eduKateSG · Kinematics · Worked learning guide
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Learn how to reconstruct motion when acceleration changes. Follow the integral, retain the initial conditions and join stages without resetting the object by accident.
1. Start with the derivative chain and its units
3. Explain why average acceleration cannot supply every result
5. Add a braking stage and stop the model at rest
7. Recognise time-dependent and state-dependent laws
9. Practise with complete state checks
Full chapter index · Practice and checking · Kinematics subject guide
The familiar constant-acceleration equations work because acceleration is constant over the interval being modelled. When acceleration ramps, switches or depends on time, the same definitions still apply, but the calculation must preserve the changing history. The gold standard of kinematics is choosing a method that fits that history.
This guide explains variable acceleration, jerk and piecewise motion through original worked examples. It is written for students who can differentiate and integrate simple polynomials, with geometric checks that make the mathematics visible. It does not prescribe a vehicle braking or machinery control programme.
Begin with The Gold Standard of Kinematics for the core quantities. The task here is to move beyond one constant value of a, retain the correct initial conditions and show that position, velocity and acceleration describe the same motion.
Chapter index
Chapters 1–4
Chapters 5–8
Chapters 9–10
Position x(t), velocity v(t), acceleration a(t) and jerk j(t) describe successive rates of change along a fixed one-dimensional axis. Velocity is dx/dt, acceleration is dv/dt and jerk is da/dt where those derivatives exist.
Their SI units are metres, metres per second, metres per second squared and metres per second cubed. The increasing power of time reflects repeated differentiation. Jerk is not another name for a large acceleration: a large constant acceleration has zero jerk within its constant stage.
A changing acceleration means that velocity does not have one constant slope. If acceleration increases linearly, velocity curves quadratically and position curves cubically. The shapes follow from integration, not from a new collection of unrelated formulas.
Suppose acceleration is 0 at t = 0 s and increases steadily to 2 m/s² at t = 4 s. Its jerk is 0.5 m/s³ during that ramp. The increase in velocity is the area under the acceleration-time graph: a triangle with area one half × 4 × 2 = 4 m/s.
That area determines the velocity change, not the total distance. Position requires integrating velocity after the initial velocity has been included. Confusing these two areas is a common source of answers with the wrong units.
A definition check
Ask which graph is being used and what its area measures. Area under acceleration versus time has units m/s and gives change in velocity. Area under velocity versus time has units m and gives signed displacement. Area under speed versus time gives distance.
A changing acceleration does not make the definitions unreliable. It makes assumptions more visible. Keep the axis, time interval and starting values beside the calculation, then reconstruct each level in the correct order.
For an acceleration function a(t) and initial velocity v₀ at time t₀, velocity is v(t) = v₀ + ∫ from t₀ to t of a(τ)dτ. Position is x(t) = x₀ + ∫ from t₀ to t of v(τ)dτ.
The dummy variable τ keeps the upper time limit distinct from the integration variable. A definite integral automatically starts the accumulated change at zero when t = t₀. This makes it especially useful for joining stages.
Take a(t) = 0.5t m/s² for 0 ≤ t ≤ 4 s, where t is measured in seconds and the coefficient is 0.5 m/s³. Let v(0) = 1 m/s and x(0) = 0 m.
Integration gives v(t) = 1 + 0.25t² m/s. A second integration gives x(t) = t + t³/12 m. At 4 s, velocity is 5 m/s and position is 28/3 m, approximately 9.333 m.
Check by differentiation: dx/dt = 1 + t²/4, dv/dt = t/2, and da/dt = 1/2 with the appropriate units. Check the initial values by substituting t = 0. Both checks should pass before the final number is accepted.
Why the integration constants matter
An indefinite integral of a(t) supplies a family of possible velocities. The initial velocity selects one member of that family. Integrating again supplies a family of possible positions; the initial position selects the trajectory.
Dropping the constants does not merely lose a decorative term. It replaces the specified starting motion with another one. In this example, omitting v₀ changes the final velocity from 5 to 4 m/s and the displacement from 9.333 to 5.333 m. Both mistakes persist even if every later integration is performed correctly.
