Solving Velocity-Dependent Acceleration Problems

Hello! In our last lesson, we tackled problems with time-dependent acceleration, a(t), using the method of direct integration. Today, we'll continue our exploration of non-uniform acceleration by investigating a different, more physically nuanced scenario: when acceleration depends on an object's velocity, a(v).

This situation is common in the real world, especially when dealing with motion through fluids like air or water, where resistive forces (drag) change with speed. Your mathematical background will be crucial here, as we'll need to employ a powerful technique from calculus—the method of separation of variables—to solve the resulting differential equations.

This lesson directly addresses the learning outcome: to solve problems with velocity-dependent acceleration, a(v), using the method of separation of variables.

1. The Challenge of `a(v)`

Recall from the last lesson that for a(t), we could find velocity by directly integrating with respect to time:
$v (t) = v_{0} + \int_{t_{0}}^{t} a (t^{'}) d t^{'}$
This worked because the integrand a(t') was an explicit function of the integration variable, t'.

Now, consider the case where a = a(v). If we try the same approach, we get:
$\int d v = \int a (v) d t$
We are faced with a problem: the integrand a(v) is a function of velocity, but we are asked to integrate with respect to time. We cannot directly evaluate this integral without first knowing how v depends on t, which is the very thing we are trying to find! This circularity means we need a new strategy.

2. The Mathematical Tool: Separation of Variables

The solution lies in rearranging the fundamental differential equation a = dv/dt. Since a is a function of v, we write:
$a (v) = \frac{d v}{d t}$
Instead of trying to integrate with respect to t immediately, we can algebraically rearrange—or "separate"—the variables. We gather all terms involving v on one side with dv, and all terms involving t on the other side with dt.
$d t = \frac{1}{a ( v )} d v$
Now, the equation is "separated." The left side is purely a function of t (in this case, just dt), and the right side is purely a function of v. This structure allows us to integrate both sides independently.

To formalize this, let's consult a resource.

Freshman Physics

The following text from 'Freshman Physics' provides a concise mathematical definition of a separable Ordinary Differential Equation (ODE), which is the formal name for the type of equation we are solving.

Please read the short section '13.3 Separable ODE'. It presents the general form and the method of solving by rearranging and integrating. This formally captures the technique we just outlined.

The general procedure to find the relationship between velocity and time is therefore:
$\int_{t_{i}}^{t_{f}} d t = \int_{v_{i}}^{v_{f}} \frac{1}{a ( v )} d v$
Executing these integrals gives us an equation relating t and v. While this often gives t as a function of v, i.e., t(v), we can then algebraically rearrange to find the more familiar v(t).

3. Physical Context: Motion with Resistive Forces

The most common physical application of velocity-dependent acceleration is motion with drag. When an object moves through a fluid (like air or water), it experiences a resistive force that opposes its motion and whose magnitude depends on the object's speed.

Terminal Velocity Concept and Formulas — This image illustrates the forces on a falling object subject to air resistance. The downward weight `W` is constant, while the upward air resistance `f` increases with velocity. This changing force leads to a velocity-dependent acceleration.

There are two primary models for this drag force:

Linear Drag: $F_{d r a g} = b v$ . This applies to objects moving at low speeds or in viscous fluids. Here, b is a constant that depends on the fluid's viscosity and the object's shape and size.
Quadratic Drag: $F_{d r a g} = c v^{2}$ . This is a better approximation for objects moving at higher speeds in less viscous fluids, like a car or a skydiver in air. Here, c is a different constant.

Let's apply our method to solve problems involving both models.

4. Case 1: Linear Drag and Terminal Velocity

Consider an object of mass m dropped from rest in a fluid where the drag force is linear, $F_{d r a g} = b v$ . We'll choose the downward direction as positive.

Applying Newton's Second Law:
$F_{n e t} = ma$
$m g - b v = m \frac{d v}{d t}$
From this, we get our expression for a(v):
$a (v) = g - \frac{b}{m} v$
Now we have a differential equation ready to be solved.

