Join constant acceleration stages without resetting velocity
Carry final velocity and position into the next stage, use local stage durations and check a complete journey with velocity–time areas.
Section 4.2 · Kinematics of motion in a straight line
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.2. Motion is restricted to one dimension. Choose a positive direction; distance and speed are scalars, while displacement, velocity and acceleration carry a direction. Deceleration means decreasing speed, not necessarily negative acceleration. Calculus uses Paper 1 techniques only. Constant-acceleration formulae require constant acceleration on the stated interval; use a new interval and its initial velocity when conditions change. Where vertical motion occurs, use the supplied g = 10 m s⁻² and the stated no-air-resistance model. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Use the constant-acceleration equations.
- Calculate signed areas and split at velocity zeros.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A particle reaches 6 m s⁻¹ after 2 s, then continues uniformly for 4 s. What displacement belongs to that uniform stage?
What you will learn
- Solve a journey with successive constant-acceleration and uniform-velocity stages.
- Track velocity continuity and account for a reversal inside a stage.
Each stage has its own duration and initial velocity
When acceleration changes, begin a new interval with the previous stage’s final velocity. Keep an elapsed-time record separately from each stage’s local duration. A particle starts from rest, accelerates at 2 m s⁻² for 4 s, travels at its reached velocity for 3 s, then accelerates at −4 m s⁻² until it first stops. The first stage ends with v = 8 m s⁻¹ and s = 16 m; the uniform stage adds 8 × 3 = 24 m.
The stopping stage starts with u = 8, not zero, and 0 = 8 − 4τ gives duration τ = 2 s. Its displacement is ½(8 + 0) × 2 = 8 m. Total time is 4 + 3 + 2 = 9 s and displacement is 48 m. All velocities are non-negative, so distance is also 48 m. A velocity graph joins (0, 0), (4, 8), (7, 8) and (9, 0); the two triangles and rectangle provide a separate check.
A stage need not end at the first stop unless the problem says so
If a particle starts with u = 6 and a = −2 for a fixed 4 s interval, it stops at 3 s but then reverses during the remaining second. Its final velocity is −2 m s⁻¹, its displacement is 8 m and its distance is 9 + 1 = 10 m. A later uniform stage must start at that negative velocity; do not silently stop the model at the first velocity zero when the duration is fixed.
Use a supplied first-stop condition to select a stopping time; use a supplied duration to continue through any reversal. Add stage displacements with signs to obtain the whole-journey displacement, and split inside stages when adding distance. Idealised acceleration changes give corners in a continuous velocity graph; they do not reset velocity instantaneously. The model describes the stated intervals only, and extending a deceleration past a stop requires explicit ongoing-motion assumptions.
Pause and explain
For u = 6 and constant a = −2 over a fixed 4 s, what final velocity must pass into the next stage?
Worked example
From rest, a particle accelerates at 2 m s⁻² for 4 s, continues uniformly for 3 s, then has a = −4 m s⁻² until its first stop. Find total time and distance.
Show the worked solution
- Stage 1 gives v = 8 m s⁻¹ and displacement 16 m; stage 2 adds 24 m in 3 s.
- Stage 3 starts at 8 m s⁻¹ and stops after 2 s, adding 8 m.
- Total time is 9 s and distance is 16 + 24 + 8 = 48 m, matching the non-negative velocity-graph area.
Answer Total time 9 s; distance and displacement 48 m.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
From rest, a particle accelerates at 3 m s⁻² for 2 s, travels uniformly for 4 s, then has a = −2 m s⁻² until it first stops. Find total time and distance.
Give me a hint
Find each stage’s reached velocity, duration and area before adding totals.
Compare my reasoning
- The first stage reaches 6 m s⁻¹ and adds ½ × 3 × 2² = 6 m; uniform travel adds 6 × 4 = 24 m.
- The stopping stage takes 6/2 = 3 s and adds ½ × 6 × 3 = 9 m.
- Total time is 2 + 4 + 3 = 9 s and distance is 6 + 24 + 9 = 39 m.
Total time 9 s; distance and displacement 39 m.
Look for these in your work
- Carry the velocity 6 m s⁻¹ into both later stages.
- Add durations and non-negative areas separately.
A particle initially moves at 10 m s⁻¹, has a = −2 m s⁻² for 3 s, then continues at its reached velocity for 5 s. Find final velocity, total time and total displacement.
Give me a hint
The first stage does not last long enough to stop the particle.
Compare my reasoning
- After 3 s, v = 10 − 2 × 3 = 4 m s⁻¹, and displacement is 10 × 3 − 3² = 21 m.
- The uniform stage adds 4 × 5 = 20 m and retains v = 4 m s⁻¹.
- Total time is 8 s and displacement is 21 + 20 = 41 m; distance is also 41 m because no reversal occurs.
Final velocity +4 m s⁻¹; total time 8 s; displacement and distance 41 m.
Look for these in your work
- Use 4 m s⁻¹ for the second stage rather than the original 10.
- Check velocity remains positive through the first stage.
A particle starts with v = +6 m s⁻¹, has constant a = −2 m s⁻² for 4 s, then continues uniformly for another 3 s. Find final velocity, whole-journey displacement and distance.
Give me a hint
The first stage contains a reversal, so split its distance at the zero velocity.
Compare my reasoning
- First-stage velocity reaches zero at 3 s and ends at −2 m s⁻¹ after 4 s.
- Its displacement is 6 × 4 − 4² = 8 m; its distance is 9 + 1 = 10 m.
- The uniform stage adds displacement −2 × 3 = −6 m and distance 6 m, giving total displacement +2 m and distance 16 m.
Final velocity −2 m s⁻¹; displacement +2 m; distance 16 m over 7 s.
Look for these in your work
- Continue through rest because the first-stage duration is fixed.
- Preserve negative velocity into uniform travel and add path lengths separately.
Common mistakes
- Use stage duration rather than total elapsed time in a stage’s equations.
- Carry the reached velocity into the next stage.
- A fixed acceleration interval can continue through rest and reverse direction.
What can you explain now?
If stage 1 ends with v = −2 m s⁻¹ and stage 2 is uniform motion, what is stage 2’s initial velocity?
Compare with the explanation
−2 m s⁻¹.
Uniform motion retains the signed velocity unless an explicitly modelled event changes it.
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Make it stick
Carry velocity between stages, use local durations and split reversals before adding whole-journey distance.
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