Use velocity vectors to find positions and collisions
Model straight-line motion with position vectors, compare particles at the same time, and solve a collision condition component by component.
Cambridge IGCSE Additional Mathematics 0606, 2028–2030 syllabus, Version 1. Paper 1 is non-calculator; Paper 2 requires a scientific calculator. These original lessons provide self-checked topic practice, not a Cambridge grade or a timed paper. Full-paper assessment and human specialist review are still outstanding.
Before you begin
- Cambridge assumes knowledge of IGCSE Mathematics or an equivalent course. Check the domain before using an inverse or logarithm, and retain exact values where the question asks for them.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the position after time t for initial position p₀ and constant velocity v?
What you will learn
- Find a position vector from an initial position and constant velocity
- Solve a same-time meeting or collision condition
- Check that coordinate equations describe the same time and position
Constant velocity gives a linear position vector
If a particle starts at position p₀ and has constant velocity v, then after time t its position is p(t) = p₀ + tv. Use matching time and distance units so the product tv has the correct units of displacement.
For two particles, write both position vectors as functions of the same time. They meet only when both component equations are true at the same t.
A collision requires a shared position and time
Set horizontal positions equal and vertical positions equal. Solve the resulting equations together; a crossing of one coordinate alone is not a collision.
Substitute the candidate time into both position vectors. If the positions match and the time lies in the allowed interval, the model predicts a meeting.
Pause and explain
What must match for two particles to collide?
Worked example
Particle A starts at (0, 0) with velocity (3, 2) m/s. Particle B starts at (10, 0) with velocity (1, 2) m/s. Do they meet, and where?
Show the worked solution
- At time t seconds, A is at (3t, 2t) and B is at (10 + t, 2t).
- The vertical coordinates match for every t; equate horizontal coordinates: 3t = 10 + t.
- Then t = 5 and both positions are (15, 10).
Answer They meet after 5 seconds at position (15, 10) metres.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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 particle starts at (−2, 5) and moves with velocity (4, −1) m/s. Find its position after 3 seconds.
Give me a hint
Multiply each velocity component by elapsed time, then add the initial position.
Compare my reasoning
- Displacement is 3(4, −1) = (12, −3).
- Add to (−2, 5).
- Evaluate each component.
Position (10, 2) metres.
Look for these in your work
- Time multiplied by velocity
- Initial position added
- Both coordinates calculated
A starts at (0, 0) with velocity (4, 1); B starts at (15, 3) with velocity (1, 0), all in metres and seconds. Find whether and when they meet.
Give me a hint
Write both position vectors and equate both coordinates at the same t.
Compare my reasoning
- A(t) = (4t, t); B(t) = (15 + t, 3).
- Vertical equality gives t = 3. Horizontal equality then requires 12 = 18, which is false.
- There is no time satisfying both coordinates, so they do not meet.
They do not meet: the vertical coordinates match at t = 3 s, but the horizontal positions are then 12 m and 18 m.
Look for these in your work
- Both motion vectors written
- Coordinate conditions checked together
- One-coordinate false match rejected
A cyclist at (2, 0) travels with velocity (5, 2) m/s. A runner at (20, 0) travels with velocity (2, 2) m/s. Find when and where the cyclist catches the runner.
Give me a hint
The vertical positions remain equal; compare horizontal positions at the same time.
Compare my reasoning
- Cyclist: (2 + 5t, 2t); runner: (20 + 2t, 2t).
- Set 2 + 5t = 20 + 2t, giving t = 6 seconds.
- At that time both are at (32, 12).
They meet after 6 seconds at (32, 12) metres.
Look for these in your work
- Initial positions retained
- Relative closing rate used
- Meeting position checked
Common mistakes
- Adding velocity directly to position without multiplying by time
- Comparing positions at different times
- Solving only one coordinate equation
- Forgetting to verify the meeting point in both position vectors
What can you explain now?
A particle starts at (2, 1) and has velocity (3, −2). What is its position after 4 seconds?
Compare with the explanation
(14, −7)
Add 4(3, −2) = (12, −8) to the initial position.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
How can position vectors turn a motion problem into simultaneous component equations?
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.