01 · Who is ahead?
Compare positions at t = 3 s.
Hint
Evaluate each position.
Worked solution
A is at12 m and B at9 m. A is3 m ahead.
Understand · explore · practise
Meeting and overtaking on a shared clock
Find meeting times by equating positions on common axes, interpret initial coincidence and check candidate times against piecewise motion stages.
Before you startSUVAT, position versus displacement and quadratic equations.
01 / Meeting means equal positions at the same time
Equal velocities answer a different question.
Use the same origin, positive direction and clock for both particles.
Write each position as initial coordinate plus signed displacement. Set the positions equal and solve for time. Then check whether the motion laws are valid at each candidate time.
Compare positions on one clockExplore
Two particles start together at x = 0. A moves at 4 m/s, so xA = 4t. B starts from rest with a = 2 m/s², so xB = t². Choose an overall time.
02 / Keep initial coincidence separate from later meeting
Factoring out time can hide a meaningful root.
For xA = 4t and xB = t², find meetings.Worked example
4t = t²
Meeting means the same coordinate.
t(t − 4) = 0
Do not divide by t and lose the initial root.
t = 0 or t = 4 s
The first is the start; the second is a later meeting.
x = 4 × 4 = 16 m
Check the common coordinate in both expressions.
02 · After the meeting
Compare positions at t = 5 s.
Hint
Use the same clock reading.
Worked solution
A is at20 m and B at25 m. B is now5 m ahead.
03 / Equal velocity does not imply a meeting
It can identify a maximum or minimum separation.
Watch: equal speeds before equal positions
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 · Equal velocities
At what time do the two particles have equal velocity?
Hint
vA = 4 and vB = 2t.
Worked solution
2t = 4 gives t = 2 s. Their positions are8 m and4 m, so they are not together.
04 · Greatest lead
Find the greatest lead of A over B before the later meeting.
Hint
The lead is4t − t² = 4 − (t − 2)².
Worked solution
The greatest lead is4 m at t = 2 s.
05 · Velocities when meeting
Find each velocity at t = 4 s.
Hint
Meeting does not require equal slopes of position graphs.
Worked solution
vA = 4 m/s and vB = 8 m/s. B passes A while moving faster.
04 / Include initial coordinates
Equal displacements need not give equal positions.
06 · Head start
A starts12 m ahead and moves at2 m/s. B starts at0 from rest with acceleration4 m/s². Write positions.
Hint
Add the head start only to A.
Worked solution
xA = 12 + 2t and xB = 2t², using the same t ≥ 0.
07 · Catch-up time
Find the physical meeting time for question6.
Hint
Solve 2t² = 12 + 2t.
Worked solution
t² − t − 6 = (t − 3)(t + 2) = 0. Only t = 3 s is nonnegative. Both positions are18 m.
08 · Displacement trap
Why would setting 2t = 2t² give the wrong catch-up time?
Hint
It ignores the initial separation.
Worked solution
Those are the two displacements, not the two positions. A began12 m ahead; the missing12 changes the meeting equation.
05 / A candidate must belong to the stage used
An accelerating particle may start cruising before it catches up.
09 · Acceleration ends
Return to A at4 m/s and B with a = 2 m/s² from rest. Now B accelerates for only3 s, then cruises at its attained speed. Can the earlier t = 4 answer be used unchanged?
Hint
The acceleration formula xB = t² applies only up to3 s.
Worked solution
No. The nonzero candidate t = 4 lies outside that stage. At t = 3, B is at9 m with velocity6 m/s, while A is at12 m.
10 · Cruise position
Write B’s position for t ≥ 3 in question9.
Hint
Its cruise duration is t − 3.
Worked solution
xB = 9 + 6(t − 3) = 6t − 9. The initial9 m belongs to the acceleration stage.
11 · Correct meeting
Find the later meeting for that piecewise journey.
Hint
Set 4t = 6t − 9.
Worked solution
t = 4.5 s, which satisfies t ≥ 3. Both positions are18 m. At the start of cruise B is3 m behind and closes at2 m/s, needing another1.5 s.
06 / Keep signed differences distinct from distance apart
Distance apart is the absolute difference of coordinates.
12 · Signed difference
In the original uninterrupted model, find xB − xA and distance apart at t = 1 and t = 5.
Hint
Use t² − 4t.
Worked solution
At1 s, the signed difference is−3 m and distance apart3 m. At5 s, the signed difference is+5 m and distance apart5 m.
If a question asks when particles are a given distance apart, consider both signs of the coordinate difference. The sign tells you which particle is ahead; the magnitude tells you the separation.
07 / Check the physical interaction assumption
A meeting calculation does not determine what happens in a collision.
13 · Passing or colliding
Does solving xA = xB automatically allow both formulae to continue afterwards?
Hint
Point-particle paths may cross, but a collision changes motion.
Worked solution
No. Continuation depends on the model. Independent ideal particles may pass, while interacting bodies may collide and require new velocities or a new phase.
14 · Never meeting again
Two particles start together and move with different constant positive velocities on the same line. Can they meet again if those velocities stay unchanged?
Hint
Their position difference is a nonzero constant times t.
Worked solution
No. Their only equal-position time is t = 0. The faster particle’s lead grows after departure.
08 / Equate positions and validate every candidate
Check origins, clocks, stages and direction.
Use one coordinate system and one clock. Include initial offsets, retain initial coincidence separately from later meetings, and reject roots outside the stage that produced them. Equal velocities are not the meeting condition.
Section 1 of 8 · Meeting means equal positions at the same time