Modeling Motion with Velocity-Dependent Forces

Hello! Welcome to your second lesson in Foundations of Calculus-Based Mechanics.

In our last lesson, we established how to use integration to solve kinematics problems where acceleration was a given function of time, position, or velocity. We saw that the case of a(v) is particularly important. Today, we will explore the physical origin of such acceleration by applying Newton's second law to systems where forces depend on velocity.

This lesson directly addresses the learning outcome: Formulate and solve differential equations of motion for systems with velocity-dependent forces, such as air resistance. We will see how a physical principle (F=ma) translates into a differential equation and then use your calculus skills to solve it, revealing the complete motion of the object.

Our journey today will cover:

The two primary physical models for drag force: linear and quadratic.
Formulating and solving the first-order differential equation for an object moving under gravity and linear air resistance.
Analyzing the concept of terminal velocity.
Extending the analysis to the more complex case of 2D projectile motion with air resistance.
A brief look at why quadratic air resistance poses a much greater mathematical challenge.

1. The Physics of Drag Force

When an object moves through a fluid (like air or water), it experiences a resistive force called drag. This force always opposes the object's velocity. Unlike the simple kinetic friction you've studied, drag is not constant; it depends on the object's speed.

There are two main models used in physics, depending on the conditions:

Linear Drag (Low Speed Regime): For small objects moving at low speeds, the drag force, $F_{D}$ , is approximately proportional to the speed, $v$ . This is described by Stokes' Law:
$F_{D} = - b v$
The constant b depends on the fluid's viscosity and the object's size and shape. This model applies to things like dust particles in the air or a small bead falling through oil.
Quadratic Drag (High Speed Regime): For most macroscopic objects moving at moderate to high speeds in air (like a car, a baseball, or a skydiver), the drag force is proportional to the square of the speed, $v^{2}$ .
$F_{D} = \frac{1}{2} C ρ A v^{2}$
Here, $ρ$ is the fluid density, $A$ is the object's cross-sectional area, and $C$ is the dimensionless drag coefficient, which depends on the object's shape. This model is critical in many engineering applications, from automotive design to aeronautics.

To get a better feel for these concepts and see how they are applied, please read the following introductory sections.

6.7: Drag Force and Terminal Speed

This text, 'Drag Force and Terminal Speed' from OpenStax University Physics, introduces the mathematical forms of both quadratic and linear drag forces and defines the relevant physical quantities.

Please read the first main section, 'Drag Forces', and the section titled 'Stokes’ law'. Focus on understanding the difference between the F_D ∝ v² and F_D ∝ v models and the physical conditions under which each is applicable.

2. 1D Motion with Linear Air Resistance

Let's now analyze the classic problem of an object falling from rest under the influence of gravity and linear air resistance. This is where we formulate and solve our first true differential equation of motion.

Consider an object of mass $m$ falling downwards. We'll define the downward direction as positive. The forces acting on it are:

Gravity: $F_{g} = m g$ (downwards, so positive)
Linear Drag: $F_{D} = - b v$ (upwards, so negative, since $v$ is positive downwards)

Applying Newton's Second Law, $Σ F = ma$ :

m g - b v = m \frac{d v}{d t}

This is a first-order linear ordinary differential equation for the velocity $v (t)$ . Solving it will give us the object's velocity at any time.

The following video provides an excellent, rigorous derivation and solution of this exact problem.

Terminal velocity differential equation | Lecture 8 | Differential Equations for Engineers

In 'Terminal velocity differential equation', Prof. Jeffrey Chasnov masterfully sets up the differential equation for a falling skydiver (assuming linear drag for this model), analyzes the equilibrium condition to find terminal velocity, and solves the full equation using the method of integrating factors.

Please watch the entire video (about 11 minutes). Pay close attention to: How Newton's Second Law is used to set up the differential equation. How 'terminal velocity' is found by setting the acceleration (dv/dt) to zero. The step-by-step application of the integrating factor method to solve for v(t). The physical interpretation of the final solution.

Summary of the Solution

As demonstrated in the video, we can rearrange the equation into the standard form for a linear first-order ODE:

\frac{d v}{d t} + \frac{b}{m} v = g

Solving this (with the initial condition $v (0) = 0$ ) yields the velocity as a function of time:

v (t) = \frac{m g}{b} (1 - e^{- \frac{b}{m} t})

Recognizing that the term $\frac{m g}{b}$ is the terminal velocity, $v_{T}$ , which is the constant velocity reached when the drag force perfectly balances gravity, we can write the solution more elegantly:

v (t) = v_{T} (1 - e^{- t / τ})

Here, $τ = m / b$ is the time constant of the system.

Note on Solution Method: The video uses an integrating factor. The equation is also separable, and can be solved by rearranging to $\frac{d v}{g - ( b / m ) v} = d t$ and integrating, as shown in the resource LINK (Section "The Calculus of Velocity-Dependent Frictional Forces") and LINK. Both methods are valid and lead to the same result.

Test your understanding!

What is the physical significance of the time constant $τ = m / b$ ? What is the object's velocity at $t = τ$ ? What about at $t = 3 τ$ ?

