such that the origin is on the ground and the positive direction is up, or opposing velocity, we then have:
$$\frac{dx}{dt}=250\left(e^{-\frac{t}{25}}-1 \right)$$
Separating variables, we obtain:
$$dx=250\left(e^{-\frac{t}{25}}-1 \right)\,dt$$
and integrating:
$$\int\,dx=250\int e^{-\frac{t}{25}}-1\,dt$$
$$x(t)=250\left(-25e^{-\frac{t}{25}}-t \right)+C$$
Now using the fact that $x(0)=100000$, we may use this to determine the parameter $C$:
$$x(0)=250\left(-25 \right)+C=100000\,\therefore C=106250$$
And so the solution satisfying the IVP is:
$$x(t)=250\left(-25e^{-\frac{t}{25}}-t \right)+106250=250\left(-25e^{-\frac{t}{25}}-t+425 \right)$$
Now we may answer the questions:
a) Find Luke's velocity and displacement.
$$v(t)=250\left(1-e^{-\frac{t}{25}} \right)$$
$$x(t)=250\left(-25e^{-\frac{t}{25}}-t+425 \right)$$
b) Find Luke's limiting velocity.
$$\lim_{t\to\infty}v(t)=250\,\frac{\text{m}}{\text{s}}$$
c) Suppose it would take 200 seconds for the millennium falcon to shake of the pursuing TIE fighters and turn around for a second rescue attempt. How far from the ground would the rescue be made?
$$x(200)\approx56247.9033585756\text{ m}$$
