Solve two particle meetings with a shared position and clock
Write both positions in one coordinate system, model delayed starts with the correct time shift and test all candidate meeting times.
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 position equations for constant acceleration.
- Solve quadratics and interpret their roots in a time domain.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
B starts at global t = 2 s from x = 5 m and moves at +3 m s⁻¹. Which position applies for t ≥ 2?
What you will learn
- Set up simultaneous motion equations for two particles.
- Check meeting times against starting times and explain which particle catches the other.
Equal position at the same time defines a meeting
Choose one origin, one positive direction and a common clock. Particle A starts from rest at x = 0 with a = 2 m s⁻², so xₐ = t². At t = 0, particle B is at x = 8 m and moves positively at constant 2 m s⁻¹, so xᵦ = 8 + 2t. A meeting requires t² = 8 + 2t, giving (t − 4)(t + 2) = 0. Only t = 4 is within the t ≥ 0 model.
At 4 s both particles are at x = 16 m, but their velocities are different: vₐ = 2t = 8 m s⁻¹ and vᵦ = 2 m s⁻¹. Equal positions do not imply equal velocities. On a displacement graph, a meeting is an intersection, while equal instantaneous velocities correspond to equal tangent slopes. If particles approach from opposite directions, their signed velocity equations must reflect that rather than assigning both a positive speed.
Delayed starts can produce two valid meetings
Suppose A still has xₐ = t² from global t = 0, while B starts from the origin at t = 1 s and moves positively at 8 m s⁻¹. For t ≥ 1, xᵦ = 8(t − 1), because B has been moving for t − 1 seconds, not t seconds. Equating positions gives t² − 8t + 8 = 0, with roots t = 4 ± 2√2. Both are after B starts, so both are valid meetings within the ongoing-motion model.
At the first meeting, B is faster than A and catches it; at the second, A is faster and catches B again. The positions are 24 ± 16√2 m, using the same sign as the respective time root. Check roots against all interval assumptions, such as delayed starts or a specified first stop. A negative root can describe a mathematical extension of an equation rather than a physical event, and a real root before B starts cannot represent a meeting of the modelled moving particles.
Pause and explain
For t² = 8 + 2t with the motion model beginning at t = 0, which meeting time is valid?
Worked example
A starts from rest at x = 0 with constant a = 2 m s⁻². B starts at x = 8 m at the same time and moves positively at 2 m s⁻¹. Find their meeting time, position and velocities.
Show the worked solution
- Using the shared clock, xₐ = t² and xᵦ = 8 + 2t.
- Equate them: t² − 2t − 8 = 0, giving t = 4 or −2; retain t = 4 s for the model.
- Both positions are 16 m; their velocities are vₐ = 8 m s⁻¹ and vᵦ = 2 m s⁻¹.
Answer Meeting at 4 s and x = 16 m; A velocity +8 m s⁻¹, B velocity +2 m s⁻¹.
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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.
A starts from rest at x = 0 with constant a = 2 m s⁻². B starts simultaneously at x = 15 m and moves positively at 2 m s⁻¹. Find their meeting time and position.
Give me a hint
Include B’s 15 m head start before equating the positions.
Compare my reasoning
- Write xₐ = t² and xᵦ = 15 + 2t.
- The meeting equation t² − 2t − 15 = 0 gives t = 5 or −3.
- Only t = 5 s is in the model; both positions then equal 25 m.
Meeting at 5 s and x = 25 m; negative time −3 s excluded.
Look for these in your work
- Keep the head start in B’s position equation.
- Substitute the retained time into both positions.
A starts at x = 0 moving at +6 m s⁻¹, while B starts simultaneously at x = 20 m moving at −4 m s⁻¹. Both velocities remain constant. Find their meeting time and position.
Give me a hint
B approaches the origin, so its position decreases on the same signed axis.
Compare my reasoning
- Write xₐ = 6t and xᵦ = 20 − 4t.
- Equating positions gives 10t = 20, so t = 2 s.
- Both positions equal 12 m; velocities remain +6 and −4 m s⁻¹ rather than becoming equal.
Meeting at 2 s and x = 12 m; velocities +6 and −4 m s⁻¹.
Look for these in your work
- Use B’s negative directed velocity.
- Check the same time gives one common position.
A starts from rest at x = 0 at global t = 0 with a = 2 m s⁻². B starts at the same origin at global t = 1 s and moves at +8 m s⁻¹ thereafter. Find all meetings for t ≥ 1 and explain the catching order.
Give me a hint
Shift only B’s elapsed moving time and test both quadratic roots.
Compare my reasoning
- For t ≥ 1, xₐ = t² and xᵦ = 8(t − 1); the meeting equation is t² − 8t + 8 = 0.
- Roots are t = 4 ± 2√2 s, both greater than 1; positions are x = 24 ± 16√2 m respectively.
- At the first meeting 2t < 8 so B catches A; at the second 2t > 8 so A catches B. Both events are valid while the stated motion continues.
Meetings at 4 − 2√2 and 4 + 2√2 s, at positions 24 − 16√2 and 24 + 16√2 m respectively; B catches A first, then A catches B.
Look for these in your work
- Use t − 1 for B and retain both domain-valid roots.
- Compare the actual velocities at each meeting to establish the catching order.
Common mistakes
- Use a shared origin and clock for both position equations.
- Replace t by elapsed time since a delayed start.
- Check all roots against the model and do not confuse a meeting with equal velocities.
What can you explain now?
At a displacement-graph intersection, must the two particles have equal velocities?
Compare with the explanation
No; only their positions are equal at that time.
Their velocities are the respective tangent gradients, which can differ.
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Make it stick
Use one coordinate system and clock, include each start position and delay, equate positions and verify every event root.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.