I have a question over the calculation of air resistance. When you're calculating the air resistance of a bullet, it's velocity is decreasing over time. The air resistance is also dependent on the velocity and thus, the air resistance is decreasing as the velocity decreases. Overall, the velocity is dependent on the acceleration which is dependent on the velocity. Is there a way to overcome this without using multivariable calculus?
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2$\begingroup$ Well, If you want to model anything in real life, you got to use calculus. $\endgroup$– QuIcKmAtHsDec 17, 2018 at 13:30
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$\begingroup$ It's not multivariable calculus, but it's second-order non-linear calculus, which isn't easy to solve. $\endgroup$– AetolDec 17, 2018 at 13:41
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$\begingroup$ Acceleration is a double derivative of position. Speed a single-derivative of position. Combine those into one equation and you have a differential equation - there is unfortunately not really a way around that. Learning differential equations, with all its calculus, is essential. $\endgroup$– SteevenDec 17, 2018 at 14:31
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1$\begingroup$ Your job may be more difficult than you realize. Most bullets start out supersonic, then go sub-sonic as they lose speed. I don't know the equations, but I expect that you need different equations when dealing with supersonic, trans-sonic, and subsonic velocities. $\endgroup$– David WhiteMar 31, 2022 at 22:09
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$\begingroup$ You don't need multivariate calculus, just plain calculus as everything is a function of velocity. $\endgroup$– John AlexiouSep 16, 2023 at 4:40
2 Answers
In a problem where the only force is air resistance (neglecting gravity), the (one-dimensional) equation of motion is $$F=m\ddot{x}=f(v),$$ where the $|f(v)|$ is the magnitude of the velocity-dependent aerodynamic friction, and the sign of $f(v)$ is always opposite that of $v$, so that the friction works to decrease the speed.
Rewriting $v=\dot{x}$, the equation becomes $$m\dot{v}=m\frac{dv}{dt}=f(v),$$ which is a first-order, separable, ordinary differential equation. The solution can be reduced to performing integrals and inverting the resulting functions. An analytic solution may be found if the nontrivial integral in $$m\int\frac{dv}{f(v)}=\int dt+C$$ can be done in closed form. This yields $v$ as a function of $t$, and $v(t)$ can be integrated one more time to give $x(t)$.
Any time you can describe the acceleration (or deceleration) of an object as a function of speed $v$ only, then plain old calculus allows for direct integration to estimate time distance and time needed to vary the speed.
Given $a(v)$ use
$$ \Delta t = \int_{v_1}^{v_1+\Delta v} \frac{1}{a}\, {\rm d}v \tag{1}$$
and
$$ \Delta x = \int_{v_1}^{v_1+\Delta v} \frac{v}{a}\, {\rm d}v \tag{2}$$
where $v_1$ is the initial speed, $\Delta v$ is the step in speed that is being investigated, $\Delta t$ is the time span needed for this change in speed, and $\Delta x$ is the distance span needed for $\Delta v$ to occur.
Specifically with the problem of air resistance, if we assume $a = -\beta\, v^2$ with $\beta$ some constant, then
$$ \Delta t = \int_{v_1}^{v_1+\Delta v} \frac{1}{-\beta v^2}\, {\rm d}v =- \frac{\Delta v}{\beta v_1 (v_1 + \Delta v)} \tag{3} $$
and
$$ \Delta x = \int_{v_1}^{v_1+\Delta v} \frac{v}{-\beta v^2}\, {\rm d}v = - \frac{1}{\beta} \ln \left( \frac{v_1+\Delta v}{v_1} \right) \tag{4}$$
Notice the negative sign in front of the expressions means that only negative values for $\Delta v$ yield positive values for $\Delta x$ and $\Delta t$.
Since the problem presents itself a bit backwards (as time is a function of speed), you can invert (3) to get
$$\Delta v = -\frac{\beta v_1^2 \Delta t}{1+\beta v_1 \Delta t} \tag{5} $$
and then plug into (4) to get
$$\Delta x = \frac{1}{\beta} \ln \left( 1 + \beta v_1 \Delta t \right) \tag{6}$$
The above expressions have now time as the independent variable, but caution is needed as this inversion is not always possible for every $a(v)$.
An auxiliary kinematic expression is
$$\Delta v = v_1 \left( \hat{e}^{-\beta \Delta x}-1 \right) \tag{7}$$
which directly relates speed and time.
Proof of (1) follows from ${\rm d}v = a \, {\rm d}t$ of which you divide both side by $a$ and integrate the equation
Proof of (2) follows from $ {\rm d}x = v \, {\rm d} t = \frac{v}{a} \, {\rm d}v$ by using the rhs of the above expression to substitute for ${\rm d}t$.