IB Maths AI HL Modelling with Vectors Paper 1 & 2 Kinematics ~9 min read

Kinematics with Vectors

Moving objects in 2D or 3D have a position vector that depends on time, usually written r(t) = r₀ + tv for constant velocity. Two key kinematics questions: do two moving objects ever meet? (set the position vectors equal, find a common t) and what’s the closest they ever get? (minimise the magnitude of the displacement vector over time).

📘 What you need to know

Key terms:
· Displacement = position from a fixed starting point (vector)
· Velocity = rate of change of displacement (vector)
· Speed = |velocity| (scalar, always positive)
· Acceleration = rate of change of velocity (vector)
Same parameter t for both objects when checking intersection — t is real time, ticking simultaneously for both.
Do they intersect? Set r_A(t) = r_B(t), solve one component for t, check the other two components give the same t.
Shortest distance between them: form d(t) = r_B(t) − r_A(t), then minimise |d|² over t (avoid the square root by working with |d|²).
Three ways to minimise |d|²: complete the square, differentiate and set to 0, or GDC minimum-finder.
Initial position: substitute t = 0 into r(t).

Do two moving objects intersect?

Both objects share the same clock. If they meet, there’s one value of t that puts them at the same point. Set the position vectors equal, get three component equations, solve one for t, then check the other two. All three must agree.

Intersection condition r_A(t) = r_B(t) ⇒ same t works in all components if components give different t values, the paths cross in space but the objects don’t meet

Left: the paths cross AND the objects arrive at the crossing point at the same moment (t = 2) — they meet. Right: paths may cross in space but the objects pass through at different times; instead minimise |d(t)| over all t.

Shortest distance between two moving objects

Build the displacement d(t) = r_B(t) − r_A(t). The squared magnitude |d|² is a quadratic in t (sum of squares of linear terms). Minimise that quadratic, then take the square root.

Closest approach |d(t)|² = quadratic in t ⇒ minimum at t = t* ⇒ closest distance = √(|d(t*)|²) use d/dt |d|² = 0, or complete the square, or GDC minimum

Why minimise |d|², not |d|? Differentiating a square root involves the chain rule and a messy expression; differentiating a quadratic gives a single linear equation. Both give the same t* because |d| and |d|² are minimised at the same point (square root is monotonic).

🧭 Recipe — kinematics with two moving objects

Initial positions — substitute t = 0 into r_A and r_B.
Do they meet? Set r_A(t) = r_B(t). Solve one component for t, then verify the other components give the same t.
If they meet, substitute the value of t into either position vector to get the meeting point.
If they don’t meet, form d(t) = r_B(t) − r_A(t).
Minimise |d(t)|² — differentiate and set to 0, complete the square, or use GDC.
Substitute t* back into d, take magnitude → shortest distance.

Worked examples

WE 1

Initial positions of two objects (from the PDF)

Two objects A, B have position vectors r_A = (3−1) + t(−24) and r_B = (25) + t(3−1). Find the initial positions of the two objects.

substitute t = 0 r_A(0) = (3, −1) + 0·(−2, 4) = (3, −1) r_B(0) = (2, 5) + 0·(3, −1) = (2, 5) A starts at (3, −1); B starts at (2, 5) “initial” always means t = 0.

WE 2

Shortest distance between two moving objects (from the PDF)

Using A, B from WE 1, find the shortest distance between the two objects and the time when this occurs.

positions at time t A: (3 − 2t, −1 + 4t) B: (2 + 3t, 5 − t) displacement vector d = B − A d = (2+3t − (3−2t), 5−t − (−1+4t)) d = (−1 + 5t, 6 − 5t) |d|² = sum of squares |d|² = (−1+5t)² + (6−5t)² = 1 − 10t + 25t² + 36 − 60t + 25t² = 50t² − 70t + 37 minimise: d/dt(|d|²) = 0 100t − 70 = 0 → t = 0.7 substitute t = 0.7 |d|² = 50(0.49) − 70(0.7) + 37 = 24.5 − 49 + 37 = 12.5 |d| = √12.5 d ≈ 3.54 m at t = 0.7 min PDF’s exact answer.

