IB Maths AI HLModelling with VectorsPaper 1 & 2Calculus + vectors~9 min read
Constant & Variable Velocity
Two cases of motion. Constant velocity: straight-line motion with r = r0 + tv. Variable velocity: v depends on t, so calculus enters — differentiate position to get velocity, velocity to get acceleration; integrate to go back. One constant of integration per component.
📘 What you need to know
Constant velocity formula: r(t) = r0 + tv · r0 = position at t = 0
· v = constant velocity vector
· same form as the vector equation of a line: r0 ≡ a, v ≡ b
Variable velocity: v(t), a(t), r(t) all functions of t.
Differentiation chain: v = drdt, a = dvdt = d²rdt²
Integration chain: v = ∫a dt, r = ∫v dt
Component-wise: differentiate or integrate each component separately. One constant C per component.
To find C: use a known position or velocity at a specific time (boundary condition).
Speed = |v|; this is a scalar function of time.
Constant velocity — straight-line motion
If v doesn’t change with time, the object moves in a straight line. The position at time t is just the starting point plus t times the velocity.
Constant velocityr(t) = r0 + tvr0 = initial position · v = velocity (constant) · same form as r = a + λb
When velocity depends on time, treat each component of the vector as its own function of t. Differentiate to climb the ladder (position → velocity → acceleration), integrate to descend (acceleration → velocity → position).
Climb the ladder (position → velocity → acceleration) by differentiating each component. Descend (acceleration → velocity → position) by integrating each component — and remember to add a constant for every component.
Variable velocity — calculus linksv = drdt, a = dvdt = d²rdt²reverse: v = ∫a dt + C, r = ∫v dt + Cone constant per component · find each using a known position/velocity
Why component-wise? Each component (i, j, k) evolves independently. Differentiating or integrating a vector just means doing it to each component, with its own constant when integrating.
🧭 Recipe — motion with vectors
Constant velocity — use r = r0 + tv. Read off r0 and v; substitute t.
Find r0 and v from two points: r0 = position at t = 0; v = (position at later t − r0) / t.
Variable: given v(t), find a — differentiate each component.
Variable: given v(t), find r — integrate each component, add constants C, D (one per component). Use r at a known t to solve for the constants.
Variable: given a(t), find v then r — integrate twice; need two sets of boundary conditions (one for v, one for r).
Speed at time t = |v(t)|.
Worked examples
WE 1
Constant velocity from two points (from the PDF)
A car moves at constant speed in a straight line from A(−4, 3) to B(6, −5) in 2 minutes. Position vector p(t) = a + tb. Find a and b.
a = position at t = 0 (point A)a = (−4, 3)at t = 2, position = Ba + 2b = (6, −5)(−4, 3) + 2b = (6, −5)2b = (10, −8)b = (5, −4)a = (−4, 3); b = (5, −4)PDF’s exact answer. b is the velocity in metres per minute.
WE 2
Acceleration & position from variable velocity (from the PDF)
A ball rolls down a hill with velocity v = (53) + t(0−0.8). At t = 0 the ball is at position 3i − 2j. (a) Find a. (b) Find r(t).
v(t) = (5, 3 − 0.8t)(a) acceleration = dv/dta = (0, −0.8)a = −0.8 j(b) r = ∫ v dtr = (5t + C, 3t − 0.4t² + D)use r(0) = (3, −2)(5(0) + C, 3(0) − 0.4(0)² + D) = (3, −2)C = 3, D = −2r = (5t + 3) i + (3t − 0.4t² − 2) jPDF’s exact answer. Always add one constant per component, then use boundary data.
WE 3
Speed at a specific time
An object has velocity v(t) = (2t, 3 − t, 1) m/s. Find the speed when t = 2 s.
speed = |v(t)|substitute t = 2v(2) = (4, 1, 1)magnitude|v(2)| = √(16 + 1 + 1) = √18speed = √18 ≈ 4.24 m/sspeed is always a positive scalar — magnitude of the velocity vector.
WE 4
Integrate acceleration TWICE
A particle has acceleration a(t) = (6t, 4) m/s². At t = 0: v = (1, 0), r = (0, 0). Find v(t) and r(t).
v = ∫ a dtv = (3t² + C, 4t + D)use v(0) = (1, 0)C = 1, D = 0v(t) = (3t² + 1, 4t)r = ∫ v dtr = (t³ + t + E, 2t² + F)use r(0) = (0, 0)E = 0, F = 0r(t) = (t³ + t, 2t²)two integrations need two sets of boundary conditions — one for v, one for r.
WE 5
Find when velocity is perpendicular to a vector
A particle’s velocity is v(t) = (3 − t, 2t). For what value of t is v perpendicular to u = (4, 1)?
perpendicular ⇒ v · u = 0(3 − t)(4) + (2t)(1) = 012 − 4t + 2t = 012 − 2t = 0 → t = 6t = 6 sapplies same perpendicular trick as static vectors — dot product = 0.
WE 6
Position at later time + minimum speed
Velocity v(t) = (t − 2, 3). At t = 0, position is (1, −4). (a) Find r(t). (b) Find the minimum speed.
(a) integrate vr = (½t² − 2t + C, 3t + D)use r(0) = (1, −4)C = 1, D = −4r(t) = (½t² − 2t + 1, 3t − 4)(b) speed² = |v|² = (t−2)² + 9speed² = (t − 2)² + 9minimum when (t − 2)² = 0t = 2, min speed² = 9min speed = 3 (at t = 2 s)minimum of a sum of squares: zero out the variable square term.
💡 Top tips
Component-wise everything — differentiate and integrate each component separately.
One constant per component — when integrating a 2D velocity, you get C and D (two separate constants). 3D needs three.
Always use boundary data to fix integration constants — usually r(0) or v(0).
Speed vs velocity: speed is a scalar (|v|); velocity is a vector. Speed is always positive.
Constant velocity = vector line equation: r = r0 + tv is just r = a + λb with t as the parameter.
To find velocity from two positions: v = (later position − earlier position) / (time elapsed).
⚠ Common mistakes
One constant for the whole vector — wrong. You need one constant per component.
Forgetting to use boundary conditions — without them, C and D stay unknown and the answer is incomplete.
Differentiating only one component — if v = (5, 3 − 0.8t), the i component differentiates to 0, not “nothing”. Acceleration = (0, −0.8), not just −0.8j with no i.
Confusing speed and velocity — speed is |v|, a scalar. Don’t write “speed = (5, 3)”.
Treating t as a vector — t is a scalar; multiply it by a vector to get a displacement.
For double integration (a → r), forgetting that you need two sets of boundary conditions (one for v, one for r).
Chapter complete — you now have both Modelling with Vectors sub-topics: Kinematics with Vectors (intersection of paths, shortest distance between moving objects) and Constant & Variable Velocity (constant-velocity formula r = r0 + tv, plus the calculus chain a = dv/dt, v = dr/dt, and their integrals). Together they cover every Paper 1 & 2 modelling-with-vectors question on the AI HL syllabus.
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