Kinematics with Non-Uniform Acceleration: An Integration Approach

Hello! Welcome to the first lesson in our course on calculus-based mechanics. Given your goal to build a deep, mathematically rigorous understanding of physics, this module will establish the essential calculus tools we'll use throughout our study.

Today, we'll focus on the first learning outcome: applying integration techniques to solve kinematics problems with non-uniform acceleration.

In your previous studies, you've likely worked with constant acceleration using the standard kinematic equations. However, in most real-world scenarios—from a rocket launch to the motion of a particle in a complex field—acceleration is not constant. This is where your calculus knowledge becomes indispensable. Since you're already familiar with differentiation and the basics of integration, we'll now see how to use integration as the "reverse" of differentiation to move from acceleration back to velocity and position.

This lesson will cover:

The fundamental relationships between position, velocity, and acceleration using integration.
How to solve problems where acceleration is a function of time, $a (t)$ .
Techniques for cases where acceleration depends on position, $a (s)$ , or velocity, $a (v)$ .
Application of these concepts to advanced, competitive-level problems.

Let's begin.

1. From Differentiation to Integration in Kinematics

You know the fundamental definitions of instantaneous velocity and acceleration in one dimension:

v (t) = \frac{d s}{d t} and a (t) = \frac{d v}{d t}

where $s$ is position, $v$ is velocity, and $a$ is acceleration.

These are differential relationships. To solve problems where acceleration is a known function, we need to use the inverse process: integration. This allows us to determine velocity from a changing acceleration, and position from a changing velocity.

To understand the formal basis for this, please read the following sections from a chapter on one-dimensional kinematics provided by MIT OpenCourseWare.

8.01SC S22 Chapter 4: One Dimensional Kinematics

This reading from MIT OCW, 'One Dimensional Kinematics and Integration', formally introduces how integration is used when acceleration is not constant. It connects the physical concepts of changing velocity and displacement to the mathematical tools of indefinite and definite integrals.

Please read sections 4.6, 4.6.1, 4.6.2, 4.6.3, and 4.6.4. Focus on how the change in velocity is the definite integral of acceleration over time, and similarly, how displacement is the definite integral of velocity. Pay attention to how initial conditions (like v₀ and x₀) are incorporated.

As the text explains, we can find the change in velocity and position over a time interval from $t_{i}$ to $t_{f}$ using definite integrals:

Δ v = v (t_{f}) - v (t_{i}) = \int_{t_{i}}^{t_{f}} a (t) d t

Δ s = s (t_{f}) - s (t_{i}) = \int_{t_{i}}^{t_{f}} v (t) d t

Alternatively, we can use indefinite integrals if we solve for the constant of integration, $C$ , using initial conditions. For example, given $a (t)$ and an initial velocity $v_{0}$ at $t = 0$ :

v (t) = \int a (t) d t = V (t) + C

where $V (t)$ is the antiderivative of $a (t)$ . We find $C$ by using the initial condition: $v (0) = V (0) + C = v_{0}$ .

2. Solving Kinematics Problems with Integration

The method you use depends on the functional form of the acceleration. Let's explore the three common cases.

Case 1: Acceleration is a function of time, $a = f (t)$

This is the most straightforward case. We integrate with respect to time twice.

Find Velocity: Separate variables in $a = d v / d t$ to get $d v = a (t) d t$ . Integrate both sides: $\int_{v_{0}}^{v (t)} d v^{'} = \int_{0}^{t} a (t^{'}) d t^{'} ⟹ v (t) - v_{0} = \int_{0}^{t} a (t^{'}) d t^{'}$
Find Position: Use the resulting $v (t)$ in $v = d s / d t$ . Separate variables to get $d s = v (t) d t$ and integrate: $\int_{s_{0}}^{s (t)} d s^{'} = \int_{0}^{t} v (t^{'}) d t^{'} ⟹ s (t) - s_{0} = \int_{0}^{t} v (t^{'}) d t^{'}$

The following video provides a clear, step-by-step example of this entire process, including finding maximum velocity (a concept from your "Applications of Derivatives" study).