In the ramp example, acceleration rises from 0 to 2 m/s² over 4 s. Its time average is 1 m/s². Applying this average over four seconds gives the correct velocity change of 4 m/s, because the average was defined to preserve that integral.
But replacing the varying acceleration with a constant 1 m/s² gives displacement 1×4 + 0.5×1×4² = 12 m. The actual displacement is 28/3 m. The average acceleration preserved final velocity but changed the timing of the velocity gain.
The constant model adds velocity earlier than the rising ramp does. That earlier gain contributes to displacement for longer, so the constant model predicts a greater displacement.
This can be expressed directly. Over 0 ≤ t ≤ T, displacement is v₀T + ∫ from 0 to T of (T − τ)a(τ)dτ. An acceleration contribution near the start carries a larger time weight than an equal contribution near the end.
Two histories, one velocity change
Consider a 4 s interval starting from rest. Model A applies 2 m/s² during the first 2 s and zero afterwards. Model B applies zero during the first 2 s and 2 m/s² during the last 2 s. Both gain 4 m/s.
Model A travels 4 m during acceleration and 8 m while coasting, totalling 12 m. Model B travels zero during the first stage and 4 m during the second, totalling 4 m. Same duration and same final velocity do not determine displacement.
The correct general principle is precise: average acceleration determines the total velocity change over its averaging interval. It does not determine the whole velocity history. A method that preserves one aggregate need not preserve another quantity that depends on when the changes occurred.
Continue the ramp example after t = 4 s. At the join, the object is at x = 28/3 m with velocity 5 m/s. Suppose it now coasts with zero acceleration for 2 s.
Introduce local stage time u = t − 4 s. During 0 ≤ u ≤ 2 s, velocity is 5 m/s and position is x = 28/3 + 5u metres. At t = 6 s, position is 58/3 m, approximately 19.333 m.
Local time makes the initial conditions easy to see. At u = 0, the new expression returns exactly the preceding stage’s final position and velocity. The object does not restart from the coordinate origin simply because a new formula begins.
Continuity is quantity-specific
Position remains continuous at the join. Velocity also remains continuous because there is no impulse represented in this ideal piecewise model. Acceleration changes from 2 m/s² immediately before the join to zero immediately after it.
That jump means ordinary finite jerk is not defined at the switching instant. Jerk is 0.5 m/s³ during the ramp and zero during the coasting stage, but those within-stage values do not describe the acceleration jump itself.
A piecewise-constant or piecewise-linear acceleration model can still be useful. Its ideal switching points should simply be identified. Real actuators often change over a nonzero transition time, so the sharp switch may be an approximation to a smoother process.
A stage ledger
Keep a small table with stage start time, start position, start velocity, acceleration rule and end time. Compute the ending state and copy it into the next stage. This is a repeatable method for any number of stages.
A student who struggles with one long expression can solve correctly using local time and a state ledger. The physics does not require a single giant polynomial spanning the entire journey.
At t = 6 s, let the object begin an ideal constant deceleration of −1 m/s² from velocity 5 m/s. Define w = t − 6 s. For the braking stage, v = 5 − w m/s and x = 58/3 + 5w − 0.5w² metres.
The first zero of velocity occurs at w = 5 s, so stopping time is t = 11 s. Braking displacement is 25 − 12.5 = 12.5 m. Final position is 58/3 + 25/2 = 191/6 m, approximately 31.833 m.
If the intended model is “brake to rest, then remain at rest,” the braking expression ends at w = 5 s. Afterwards v = 0 and x = 191/6 m. Continuing v = 5 − w beyond that instant makes the object reverse direction, which is a different model.
Distinguish a root from a stage condition
A polynomial may have several roots, but only roots inside the current stage are candidates for a stated event. A stopping time found before the stage begins is not an event within that stage. A later root after a change of rule may be irrelevant.
