01 · Early position
Find displacement after 2 s of the first stage.
Hint
Use t = 2 with the first acceleration.
Worked solution
s = 3 × 2 + ½ × 2 × 2² = 10 m.
Understand · explore · practise
Linked stages of constant acceleration
Join constant-acceleration stages using shared boundary velocities, local durations and cumulative displacement.
Before you startSUVAT equation selection, signed velocity and displacement.
01 / Give each stage its own quantities
Keep one position reference and one overall clock.
At an ordinary boundary, final velocity becomes the next initial velocity.
The acceleration may change between stages. Use a fresh local duration in each stage; do not put the total journey time into an equation for only the second stage. Add signed displacements to obtain the overall displacement.
Carry velocity into the next stageExplore
A particle starts at 3 m/s, accelerates at 2 m/s² for 4 s, then accelerates at −1 m/s² for up to 6 s. Choose the elapsed time within the second stage.
02 / Solve the first stage before starting the second
It supplies the initial velocity for the next calculation.
The first stage starts at u = 3 m/s with a = 2 m/s² for 4 s.Worked example
v₁ = 3 + 2 × 4 = 11 m/s
This will be the initial velocity of stage 2.
s₁ = 3 × 4 + ½ × 2 × 4² = 28 m
This is displacement measured from the original start.
02 · Boundary values
What are velocity and displacement at the start of stage 2?
Hint
Carry over the final values of stage1.
Worked solution
Velocity = 11 m/s and cumulative displacement = 28 m. The second-stage elapsed time is zero there.
03 / Restart elapsed time, not velocity
Let τ be the duration since the second stage began.
Watch: overall time and second-stage duration
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 · Full second stage
Find its displacement and final velocity after 6 s with a = −1 m/s².
Hint
Use u₂ = 11 and τ = 6.
Worked solution
s₂ = 11 × 6 − ½ × 6² = 48 m. v₂ = 11 − 6 = 5 m/s.
04 · Whole journey
Find total time, displacement and average velocity.
Hint
Add stage durations and signed displacements.
Worked solution
Total time = 4 + 6 = 10 s. Displacement = 28 + 48 = 76 m. Average velocity = 7.6 m/s. Velocity stays positive, so distance and average speed have the same magnitudes.
04 / Convert a clock reading to a local duration
Subtract the time when that stage started.
05 · At seven seconds
Find velocity and cumulative displacement at overall t = 7 s.
Hint
The second stage has lasted τ = 7 − 4 = 3 s.
Worked solution
v = 11 − 3 = 8 m/s. Second-stage displacement = 33 − 4.5 = 28.5 m, giving total displacement = 56.5 m.
06 · Wrong substitution
Why is 11 × 7 − ½ × 7² not the displacement in stage2 by overall t = 7?
Hint
Seven is the overall clock reading.
Worked solution
Stage2 has run for only3 seconds. Its equation requires τ = 3. After finding its displacement, add the28 m already covered in stage1.
05 / Write simultaneous relations when a boundary speed is unknown
One shared speed links the stages.
07 · Unknown peak velocity
A particle starts from rest, accelerates uniformly for5 s to V, then travels at V for4 s. Total displacement is130 m. Find V.
Hint
The two areas are ½ × 5V and4V.
Worked solution
6.5V = 130, so V = 20 m/s.
08 · Recover acceleration
Find the first-stage acceleration in question7.
Hint
It changes from0 to20 m/s in5 s.
Worked solution
a = 20/5 = 4 m/s².
09 · Related accelerations
A particle starts at2 m/s, accelerates at a for4 s, then at a/2 for6 s. Total displacement is102 m. Find a.
Hint
First boundary speed is2 + 4a. Add both displacements.
Worked solution
s₁ = 8 + 8a. s₂ = 6(2 + 4a) + ½(a/2) × 36 = 12 + 33a. Thus20 + 41a = 102, so a = 2 m/s².
06 / Convert duration ratios into explicit local times
Twice as long does not mean the same acceleration.
10 · Two durations
A particle starts from rest and accelerates at2 m/s² for3 s, then at−1 m/s² for6 s. Find its final velocity.
Hint
The second duration is twice the first.
Worked solution
Stage1 ends at6 m/s. Stage2 ends at6 − 6 = 0 m/s.
11 · Distance and a pause
For question10 find total distance, then average speed if it remains at rest for a further2 s.
Hint
The velocity graph consists of two positive triangles and a rest interval.
Worked solution
Distance = ½ × 3 × 6 + ½ × 6 × 6 = 27 m. Total time including the pause is11 s, so average speed = 27/11 m/s, approximately2.45 m/s.
07 / Check boundaries and model intervals
An acceleration change need not cause a velocity jump.
12 · At the boundary
In the main example, what happens to velocity and acceleration at overall t = 4?
Hint
The two stages meet at11 m/s.
Worked solution
Velocity is continuous at11 m/s. Acceleration changes from+2 to−1 m/s²; acceleration, as the ordinary derivative of velocity, is not uniquely defined at the idealised sharp corner.
13 · One equation for all ten seconds
Why can neither a = +2 nor a = −1 be used for the whole main journey?
Hint
Neither is the acceleration throughout.
Worked solution
The acceleration changes after4 s. Use a separate equation for each stage or the full piecewise velocity-time graph.
14 · A later reversal
If the second stage instead continued beyond11 s of its own elapsed time, what extra care would distance require?
Hint
Its velocity is11 − τ.
Worked solution
A reversal occurs at τ = 11 s, which is overall t = 15 s. Split the second-stage path there and add absolute changes for distance. This extension is outside the original six-second second stage.
08 / Join stages using boundary data
Local times and cumulative positions play different roles.
Finish one stage, carry its velocity forward and reset that stage’s elapsed-time variable. Add stage displacements using a shared axis. Convert overall clock readings into local durations, and check for any reversal before adding total distance.
Section 1 of 8 · Give each stage its own quantities