Application of calculus to kinematics / acceleration,velocity, displacement

This video, 'Application of calculus to kinematics', demonstrates the complete workflow for a problem where acceleration is given as a polynomial function of time. It's a perfect practical demonstration of the theory we just discussed.

Watch the full video. Observe how the presenter: Integrates a(t) to get v(t) and uses the initial velocity to find the constant of integration. Uses a definite integral of v(t) to find displacement over an interval. Finds maximum velocity by setting acceleration to zero. Finds the time when the particle is at rest (v=0) and calculates displacement up to that time.

Case 2: Acceleration is a function of position, $a = f (s)$

When acceleration depends on position (e.g., a particle attached to a spring or in a non-uniform gravitational field), integrating with respect to time is not direct. We need a different relation. We can derive it by eliminating $d t$ from our basic definitions:
From $v = d s / d t$ , we have $d t = d s / v$ .
Substituting this into $a = d v / d t$ , we get $a = d v / (d s / v) = v \frac{d v}{d s}$ .

Rearranging gives us the crucial work-energy integral precursor:

a d s = v d v

Now we can integrate, relating velocity and position directly:

\int_{s_{0}}^{s} a (s^{'}) d s^{'} = \int_{v_{0}}^{v} v^{'} d v^{'} = \frac{1}{2} v^{2} - \frac{1}{2} v_{0}^{2}

Case 3: Acceleration is a function of velocity, $a = f (v)$

This case is common in physics, especially when dealing with drag or friction forces. We can relate velocity to either time or position.

To find v(t): Start with $a = d v / d t$ , separate variables as $d t = \frac{d v}{a ( v )}$ , and integrate: $\int_{0}^{t} d t^{'} = t = \int_{v_{0}}^{v} \frac{d v ^{'}}{a ( v ^{'} )}$
To find v(s): Start with $a d s = v d v$ , separate as $d s = \frac{v d v}{a ( v )}$ , and integrate: $\int_{s_{0}}^{s} d s^{'} = s - s_{0} = \int_{v_{0}}^{v} \frac{v ^{'} d v ^{'}}{a ( v ^{'} )}$

These last two cases are demonstrated clearly in the following examples from a standard engineering dynamics textbook.

Kinematics of a Particle

This reading from 'Kinematics of a Particle' provides excellent worked examples for handling situations where acceleration is a function of position or velocity.

Please study Example 12.2 and Example 12.4. In Example 12.2, notice how they solve for v(t) and s(t) when acceleration is a function of velocity, a = f(v). In Example 12.4, see the direct application of the 'v dv = a ds' relation when acceleration is a function of position, a = f(s).

Test your understanding!

A particle's acceleration is given by $a (s) = - k / s^{2}$ , where $k$ is a positive constant. This is the form of gravitational acceleration from a point mass. If the particle starts from rest at a position $s = s_{0}$ , find its velocity $v$ as a function of its position $s$ .

Show answer

Since acceleration is a function of position, we use the relation $v d v = a (s) d s$ . We integrate from the initial state ( $v = 0$ at $s = s_{0}$ ) to the final state ( $v$ at $s$ ).

\int_{0}^{v} v^{'} d v^{'} = \int_{s_{0}}^{s} - \frac{k}{s ^{'2}} d s^{'}

[\frac{1}{2} v^{'2}]_{0}^{v} = [\frac{k}{s ^{'}}]_{s_{0}}^{s}

\frac{1}{2} v^{2} - 0 = \frac{k}{s} - \frac{k}{s _{0}}

v^{2} = 2 k (\frac{1}{s} - \frac{1}{s _{0}})

v = 2 k (\frac{1}{s} - \frac{1}{s _{0}})

We take the positive root assuming the particle moves in the direction of increasing $s$ , but the direction would depend on the coordinate system setup. The key is the relationship derived from integration.

3. Advanced Applications and Problem Solving

Your goal includes tackling high-level problems. Let's analyze a few examples that require a deeper conceptual link between the physics and the mathematics. The following problems are from a JEE Advanced preparation session, matching the level of rigor you're looking for.

Problem 1: Conceptual Analysis of Motion

Scenario: A particle starts from rest at the origin ( $s = 0, t = 0$ ) and comes to rest again at $s = 1$ m after $t = 1$ s. What can we say about its instantaneous acceleration, $α$ ?