WE 3

Do two objects meet?

Objects move with r_A = (7−1−6) + t(−215) and r_B = (154) + t(1−20). Do they collide? If yes, where?

set r_A(t) = r_B(t) x: 7 − 2t = 1 + t → t = 2 y: −1 + t = 5 − 2t → 3t = 6 → t = 2 ✓ z: −6 + 5t = 4 → t = 2 ✓ all three give t = 2 meeting point: substitute t = 2 into r_A r_A(2) = (7−4, −1+2, −6+10) = (3, 1, 4) they meet at (3, 1, 4) when t = 2 all three components must agree on the same t.

WE 4

Paths cross but objects don’t meet

Do the objects with r_A = (00) + t(21) and r_B = (40) + t(01) collide?

equate components x: 2t = 4 → t = 2 y: t = t ✓ (any t) but at t = 2, A is at (4, 2) and B is at (4, 2) A(2) = (4, 2), B(2) = (4, 2) they DO collide at (4, 2) when t = 2 ✓ when y-equation gives “any t”, the x-equation alone fixes t. Always plug back to verify.

WE 5

3D shortest distance — complete the square

Find the minimum distance between objects with r_A = (90−7) + t(−215) and r_B = (06−1) + t(1−20).

d = r_A − r_B d = (9−3t, −6+3t, −6+5t) |d|² = (9−3t)² + (−6+3t)² + (−6+5t)² = 81−54t+9t² + 36−36t+9t² + 36−60t+25t² = 43t² − 150t + 153 d/dt(|d|²) = 86t − 150 = 0 t = 150/86 = 75/43 minimum |d|² |d|²_min = 153 − (150)²/(4·43) = 954/43 d_min = √(954/43) ≈ 4.71 units PDF’s exact answer.

WE 6

Two boats — meet or pass close?

Boat P starts at (0, 0) with velocity (4, 3) km/h. Boat Q starts at (10, 0) with velocity (1, 2) km/h. Will they collide? If not, what’s the closest they get and when?

r_P(t) = (4t, 3t), r_Q(t) = (10+t, 2t) collide? equate x: 4t = 10 + t → t = 10/3 y: 3t = 2t → t = 0 ✗ no collision → find d_min d = r_Q − r_P = (10 − 3t, −t) |d|² = (10−3t)² + t² = 100 − 60t + 10t² d/dt = 20t − 60 = 0 → t = 3 substitute |d|² = 100 − 180 + 90 = 10 |d| = √10 ≈ 3.16 km no collision; closest = √10 km at t = 3 h

💡 Top tips

Same parameter t in both position vectors when checking collision — t is the same clock for both objects.
All 3 components must match for a collision — one matching component is not enough.
Minimise |d|², not |d| — same t*, much cleaner algebra (quadratic vs square root).
Three methods to minimise: complete the square, differentiate & set to 0, or GDC.
Initial = t = 0; speed = |velocity|; collision means same point AND same time.

⚠ Common mistakes

Using different parameters for the two objects — if you write t for A and s for B and they intersect in space at any (t, s), that doesn’t mean the objects collide. For collision, t = s.
Forgetting to verify all 3 components — solving 2 of 3 may give a “fake” collision that fails the third equation.
Minimising |d| directly — works but the algebra is messy. Use |d|².
Forgetting the square root at the end — the minimum of |d|² is the squared distance; take √ for the actual distance.
Reading “initial” as t = 1 instead of t = 0 — initial means before any time has passed.

Next up — Constant & Variable Velocity. When velocity is constant, the position is r = r₀ + tv — a straight line. When velocity varies with time, calculus enters: v = dr/dt, a = dv/dt for differentiation; integrate to go the other way.

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