--- title: "Kinematics with Calculus" description: "When the acceleration is not constant, differentiate displacement to get velocity and velocity to get acceleration, and integrate back the other way, using an initial condition to fix each constant." canonical: https://lightmysky.com/learn/mathematics/kinematics-with-calculus-mt_2kNspOOoDw source: https://lightmysky.com/learn/mathematics/kinematics-with-calculus-mt_2kNspOOoDw.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Kinematics with Calculus When the acceleration is not constant, differentiate displacement to get velocity and velocity to get acceleration, and integrate back the other way, using an initial condition to fix each constant. Subject: Mathematics · Area: Mechanics · Ages 17 to 18 Page: https://lightmysky.com/learn/mathematics/kinematics-with-calculus-mt_2kNspOOoDw ## Ready when they can - Differentiate s(t) twice and give the units of each result - Integrate a(t) to get v(t) and use v(0) to find the constant - Find when a particle is instantaneously at rest by solving v(t) = 0 ## Lesson: Kinematics with Calculus You have already learned to run a derivative backwards, and motion uses both directions. A board rolls down the Ridgeway ramp, and its displacement s is a rule in the time t. The constant-acceleration formulas you met before only hold while the acceleration stays fixed, and on a curved ramp it does not: the board picks up speed as the slope steepens. Calculus links displacement, velocity and acceleration whatever they are doing: differentiation goes one way and integration the other. Think of it as a ladder with three rungs. Displacement s, in metres. Velocity v = ds/dt, in metres per second. Acceleration a = dv/dt, in metres per second squared. Going down the ladder you differentiate, and each step divides by another second, which is where the units come from. Going up you integrate, and each step needs a constant. That constant is the starting value, fixed by an initial condition such as v(0) or s(0). *(drawing: Down the ladder by differentiating, up it by integrating.)* **Example.** Going down. A particle has s = 2t³ - 15t² + 24t metres after t seconds. Differentiating gives v = 6t² - 30t + 24 m/s, and differentiating again gives a = 12t - 30 m/s². Now ask when it is at rest. That means v = 0, so factorise: 6(t - 1)(t - 4) = 0, giving t = 1 and t = 4. At t = 1 the acceleration is 12 - 30 = -18 m/s², and at t = 4 it is 48 - 30 = 18 m/s². The particle stops twice and accelerates opposite ways each time. **Example.** Going up. A particle has a = 6t - 4 m/s², and at t = 0 its velocity is 5 m/s and its displacement is 2 m. Integrating the acceleration gives v = 3t² - 4t + c. The initial condition v(0) = 5 puts 5 = 0 - 0 + c, so c = 5 and v = 3t² - 4t + 5. Integrate again: s = t³ - 2t² + 5t + k, and s(0) = 2 gives k = 2. So s = t³ - 2t² + 5t + 2, and each rung needed its own starting value. *(drawing: Each step up adds an unknown constant, and each initial condition fixes one.)* Three readings to keep apart. At rest means v = 0, so that is the equation to solve. It does not mean a = 0, which is where the velocity has stopped changing while the particle may be moving fast. It does not mean s = 0 either, which is where the particle is at the origin. And a negative velocity does not mean stopped: it means moving the other way, so the board is rolling back up the ramp rather than down it. **Recap.** Displacement, velocity and acceleration are three rungs of one ladder. Differentiate s to get v and v to get a, with the units dividing by another second each time. Integrate to go back up, and each step needs an initial condition to fix its constant. At rest means v = 0, which is the equation to solve for those times. A zero acceleration and a zero displacement are different questions, and a negative velocity means moving backwards, not stopped. ## Practice 30 questions on this page, each with its working shown. ## Needs first - [Integration as the Reverse of Differentiation](https://lightmysky.com/learn/mathematics/integration-as-the-reverse-of-differentiation-mt_edtk3ArxRk) - [Vertical Motion Under Gravity](https://lightmysky.com/learn/mathematics/vertical-motion-under-gravity-mt_IgNj9bn14O) - [The Power Rule for Differentiating Polynomials](https://lightmysky.com/learn/mathematics/the-power-rule-for-differentiating-polynomials-mt_l7f4j44bR3) ## Opens up - [Velocity and Acceleration as Derivatives](https://lightmysky.com/learn/science/velocity-and-acceleration-as-derivatives-mt_RdhbiJ6CTk)