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.
Answer
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).
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