To solve this, we don't need a specific function. We need to think about the physical constraints.

To start moving from rest, the particle must have a positive acceleration.
To come to a stop from a state of motion, it must have a negative acceleration (deceleration).

Therefore, the acceleration must change sign during the motion. It cannot remain positive or negative for the entire duration.

Let's consider a simple model for this motion using a velocity-time graph, as discussed in the following video segment.

Killer 5: Kinematics | JEE Advanced 2021 | JEE Physics | Shreyas Sir | Vedantu JEE Enthuse English

This segment from 'Killer 5: Kinematics' analyzes the conceptual problem just described. It cleverly uses v-t graphs and the definition of displacement as area under the curve to deduce properties of the acceleration.

Watch the section from 00:35:59 to 00:47:22. Focus on how the presenter uses a simple triangular v-t graph as a limiting case to find a bound on the acceleration.

The analysis shows that by considering the simplest case (a triangular v-t graph), the peak velocity is 2 m/s and the magnitude of acceleration during both phases is 4 m/s². Any other v-t graph that encloses an area of 1 m² within 1 s and starts/ends at v=0 will involve instantaneous accelerations that are at some point greater than or equal to 4 m/s². This is a powerful way to reason about motion without solving explicit equations.

Problem 2: Application in Continuum Mechanics

Scenario: An expanding sphere has a constant total mass and uniform density. The rate of change of its density follows the relation $\frac{1}{ρ} \frac{d ρ}{d t} = constant$ . How is the velocity of a point on its surface related to its instantaneous radius $R$ ?

This problem connects kinematics ( $v = d R / d t$ ) with the properties of a continuous body and requires careful application of differentiation.

Killer 5: Kinematics | JEE Advanced 2021 | JEE Physics | Shreyas Sir | Vedantu JEE Enthuse English

This next segment from the same video solves the expanding sphere problem. It's a great example of applying calculus, specifically the product and chain rules for differentiation, to a physical system.

Watch from 00:48:43 to 00:59:10. Pay close attention to how the presenter starts with the basic definition of density (M/V), identifies constants and variables, and then differentiates the entire expression with respect to time to find the desired relationship.

The solution shows that $v \propto R$ . This type of relationship, where the velocity of a point in an expanding system is proportional to its distance from the center, is fundamental in cosmology—it's the basis of Hubble's Law for the expanding universe.

Final Challenge

To consolidate your understanding, here is a problem that involves a non-polynomial velocity function. This will test your ability to apply both differentiation and integration in a more complex setting.

Integration and Kinematics Problem with Velocity-Time Graph — This image shows a problem where a body's velocity is given by v = 2 sin(t/10 + π/5) csc(t/30 + π/4) for 0 ≤ t ≤ 60s. The questions involve finding properties of the motion by analyzing this function.

Try to outline the steps you would take to solve parts (c), (d), and (e):

(c) Time when body is at rest: What equation would you need to solve?
(d) Distance traveled from t=0 to t=15: What integral would you set up?
(e) Acceleration at t=10: What operation would you perform on the velocity function $v (t)$ ?

Thinking through the process is as important as finding the numerical answer.

Conclusion

In this lesson, you've bridged the gap between basic kinematics and a more powerful, calculus-based approach. We have established that integration is the key tool for working backward from acceleration to find velocity and position.

Key Takeaways:

The relationship between s, v, and a is fundamentally based on calculus: $v = d s / d t$ and $a = d v / d t$ .
By using integration, we can find changes in velocity and position: $Δ v = \int a (t) d t$ and $Δ s = \int v (t) d t$ .
The integration strategy depends on whether acceleration is a function of time $a (t)$ , position $a (s)$ , or velocity $a (v)$ .
The relation $a d s = v d v$ is a powerful tool for problems where acceleration is a function of position.

Preview of the Next Lesson:
We will dive deeper into the case where acceleration is a function of velocity, $a (v)$ . This will lead us to formulating and solving first-order differential equations of motion, which are essential for describing systems with velocity-dependent forces like air resistance and viscous drag.

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