01 · Six-second reading
At t = 6 s, how long has each particle been moving and where is each?
Hint
B’s duration is6 − 4.
Worked solution
A has moved for6 s and is at12 m. B has moved for2 s and is at4 m.
Understand · explore · practise
Delayed starts and shifted time
Write delayed-start equations using t minus the delay, reject pre-launch roots and check meetings against later changes of motion.
Before you startMeeting equations, SUVAT and quadratic roots.
01 / Choose one overall clock
The second particle has been moving for less time.
If B starts at overall time d, its elapsed moving time is t − d.
Use B’s post-start equation only for t ≥ d. Before then, describe B’s actual waiting motion separately. A negative t − d is not a negative physical duration to substitute into the launch model.
One clock, two elapsed timesExplore
A leaves x = 0 at t = 0 with constant velocity 2 m/s. B waits there until t = 4 s, then starts from rest with acceleration 2 m/s². Choose an overall clock reading.
02 / Write the position laws with their domains
A waiting interval belongs in the complete description.
For A at2 m/s and B released from rest at t = 4 s with a = 2 m/s²:Worked example
xA = 2t for t ≥ 0
A has travelled for the whole clock time.
xB = 0 for 0 ≤ t ≤ 4
B waits at the origin.
xB = (t − 4)² for t ≥ 4
Use ½ × 2 × (t − 4)² after release.
02 · Before release
What is B’s position at t = 2 s?
Hint
Use the waiting branch.
Worked solution
B is at0 m. Evaluating (2 − 4)² would give4 m, but the post-release expression is invalid at t = 2.
03 / Solve, then reject roots before the delayed start
A positive clock time may still precede launch.
Watch: the second clock starts four seconds later
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 · Candidate times
Solve the post-release meeting equation.
Hint
Set (t − 4)² = 2t.
Worked solution
t² − 10t + 16 = (t − 2)(t − 8) = 0, giving t = 2 or8 s.
04 · Physical meeting
Which candidate is the later catch-up, and where does it occur?
Hint
Require t ≥ 4 for B’s moving branch.
Worked solution
Only t = 8 s qualifies. Both are at16 m. The t = 2 root comes from extending B’s expression into its waiting interval.
05 · Two durations
How long after B starts is the catch-up?
Hint
Subtract the four-second delay.
Worked solution
8 − 4 = 4 s after B starts, or8 s after A starts. Label which duration you report.
04 / You may start the clock at B’s launch instead
Then A already has a head start.
06 · Shifted equations
Let τ = 0 when B starts. Write the two positions in terms of τ.
Hint
A has already travelled8 m.
Worked solution
xA = 8 + 2τ and xB = τ², for τ ≥ 0.
07 · Solve with the shifted clock
Find catch-up using question6.
Hint
Solve τ² = 8 + 2τ.
Worked solution
(τ − 4)(τ + 2) = 0. The physical root isτ = 4 s, agreeing with overall t = 8 s.
05 / Differentiate using the correct elapsed time
Waiting is part of a whole-period average.
08 · Velocities at catch-up
Find both velocities at overall t = 8 s.
Hint
B has accelerated for4 seconds.
Worked solution
vA = 2 m/s. vB = 2(t − 4) = 8 m/s.
09 · Greatest lead
When is A’s lead over B largest after B launches?
Hint
Equal velocities occur at2(t − 4) = 2.
Worked solution
At t = 5 s, B has velocity2 m/s. The lead is xA − xB = 10 − 1 = 9 m; it then decreases.
10 · Average speed with waiting
Find B’s average speed from t = 0 to8 and from t = 4 to8.
Hint
Its distance is16 m in both intervals, but durations differ.
Worked solution
Including the wait:16/8 = 2 m/s. During motion only:16/4 = 4 m/s.
06 / Combine a delayed start with initial position and velocity
Shift only the elapsed-time terms.
11 · General launch
B starts from x = 6 m at overall t = 2 s with initial velocity3 m/s and acceleration2 m/s². Write xB afterwards.
Hint
Use τ = t − 2 and add the initial coordinate6.
Worked solution
xB = 6 + 3(t − 2) + (t − 2)² for t ≥ 2.
12 · Evaluate the launch
For question11 find position and velocity at overall t = 5.
Hint
The moving time is3 seconds.
Worked solution
Position = 6 + 9 + 9 = 24 m. Velocity = 3 + 2 × 3 = 9 m/s.
07 / Check changes to either particle’s motion
The original catch-up answer may stop being valid.
13 · A stops early
In the main example A stops at t = 6 s and remains there. B continues accelerating. Find when B reaches A.
Hint
A’s stopped coordinate is12 m; use B’s moving branch.
Worked solution
(t − 4)² = 12 gives t = 4 + 2√3 s, approximately7.46 s. The other root is before B starts. The valid time is after A stops, so the stationary branch is consistent.
14 · Initial coincidence
Were A and B ever together before the later catch-up in the original model?
Hint
They begin at the same point before A leaves.
Worked solution
Yes, at t = 0. For0 < t ≤ 4, A moves away while B waits. Initial coincidence is different from the later catch-up at8 s.
08 / Align clocks before equating positions
Check both launch times and later stage changes.
Write a waiting branch if needed, then use t minus the delay in every moving-time term for the delayed particle. Solve equal positions, reject roots outside either model’s domain and label whether the answer is an overall clock reading or time since launch.
Section 1 of 8 · Choose one overall clock