Distance equals displacement throughout this particular example because velocity stays nonnegative until the stop. If a model genuinely includes reversal, integrate the magnitude of velocity for distance, splitting at the allowed zero-velocity times.
The stage ledger now gives a complete check: the ramp adds 4 m/s, coasting adds zero and braking removes 5 m/s from the initial 1 m/s, leaving zero. The stage displacements add to 191/6 m.
These are ideal explanatory numbers. A real stopping problem would need a verified braking law, response delay, environment and operating conditions. The mathematical skill here is ending a formula when the event that defines its scope has occurred.
A constant-jerk ramp changes acceleration linearly. If jerk is J and acceleration begins at a₀, then a(u) = a₀ + Ju. Integrating gives v(u) = v₀ + a₀u + 0.5Ju² and x(u) = x₀ + v₀u + 0.5a₀u² + Ju³/6.
Consider a 2 s ramp with J = 1 m/s³, starting from rest with zero acceleration. At its end, acceleration is 2 m/s², velocity is 2 m/s and displacement is 4/3 m.
Now ramp acceleration down with jerk −1 m/s³ for another 2 s. Use local time q for the second stage. Then a = 2 − q, v = 2 + 2q − 0.5q² and displacement within this stage is 2q + q² − q³/6.
At q = 2 s, acceleration returns to zero, velocity reaches 4 m/s and the second-stage displacement is 20/3 m. Total displacement is 8 m over 4 s.
The acceleration-time plot is a triangle. Its area gives the 4 m/s velocity change. Because the acceleration history is symmetric about the interval midpoint, the average velocity for this rest-start example is 2 m/s, consistent with 8 m displacement.
What is smooth here?
Acceleration is continuous at the joins if the surrounding stages have zero acceleration and the two ramps meet at 2 m/s². Jerk changes abruptly between the levels 0, +1, −1 and 0. Thus finite within-stage jerk does not mean every higher derivative is continuous.
A smoother jerk transition would require a more detailed profile. The label smooth should identify the quantity being discussed: continuous position, continuous velocity, continuous acceleration or continuous jerk.
A profile with finite jerk can reduce abrupt acceleration changes in a model. That mathematical property alone does not establish comfort, suitability or safety for a specific passenger, vehicle or mechanism.
An acceleration function given directly as a(t) can be integrated with respect to time when the function is known. A law such as a = −kv is different: acceleration depends on velocity, which is itself unknown. It defines a differential equation.
For k > 0, dv/dt = −kv with initial velocity v₀ gives v(t) = v₀e^(−kt). Integrating gives x(t) = x₀ + (v₀/k)(1 − e^(−kt)). The velocity approaches zero as time increases but does not reach zero at a finite time in this ideal model.
Here k has units s⁻¹. If v₀ = 6 m/s and k = 0.5 s⁻¹, velocity after 2 s is 6/e m/s, approximately 2.207 m/s. Displacement is 12(1 − e⁻¹) m, approximately 7.585 m.
The limiting displacement as t tends to infinity is 12 m. That does not mean the object stops at 12 m after a particular finite number of seconds. “Approaches” and “reaches” are different mathematical claims.
A model threshold is an added rule
A simulation may declare the object stopped when speed falls below a small threshold. That can be a useful numerical convention, but the threshold belongs to the simulation specification. It does not emerge from the exponential formula as an exact stopping event.
Likewise, an ideal resistive acceleration law may cease to be adequate at very low speeds or when other forces become important. A model’s long-time limit needs physical interpretation.
This example is a bridge from elementary kinematics to differential equations. The definitions remain unchanged. What changes is how the acceleration information is supplied and which mathematical method reconstructs the motion.
| Check | Question to ask | Typical repair |
|---|---|---|
| Units | Does each term have the same physical units? | Restore coefficient units and time scales |
| Initial state | Does the formula return the stated x and v at stage start? | Retain both integration constants |
| Join state | Do adjacent stages agree in x and v? | Carry the preceding ending state forward |
| Domain | Does the event time lie inside the stage? | Reject roots outside the allowed interval |
| Derivative | Do x′, v′ and a′ match the stated functions? | Differentiate the final expressions |
| Integral | Does acceleration area equal velocity change? | Compare signed area with endpoint values |
A plotted curve can look convincing while violating one of these checks. If the position curve jumps at an ordinary stage boundary, the expressions may have been reset to zero. If velocity jumps where only acceleration changes, an integration constant may be missing.