14.2 Resistive forces - low speed case

The following video from MIT OpenCourseWare provides a rigorous derivation of the velocity-time relationship for this exact scenario. It's an excellent demonstration of the separation of variables method in a physical context.

Watch the first 8 minutes of the video (00:00 to 08:16). Professor Walter Lewin will: Set up the differential equation from Newton's Second Law. Meticulously solve it using separation of variables and integration to find v(t). Analyze the solution to define and explain the crucial concepts of terminal velocity and the time constant.

Summary of the Solution (Linear Drag)

As derived in the video, the solution to the differential equation is:
$v (t) = \frac{m g}{b} (1 - e^{- \frac{b}{m} t})$
This solution reveals two important physical concepts:

Terminal Velocity ( $v_{T}$ ): As $t \to \infty$ , the exponential term $e^{- \frac{b}{m} t}$ goes to zero. The velocity approaches a constant, maximum value called the terminal velocity.
$v_{T} = lim_{t \to \infty} v (t) = \frac{m g}{b}$
Conceptually, terminal velocity is reached when the drag force exactly balances the gravitational force ( $b v_{T} = m g$ ), causing the net force and thus the acceleration to become zero.
Time Constant ( $τ$ ): The quantity $τ = m / b$ has units of time and is called the characteristic time or time constant. It governs how quickly the object approaches terminal velocity. The equation can be written as:
$v (t) = v_{T} (1 - e^{- t / τ})$

Test your understanding!

Using the solution $v (t) = v_{T} (1 - e^{- t / τ})$ , prove that after a time t = τ has elapsed, an object starting from rest will have reached approximately 63.2% of its terminal velocity.

Show answer

We need to find the velocity at t = τ. We substitute this into the equation:
$v (τ) = v_{T} (1 - e^{- τ / τ})$
$v (τ) = v_{T} (1 - e^{- 1})$
Using the value $e \approx 2.718$ , we get $e^{- 1} \approx 1/2.718 \approx 0.368$ .
$v (τ) = v_{T} (1 - 0.368) = 0.632 \cdot v_{T}$
Thus, after one time constant, the object's velocity is 63.2% of its terminal velocity. This is a standard result for exponential growth processes.

5. Case 2: Quadratic Drag

Now, let's consider the high-speed case where the drag force is quadratic, $F_{d r a g} = c v^{2}$ . This is more challenging mathematically but essential for many real-world problems.

Again, starting with Newton's Second Law (downwards positive):
$F_{n e t} = ma$
$m g - c v^{2} = m \frac{d v}{d t}$
Our expression for a(v) is:
$a (v) = g - \frac{c}{m} v^{2}$
Let's separate the variables:
$d t = \frac{d v}{g - \frac{c}{m} v ^{2}}$
Integrating this requires more advanced techniques, specifically partial fraction decomposition, which should be familiar from your mathematics studies.

Terminal Velocity Exam Question (2 of 4: Eliminating m & k)

The following video from Eddie Woo tackles a problem with a quadratic drag force. The explanation is very thorough, focusing on the strategic thinking required to handle the more complex integration. This aligns perfectly with your goal of understanding the 'why' behind the methods.

Watch the segment from 00:08 to 10:17. Pay close attention to these key strategic steps: Separating variables and setting up the definite integral. Eliminating constants: Notice how he replaces m and k (our c) with the terminal velocity, v_T, to simplify the integrand. This is a very common and powerful technique in physics proofs. The terminal velocity is found where a=0, so mg - cv_T^2 = 0, which gives v_T^2 = mg/c. Identifying the integration method: The form of the integrand 1/(v_T^2 - v^2) signals the need for partial fractions, which he explains will lead to the logarithmic terms in the final answer.

The process shown in the video is a masterful example of applying advanced mathematical tools to a physics problem. The key is recognizing that the denominator v_T^2 - v^2 can be factored into (v_T - v)(v_T + v), allowing the use of partial fractions to break the complex integral into two simpler logarithmic integrals.

Test your understanding!

