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.
What a learner can do afterwards
- 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
1 · Read
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).
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.
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.
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.
Three equations answer three different questions, so read which one is being asked.
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.
Reading summaries adapted from OpenStax College Physics (CC BY 4.0); OpenStax College Physics 2e (CC BY-NC-SA 4.0); OpenStax Physics (CC BY 4.0); OpenStax University Physics Volume 1 (CC BY-NC-SA 4.0), openstax.org.
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Where it sits
30 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.