Inspect one-sided values at a switch. It is possible for position and velocity to agree while acceleration differs. There is no need to force all derivatives to be continuous when the ideal model explicitly includes an acceleration switch. There is a need to describe the switch honestly.
For distance, find where the actual velocity changes sign within each stage. A negative acceleration alone does not establish a reversal. During braking from positive velocity, the object continues forward until velocity reaches zero.
A productive correction strategy
Keep correct earlier stages. Identify the first boundary or derivative check that fails, repair that point and propagate the corrected state forward. Reworking an entire solution can conceal where the original misunderstanding occurred.
Then ask the learner to explain the repaired boundary in words. “The new stage begins at the old final position, with the old final velocity” is a stronger learning outcome than copying a corrected constant into a formula.
Exercise 1: a rising acceleration
Let a(t) = 2t m/s² for 0 ≤ t ≤ 3 s, with v(0) = 1 m/s and x(0) = 2 m. Find final velocity and position.
Integrating gives v = 1 + t² and x = 2 + t + t³/3. At 3 s, velocity is 10 m/s and position is 14 m. Displacement is 12 m. Differentiate both expressions and substitute the initial time to check them.
Exercise 2: coast after the ramp
Continue Exercise 1 with 2 s of coasting. Use u = t − 3 s. The new position is x = 14 + 10u m, so at t = 5 s the object is at 34 m and still moves at 10 m/s. Its acceleration has switched to zero.
Exercise 3: brake to rest
From that state, apply −2 m/s² until rest. Local braking time w gives v = 10 − 2w. The stop occurs after 5 s, with braking displacement 10×5 − 5² = 25 m. Final position is 59 m. End the braking formula at this event if the next stage is rest.
Exercise 4: same acceleration area, different travel
Starting from rest over a 2 s interval, compare constant a = 1 m/s² with a rising a = t m/s². Both gain 2 m/s. The constant case travels 2 m; the ramp travels 4/3 m. The velocity gain matches but its timing differs.
Transfer check: change the time scale
If a complete trajectory is traversed in half the time, r_new(t) = r_old(2t), velocity scales by 2, acceleration by 4 and jerk by 8 at corresponding points. These factors follow from the chain rule.
Do not apply the jerk factor simply because one instantaneous speed doubles. It requires the specified transformation of the whole time history. A transfer statement is strongest when its condition travels with it.
Frequently asked questions
Can I use v = u + at when acceleration changes? It gives the correct velocity change if a is the appropriate time-average acceleration for that interval. Using the same average in other constant-acceleration formulas does not generally preserve displacement.
Does zero jerk mean zero acceleration? No. It means acceleration is constant within the interval where the derivative exists.
Must acceleration be continuous at every stage boundary? An ideal piecewise model can include jumps. Identify them and recognise that ordinary finite jerk is not defined at those switching instants.
Why do stages use local time? It makes each stage’s initial conditions explicit. Global time works too, provided every offset and integration constant is retained consistently.
Connected reading directory
Use Kinematics for definitions and basic graphs, Calculus for differentiation and integration, and Dynamics when acceleration must be derived from forces rather than supplied directly.
The broader Learning That Transfers into Life guide explains the purpose of worked examples and checks. The present capability is to reconstruct an unfamiliar motion history while preserving its starting state, interval and physical meaning.
Sources and scope
The integration foundations were checked against OpenStax University Physics: finding velocity and displacement from acceleration. MIT OpenCourseWare’s kinematics lessons provide further study. Checked 22 October 2026. The ramp profiles, stage ledger, exponential example and exercises are original derivations. They illustrate mathematical models and do not specify operating limits or validated control systems.
Continue with connected motion guides
Reference frames and closest approach