An object falls from rest under gravity with quadratic air resistance, $F_{d r a g} = c v^{2}$ . Its equation of motion after separating variables is:
$\int_{0}^{t} d t^{'} = \int_{0}^{v} \frac{d v ^{'}}{g - \frac{c}{m} v ^{'2}}$
Let the terminal velocity be $v_{T} = m g / c$ . Rewrite the integral in terms of $v_{T}$ and use partial fraction decomposition to show that the result of the integration is:
$t = \frac{v _{T}}{2 g} ln (\frac{v _{T} + v}{v _{T} - v})$
Hint: The integral of 1/(A-x) is -ln|A-x| and the integral of 1/(A+x) is ln|A+x|.

Show answer

First, substitute $g = v_{T}^{2} \frac{c}{m}$ into the integral.
$\int_{0}^{v} \frac{d v ^{'}}{v _{T}^{2} \frac{c}{m} - \frac{c}{m} v ^{'2}} = \frac{m}{c} \int_{0}^{v} \frac{d v ^{'}}{v _{T}^{2} - v ^{'2}}$
Now, use partial fractions on the integrand:
$\frac{1}{v _{T}^{2} - v ^{'2}} = \frac{1}{( v _{T} - v ^{'} ) ( v _{T} + v ^{'} )} = \frac{A}{v _{T} - v ^{'}} + \frac{B}{v _{T} + v ^{'}}$
Solving for A and B gives $A = B = \frac{1}{2 v _{T}}$ .
So, the integral becomes:
$t = \frac{m}{c} \cdot \frac{1}{2 v _{T}} \int_{0}^{v} (\frac{1}{v _{T} + v ^{'}} + \frac{1}{v _{T} - v ^{'}}) d v^{'}$
Now integrate:
$t = \frac{m}{2 c v _{T}} [ln (v_{T} + v^{'}) - ln (v_{T} - v^{'})]_{0}^{v}$
$t = \frac{m}{2 c v _{T}} [ln (\frac{v _{T} + v ^{'}}{v _{T} - v ^{'}})]_{0}^{v}$
Substitute the limits of integration:
$t = \frac{m}{2 c v _{T}} (ln (\frac{v _{T} + v}{v _{T} - v}) - ln (\frac{v _{T} + 0}{v _{T} - 0})) = \frac{m}{2 c v _{T}} ln (\frac{v _{T} + v}{v _{T} - v})$
Finally, substitute $v_{T}^{2} = m g / c ⟹ m / c = v_{T}^{2} / g$ .
$t = \frac{v _{T}^{2} / g}{2 v _{T}} ln (\frac{v _{T} + v}{v _{T} - v}) = \frac{v _{T}}{2 g} ln (\frac{v _{T} + v}{v _{T} - v})$
This confirms the result. This expression gives the time t required to reach a specific velocity v.

Conclusion

Today, we've added a crucial technique to our kinematics toolkit. You learned how to tackle problems where acceleration is a function of velocity, a common scenario in realistic physics problems involving drag.

Key Takeaways:

When acceleration depends on velocity, a(v), direct integration with respect to time is not possible.
The correct mathematical approach is the method of separation of variables, which involves rearranging the equation a(v) = dv/dt to isolate all v terms with dv and all t terms with dt before integrating.
Linear drag ( $F \propto v$ ) leads to a differential equation whose solution involves a natural exponential function, defining the concepts of terminal velocity (v_T) and a time constant (τ).
Quadratic drag ( $F \propto v^{2}$ ) leads to a more complex integral that requires partial fraction decomposition, resulting in a solution involving natural logarithms.
The concept of terminal velocity is universal in these problems. It can be found either by taking the limit $t \to \infty$ in the function v(t) or by setting the acceleration to zero in the equation of motion (F_net = 0).

Preview of the Next Lesson:

We've now handled a(t) and a(v). There is one more important case of non-uniform acceleration to consider: when acceleration depends on position, a(x). This occurs in systems like springs and with gravitational forces that vary with distance. To solve a(x) problems, we will need to introduce yet another clever calculus identity: the chain rule formulation $a = v \frac{d v}{d x}$ . This will complete our trio of methods for one-dimensional non-uniform acceleration.

Recommended resources

1. The Challenge of a(v)