Show answer

The time constant $τ$ represents the characteristic time it takes for the system to approach its steady state (terminal velocity).

At $t = τ$ , the velocity is $v (τ) = v_{T} (1 - e^{- 1}) \approx 0.632 v_{T}$ . So, after one time constant, the object has reached about 63.2% of its terminal velocity.
At $t = 3 τ$ , the velocity is $v (3 τ) = v_{T} (1 - e^{- 3}) \approx 0.950 v_{T}$ . After three time constants, the object is at 95% of its terminal velocity.

The time constant provides a quick way to estimate how long it takes for the transient effects to become negligible.

3. Advanced Application: 2D Projectile Motion with Linear Drag

Now we'll extend this analysis to two dimensions, a problem that fits your goal of tackling more complex scenarios. This involves setting up and solving a system of coupled differential equations.

For a projectile of mass $m$ launched with initial velocity $v_{0}$ , the net force is $F = F_{g} + F_{D} = m g - b v$ . In component form, this becomes two separate equations:

m \frac{d v _{x}}{d t} = - b v_{x} ⟹ m x ¨ = - b x ˙

m \frac{d v _{y}}{d t} = - m g - b v_{y} ⟹ m y ¨ = - m g - b y ˙

A key feature of the linear drag model is that the $x$ and $y$ equations are decoupled—they can be solved independently.

The following video provides a detailed, rigorous walkthrough of this entire problem. It's advanced but highly rewarding.

GCM05: Projectile motion with air resistance

The video 'GCM05: Projectile motion with air resistance' provides a university-level analysis of this problem. It's an excellent example of applying DEs to a more complex physical system and even introduces approximation methods used in theoretical physics.

This is a long video, so we'll focus on the core parts for now. Setup (15:06 - 18:29): Watch this segment to see how the vector equation of motion is broken down into its x and y components, introducing the time constant τ = m/k (where k is used instead of b). Solution (18:29 - 30:30): Watch this part to see the separate solutions for the x-motion and y-motion. Notice that the x-motion is an exponential decay, while the y-motion is a more complex function involving both gravity and drag.

The solutions derived are:

$v_{x} (t) = v_{0 x} e^{- t / τ}$
$v_{y} (t) = (v_{0 y} + v_{T}) e^{- t / τ} - v_{T}$ , where $v_{T} = g τ = m g / b$ is the magnitude of the terminal velocity.

The asymmetry is fascinating: the time to reach the peak is no longer equal to the time to fall back down. The trajectory is not a parabola.

For a deeper dive, the section of the video from 30:30 to 52:57 shows how to find the total time of flight and range. This involves solving a transcendental equation, which can't be done analytically. The presenter uses Taylor series expansions to find an approximate solution for the low-drag case. This is a common and powerful technique in advanced physics. You are not required to reproduce this, but following the logic will give you a taste of the methods used when exact solutions are unavailable.

4. The Challenge of Quadratic Drag

What if we use the more realistic quadratic drag model, $F_{D} \propto v^{2}$ ?
The force vector is written as $F_{D} = - \frac{1}{2} C ρ A v^{2} v^= - c ∣ v ∣ v$ .

The equations of motion become:

m x ¨ = - c x ˙^{2} + y ˙^{2} x ˙

m y ¨ = - m g - c x ˙^{2} + y ˙^{2} y ˙

Notice the critical difference: the $x ˙^{2} + y ˙^{2}$ term means the equation for $x ¨$ depends on $y ˙$ , and vice-versa. These are coupled, non-linear differential equations. There is no general analytical solution. Such problems are almost always solved using numerical methods on a computer.

To appreciate this complexity, watch the final part of Dr. Viv's video.

GCM05: Projectile motion with air resistance

This final segment explains why the quadratic drag case is so much harder, setting up the coupled, non-linear equations and showing why they demand a numerical approach.

Watch from 52:57 to the end. The goal is not to solve these equations, but to understand why they are fundamentally more difficult than the linear case.

Conclusion

Today, you have taken a major step in your physics journey by using calculus to model a real-world dynamic system from first principles.

Key Takeaways:

Velocity-dependent drag forces are typically modeled as either linear ( $F_{D} \propto v$ ) or quadratic ( $F_{D} \propto v^{2}$ ).
By applying Newton's Second Law, these force laws become differential equations of motion.
For 1D linear drag, the motion is described by a first-order linear ODE, which we can solve analytically to find that velocity approaches a terminal value, $v_{T}$ , exponentially with a time constant $τ = m / b$ .
For 2D linear drag, the component equations are decoupled, allowing for independent analytical solutions for $v_{x} (t)$ and $v_{y} (t)$ .
The quadratic drag model leads to coupled, non-linear differential equations that generally cannot be solved analytically and require numerical methods.

Preview of the Next Lesson:
We've now seen how to analyze motion by focusing on forces and solving the resulting differential equations. In the next lesson, we will develop a powerful alternative framework based on the concepts of work and energy. We will derive the work-energy theorem from Newton's second law, which will often allow us to solve complex dynamics problems without directly integrating acceleration over time.

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