01 · Clock readings
For a cruise duration of 8 s, when does braking start and finish?
Hint
Add the earlier 4 seconds.
Worked solution
Braking starts at t = 12 s and ends at t = 18 s.
Understand · explore · practise
Multistage velocity–time problems
Solve accelerate–cruise–brake problems with stage durations, unknown areas, speed thresholds and halfway positions.
Before you startVelocity-time slopes and areas; linear equations.
01 / Draw each stage on one clock
A stage duration is not its end time.
Use continuity at each ordinary stage boundary.
The next stage begins at the preceding end time and velocity. An ordinary change of acceleration changes the slope, not the velocity instantly. Set out stage durations before adding areas.
Change the cruising durationExplore
A vehicle accelerates uniformly from rest to 12 m/s in 4 s, cruises, then brakes uniformly to rest in 6 s. Choose the duration of the cruise.
02 / Build a duration and area table
For this journey, every velocity is nonnegative.
Cruise for 5 seconds in the model.Worked example
Accelerate: t = 0 to 4; area = ½ × 4 × 12 = 24 m
Acceleration = 12/4 = 3 m/s².
Cruise: t = 4 to 9; area = 5 × 12 = 60 m
Acceleration = 0.
Brake: t = 9 to 15; area = ½ × 6 × 12 = 36 m
Acceleration = −12/6 = −2 m/s².
Total: 120 m in 15 s
Average speed = 120/15 = 8 m/s.
02 · Total distance
Find the total distance for that 8 s cruise.
Hint
Add two triangles and the rectangle.
Worked solution
Distance = 24 + 96 + 36 = 156 m.
03 / Turn a total distance into an equation
Only one area depends on the unknown cruise duration.
Watch: a longer cruise adds a rectangle
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 · Find the cruise
The total distance in the model is 132 m. Find the cruising duration and total journey time.
Hint
Write 24 + 12c + 36 = 132.
Worked solution
12c = 72, so c = 6 s. Total time = 4 + 6 + 6 = 16 s.
04 · Impossible target
Can this model cover a total of 48 m while retaining the same acceleration and braking stages?
Hint
Find the distance with no cruise.
Worked solution
No. The two triangles already total 60 m. Solving gives c = −1 s, which is not a physical duration. At least one model condition must change.
04 / A common speed can be the unknown
Express every area using that same speed.
05 · Peak speed
A vehicle accelerates from rest to V m/s in 5 s, cruises for 7 s, then stops uniformly in 3 s. Total distance is 176 m. Find V.
Hint
Add ½ × 5V, 7V and ½ × 3V.
Worked solution
Total area = 11V = 176, so V = 16 m/s.
06 · Rates afterwards
Find the acceleration and braking acceleration in question 5.
Hint
Divide velocity changes by each duration.
Worked solution
Acceleration = 16/5 = 3.2 m/s². Braking acceleration = −16/3 m/s², approximately −5.33 m/s².
05 / Find crossings of a speed threshold
Distance above a speed is not the area above the threshold line.
07 · Time above eight
Use the original model with c = 6 s. For how long is speed greater than 8 m/s?
Hint
Acceleration v = 3t; braking begins at t = 10 with v = 12.
Worked solution
First crossing: t = 8/3 s. Braking reaches 8 m/s after 2 s, at t = 12 s. Duration = 12 − 8/3 = 28/3 s.
08 · Distance while fast
For the same journey, find the distance travelled while speed exceeds 8 m/s.
Hint
Use full velocity heights over the interval from 8/3 to 12.
Worked solution
Accelerating portion: ½(8 + 12) × (4/3) = 40/3 m. Cruise: 72 m. Braking portion: ½(12 + 8) × 2 = 20 m. Total = 316/3 m, about 105.33 m. The area between v and the line v = 8 would answer a different question.
06 / Locate the correct stage before solving a halfway problem
Half the distance need not take half the time.
09 · Halfway in distance
For c = 6 s, when is half the total distance covered?
Hint
Half of 132 is 66 m; 24 m has been covered after acceleration.
Worked solution
A further 42 m is needed during the cruise. At 12 m/s this takes 3.5 s, so the halfway time is t = 7.5 s.
10 · Halfway in time
For c = 6 s, how far has the vehicle travelled at half the total time?
Hint
Half of 16 s is 8 s.
Worked solution
Distance by t = 8 is 24 + 12 × 4 = 72 m. This exceeds half the total distance, which is 66 m.
07 / Check the stage containing a requested event
A value must lie within the interval used to obtain it.
11 · Early checkpoint
At what time is the first 6 m covered in the original model?
Hint
This lies within the initial 24 m stage; use a triangle with height 3t.
Worked solution
½ × t × 3t = 6 gives t² = 4. The physical time is t = 2 s, within 0 to 4 s.
12 · Final checkpoint
With c = 6 s, at what time are only 9 m left?
Hint
During braking, the remaining graph is a triangle. Let r be the remaining time; current speed is 2r.
Worked solution
Remaining distance = ½ × r × 2r = r² = 9. Thus r = 3 s, so t = 16 − 3 = 13 s.
13 · After stopping
Should the braking line continue below the axis after t = 16?
Hint
That would mean reversing direction.
Worked solution
Not for this vehicle-stopping model. It ends at rest, or a new stationary stage is added. Continuing the same signed acceleration would describe a different motion.
14 · Reversal in another journey
What changes if a different multistage graph includes negative velocity?
Hint
Use signed areas for displacement and magnitudes for distance.
Worked solution
Split any zero-crossing stage first. Add signed areas for displacement and positive area magnitudes for distance; do not use net area as total distance.
08 / Use slopes locally and areas across stages
Keep clock time, duration and distance separate.
Label every boundary time, calculate stage gradients and areas, and use the total area to find an unknown. For thresholds or halfway events, locate the interval before solving. Check that every duration and event time fits the model.
Section 1 of 8 · Draw each stage on one clock