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I was given a problem for homework where we needed to calculate the time for a falling object to reach a certain velocity when accounting for drag force. I did it by setting up acceleration as a function of velocity and integrating (it was a differential equation).

However, this is an introductory physics course, with no knowledge of calculus required going in. We haven't even done derivatives yet, strictly speaking. I was fortunate enough to have taken calculus before, so I was able to recognize and solve the differential equation.

When I asked my classmates how they did it, they said they messed around with numbers until they got something that worked (it was online with no points deducted for wrong answers). For most of them, they just divided the terminal velocity by acceleration due to gravity, which makes no sense, since we weren't even asked for time taken to reach terminal velocity, but 63% of it. That method just happened to round to the same number as the correct one.

My question is, is there some way to find this value using elementary physics, or did my professor give us an unfair problem? The TAs weren't any help and I have class during her office hours.

The question itself is as follows:

The terminal velocity of a 4×10$^{-5}$ kg raindrop is about 9 m/s . Assuming a drag force $F_D=−bv$, determine the time required for such a drop, starting from rest, to reach 63% of terminal velocity.

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  • $\begingroup$ Since the answer involves an exponential/logarithm one way or another, one would have to develop some sort of solution involving an exponential/logarithm. Pick your poison... I have a feeling it's going to be some approximation of calculus. $\endgroup$ – CuriousOne Sep 18 '14 at 0:59
  • $\begingroup$ I think a solution involving logarithms would be fair game. We're pretty much expected to know that. The problem is I can't for the life of me think of any way to do this that doesn't involve a differential equation. Maybe it's because I'm used to doing problems that way after taking calculus. If someone could come up with another method, it would be greatly appreciated. $\endgroup$ – HashFail Sep 18 '14 at 2:18
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    $\begingroup$ It's possibly related that 63% is $1 - e^{-1}$ $\endgroup$ – John Rennie Sep 18 '14 at 10:00
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If the drag force is being modeled as a linear function of velocity $(\vec{F}_D=-b\vec{v})$, then the problem is straightforward. The vertical force balance for a falling droplet is $$\Sigma F_y=mg-bv=m\dot{v},$$ which gives the following differential equation for the velocity: $$\boxed{\dot{v}+\frac{b}{m}v=g}.$$ In the limiting case of the maximum velocity/zero acceleration $(\dot{v}=0)$, the force balance simplifies to $$mg=bv_{max},$$ or $$\boxed{v_{max}=\frac{mg}{b}}.$$ Going back to our differential equation, if the initial velocity $v(0)=0$, then the solution to this ODE is $$v(t)=\frac{mg}{b}\left[1-e^{-bt/m}\right].$$ By defining the time constant as $\tau=\frac{m}{b}$ and using the definition of the terminal velocity, the time evolution of the velocity simplifies to $$\boxed{v(t)=v_{max}\left[1-e^{-t/\tau}\right]}.$$ The position, if desired, is found easily enough by performing another integration: $$y(t)=\int{v}dt=v_{max}\int{\left(1-e^{-t/\tau}\right)}dt.$$ Assuming that the initial position $y(0)=0$ and simplifying, the solution for vertical position is then $$\boxed{y(t)=v_{max}t+v_{max}\tau\left[e^{-t/\tau}-1\right]}.$$ So we now have analytical solutions for the acceleration, velocity, and position of the falling object as a function of time and the system parameters, all of which are known (except for $b$). Note, however, that the requested time to reach a speed of $0.63v_{max}$ is not arbitrary. After one time-constant has passed, we will have $$\frac{v(\tau)}{v_{max}}=1-e^{-1}=0.63212=\boxed{63.212\%}.$$ Thus, we simply need to calculate the value of the time constant and the resulting value will be your answer. As regards your classmates, they are not wrong. Our goal is to calculate $\tau$, and if you look carefully at our earlier math you will see that $\tau$ does indeed equal the terminal velocity divided by $g$. Octave plots of the position, velocity, and acceleration functions are included below for reference (replace $k$ with $b$ in the second plot).

enter image description here enter image description here

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  • $\begingroup$ Yeah, we were never taught that equation you linked to. But thanks, this is pretty much exactly what I was looking for. I just wanted to know if there was a more general method to solving this question that we were supposed to be able to figure out, and it looks like the answer is no. $\endgroup$ – HashFail Sep 18 '14 at 20:27
  • $\begingroup$ @JakeChristensen There still may be another way to find your answer, but remember that Calculus (at least Newton's Calculus) was invented to solve Physics problems ;-) $\endgroup$ – Bryson S. Sep 19 '14 at 1:33
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Typically drag is proportional to speed squared, and thus the downwards acceleration is

$$ a = \dot{v} = g - \beta v^2 $$

The solution to such motion is $$\begin{aligned} x & = \int \frac{v}{a} {\rm d}v = - \frac{1}{2\beta} \ln \left( 1 - \frac{\beta v^2}{g} \right) \\ t & = \int \frac{1}{a} {\rm d}v = - \frac{1}{4\sqrt{\beta g}} \ln \left( \frac{(v\sqrt{\beta}-\sqrt{g})^2}{(v\sqrt{\beta}+\sqrt{g})^2} \right) \end{aligned} $$

So plug in the speed $v$ you want to target and it will give you the distance $x$ and $t$ to reach it.

PS. If you don't know the drag parameter $\beta$, but instead you know the top speed, then you can estimate it from the top speed, by solving $ a = g - \beta\, v_{\rm top} = 0 $.

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1) Find the force of drag at terminal velocity. 2) Multiply this force by .63 (63%) 3) Divide this new force by mass of the rain drop 4) Use the velocity acceleration time kinematics equation to solve for time $${(V)=(Vi+a(t))}$$

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  • $\begingroup$ This isn't correct. You assume that acceleration is constant (which it is explicitly not in any question involving changing speeds and air resistance). I'm assuming here that $a(t)$ means $a*t$, since if you mean $a$ as a function of $t$ that makes no sense at all. $\endgroup$ – Chris Oct 9 '18 at 23:12

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