Module 3: Kinematics · Theory · Prerequisite: K9
Motion that changes its rule is not a collection of separate problems. It is one history, described by changing equations.
Mahavakya
A physical system can change its law of evolution without losing its identity. The equation changes; the particle continues.
On this page
1. What piecewise motion is
Every part from K7 to K9 described motion governed by a single rule throughout. A real journey rarely behaves that way.
A car accelerates from rest, holds a constant speed on a motorway, brakes hard before a junction, and sits stationary at a red light. Four stages, four different accelerations, one continuous journey. The body is the same; the equation governing it changes at three moments.
Piecewise motion is motion of this kind, and the task is to describe the whole journey from a sequence of stage-by-stage descriptions.
K1 through K9 supply those stage-by-stage descriptions. This part is about connecting them.
2. State transfer between stages
Think of the particle as carrying a state — a pair (x, v) — that tells you where it is and how fast it is going at any instant. At the end of each stage the state has a value, and that value becomes the initial condition for the next stage.
(x0, v0) → Stage I → (x1, v1) → Stage II → (x2, v2) → …
The stages are not separate problems. They are the same problem — one particle, one motion — and the chain of states is what makes them one thing rather than many.
The practical consequence: do not start a new stage with arbitrary initial conditions. Every initial condition except the very first one comes from the final state of the preceding stage, which you must compute before moving on.
3. The clock never resets
This is the error that appears most reliably, and it comes in two forms.
Resetting the time. A student solves stage I over 0 to 3 s, then begins stage II by writing t = 0 again at the start of stage II. That is legitimate only if they use a local variable τ = t − 3 and are careful never to confuse it with the global t. Most of the time students carry both in the same working and mix them.
Resetting the velocity. A stage ends with v = 6 m/s, and the next stage opens with v0 = 0. That is simply a wrong initial condition. The particle had a velocity at the boundary and keeps it.
Both errors are invisible — a working with a reset clock or a reset velocity looks structurally identical to a correct one. The only check is to carry the state explicitly through each boundary and verify that it matches.
4. A complete worked example
A particle starts from rest. It accelerates at 2 m/s² for 4 s, then decelerates at 1 m/s² until it stops. Find the total distance and draw the velocity–time graph.
Stage I: 0 to 4 s, a = +2 m/s²
Initial state: (x = 0, v = 0).
v(4) = 0 + 2(4) = 8 m/s ΔxI = ½(2)(16) = 16 m
Final state: (x = 16, v = 8).
Stage II: from t = 4 s, a = −1 m/s²
Initial state: (x = 16, v = 8). The velocity from stage I carries over.
Duration until stop: v = 8 − 1 τ = 0 → τ = 8 s, so stage II ends at t = 12 s.
ΔxII = 8(8) − ½(1)(64) = 64 − 32 = 32 m
Final state: (x = 48, v = 0).
Summary
| Total time | 12 s |
| Total distance | 16 + 32 = 48 m |
| Peak velocity | 8 m/s at t = 4 s |
The velocity never goes negative, so there is no reversal and the distance equals the displacement. The v–t graph is a triangle with a kink at (4, 8): rising line to the peak, then a shallower falling line to the axis at t = 12.
5. What is continuous and what is not
At the boundary between two stages, a change in mathematical smoothness of one quantity appears one level up in the quantity below it in the chain.
Consider a sudden change in force — say, a rope that snaps. The acceleration jumps. What does that produce?
The a–t graph has a jump — the discontinuity is at the level of acceleration.
The v–t graph has a kink — velocity is continuous but its slope changes abruptly. What K6 called the impulse idealisation applies here if the jump in a is itself instantaneous; in practice, a very short interval of very large acceleration.
The x–t graph has neither — position is smooth. A particle cannot jump, and if it can sustain a kink in x–t, that kink represents an instantaneous velocity change, which is the same impulse again.
The practical rule: position is always continuous across a boundary. Velocity is continuous unless the problem describes a collision or impulse. Acceleration can be anything — it can jump, vanish, or change sign, because it is the rate of change of a rate of change and no continuity is required of it.
This pattern — a discontinuity in one quantity appearing smoothed by one derivative in the next — is the same idea K5 named for the admissibility of position–time graphs and K6 named for the impulse on a velocity–time graph. Here it appears in its most general form, and it is worth recognising as one pattern appearing in three settings.
6. Reading it off graphs
A piecewise motion is often given as a graph rather than equations, and the graph-reading skills of K5 and K6 do all the work.
From a v–t graph
- Sign of v — which way the body is moving in each stage.
- Slope — the acceleration in each stage. A kink is a boundary between stages.
- Signed area — the displacement over any interval. K6’s crossing rule applies: find where v = 0, split there, add signed areas for displacement or unsigned for distance.
- Zero crossings — reversals, and therefore where displacement and distance diverge.
Worked example — from a graph
A v–t graph shows: v rises from 0 to +6 m/s over the first 3 s (straight line), holds at +6 m/s from 3 to 7 s, then falls to 0 at 10 s (straight line). Find the total distance.
Stage I (0–3 s): triangle, area = ½(3)(6) = 9 m.
Stage II (3–7 s): rectangle, area = (4)(6) = 24 m.
Stage III (7–10 s): triangle, area = ½(3)(6) = 9 m.
Total = 9 + 24 + 9 = 42 m
v is always positive so the distance equals the displacement. Three simple areas, no equations written down. The graph-reading approach from K5 and K6 works directly.
7. Before you calculate
Every piecewise problem repays a moment of preliminary work before any number is substituted.
- Identify every stage. Mark the boundaries — the times or positions at which the rule changes.
- Decide what each stage is. Constant velocity (K7), constant acceleration (K8), or variable acceleration (K9).
- Check for reversal within any stage. If so, the boundary within that stage is where v = 0, and distance and displacement diverge there.
- Set up the state table. Write the initial state for stage I, leaving the end state blank. Carry it through each stage in sequence.
- Now write equations.
Step 3 deserves special care. Students often assume a stage with positive acceleration always produces increasing velocity — but if the stage starts with a negative velocity, the body is still moving backward and slowing down. The sign of the velocity at stage entry, not the sign of the acceleration, decides whether a reversal occurs within the stage.
Contemplation
There is something philosophically interesting in what piecewise motion assumes. The particle, during each stage, behaves as if a single equation governs it. At the boundary the equation changes, but the particle itself does not — it passes through with the same position and velocity, and the next stage begins as if the old one had always been a preparation for it.
That is a model, of course. Real forces change continuously rather than jumping at a point. But the piecewise model is exactly how engineers and physicists think about staged motion, because it separates the stages cleanly without losing the thread of continuity between them.
And keeping that thread is the whole skill. The equation changes; the particle continues.
Common misconceptions
1. Each stage is a fresh problem
Each stage inherits the final state of the previous one. Starting a stage with arbitrary initial conditions treats a single motion as multiple unrelated events.
2. The clock resets at each boundary
A local stage clock is fine if used carefully, but the global time must be tracked. Mixing the two gives wrong times for events in later stages.
3. Positive acceleration means the body speeds up throughout the stage
Only if the velocity is already positive. A stage starting with v = −4 m/s and a = +2 m/s² begins with the body moving backward and slowing.
4. Velocity is always continuous
For smooth motion, yes. For collisions and impulses, the velocity can jump discontinuously — and the position stays continuous regardless.
5. Distance equals displacement if the body ends up ahead of where it started
Not if it reversed at any point during the journey. Distance is always at least as large as the magnitude of displacement, and often larger.
Reflection
- A particle completes two stages and ends exactly where it began. What can you say about its displacement, its distance, and its final velocity?
- A stage has constant acceleration and the particle reverses within it. At what point do you split the distance calculation — at the reversal or at the stage boundary?
- The v–t graph for the worked example in section 4 is described but not drawn. Sketch it and verify the areas match.
Key takeaways
- Piecewise motion is one history described by changing equations. The particle is continuous even when its governing equation is not.
- The final state of each stage is the initial condition of the next. Carry it explicitly.
- Neither the time nor the velocity resets at a boundary.
- Position is always continuous. Velocity is continuous unless an impulse acts. Acceleration can be anything.
- A reversal within a stage splits the distance calculation — not the boundary, but the instant when v = 0 inside the stage.
- Identify stages and check for reversals before writing a single equation.
What comes next
Everything so far has been one-dimensional — one axis, one sign convention, one component. K11 lifts that restriction, and immediately the velocity is a vector with two independent components. Some things simplify; one thing becomes unexpectedly important. The next part is where kinematics meets geometry.
Prerequisite: K9 · Reading: 14 min · Practice: 30 min · Difficulty: Intermediate