IB Maths AA HL
Topic 3 β Geometry & Trigonometry
Paper 1 & 2
~6 min read
HL only
Applications to Kinematics
A line equation is also a motion equation: r = r0 + vt describes an object moving with constant velocity. The starting point becomes r0, the direction becomes the velocity v, and the parameter Ξ» becomes the time t.
π What you need to know
- Position vector at time t: r = r0 + vt (not in the formula booklet β memorise).
- r0 = position vector at t = 0 (the start).
- v = the (constant) velocity vector β direction Γ magnitude.
- Displacement: position relative to a fixed reference point.
- Velocity: rate of change of displacement (vector).
- Speed: |v|, the magnitude of velocity (scalar).
- Distance travelled (constant velocity): speed Γ time = |v| Γ t.
- Two objects collide if the same t gives equal position vectors in all components.
The motion equation
Position at time t (constant velocity)
r = r0 + vt
It’s the same equation as r = a + Ξ»b with new labels: a = r0 (starting point), b = v (velocity), Ξ» = t (time). At t = 0 you’re at the start; each unit of time moves you by v.
Velocity (vector)
v
tells you how fast AND in what direction
Speed (scalar)
|v|
just how fast β magnitude only
Common question types
| Asked for⦠| Use⦠|
|---|
| position at given time | plug t into r = rβ + vt |
| velocity from two points + time | v = (rβ β rβ) / Ξt |
| speed | |v| = β(vβΒ² + vβΒ² + vβΒ²) |
| distance travelled | distance = speed Γ time |
| time to reach a point | solve target = rβ + vt for t componentwise |
| do two objects collide? | set rβ(t) = rβ(t); same t in all components β yes |
π§ Recipe β set up a kinematics problem with vectors
- Identify r0: the position at time zero (often given as a coordinate point).
- Identify v: the velocity vector (given directly, or compute as displacement Γ· time).
- Write: r = r0 + vt.
- For position at a time β plug in t; for time at a position β solve componentwise and check all components agree.
- For speed/distance β find |v| first, then multiply by time if needed.
Worked examples
WE 1Position of a particle after a given time
A particle starts at the point (3, β1, 2) and moves with constant velocity v = 2i + 5j β k m/s. Find its position vector after 4 seconds.
Apply r = rβ + vt
r = (3, β1, 2) + 4(2, 5, β1)
= (3, β1, 2) + (8, 20, β4)
= (3+8, β1+20, 2β4)
r = (11, 19, β2)
just multiply velocity by time and add to the start position
WE 2Find velocity vector and speed
A boat travels in a straight line from P(2, 8) to Q(14, β1) in 3 hours at constant velocity. Find the velocity vector and the speed of the boat.
Step 1: Find displacement PQ
PQ = Q β P = (14β2, β1β8) = (12, β9)
Step 2: Velocity = displacement Γ· time
v = (12, β9) / 3 = (4, β3) km/h
Step 3: Speed = |v|
|v| = β(4Β² + (β3)Β²) = β(16 + 9) = β25
v = (4, β3) km/h, speed = 5 km/h
velocity per unit time is total displacement divided by time taken
WE 3Find the time to reach a given point
A particle moves with position vector r = (1, β3, 5) + t(2, 4, β3). At what time t does it pass through the point (7, 9, β4)?
Set r = (7, 9, β4) and solve each component for t
x: 7 = 1 + 2t β t = 3
y: 9 = β3 + 4t β t = 3 β
z: β4 = 5 β 3t β t = 3 β
t = 3 (seconds)
all three components must agree on the same t β otherwise the particle never passes through that point
WE 4Distance travelled in a given time
A particle has initial position vector (5, β2) and velocity v = β3i + 4j m/s. Find the distance travelled by the particle in 6 seconds.
Step 1: Find the speed |v|
|v| = β((β3)Β² + 4Β²) = β(9 + 16) = β25 = 5 m/s
Step 2: Distance = speed Γ time
distance = 5 Γ 6
distance = 30 m
starting position doesn’t matter for distance β only speed and time
WE 5Determine whether two objects collide
Two boats move with position vectors rA = (2, β1) + t(3, 4) and rB = (10, 19) + t(β1, β6), where t is in hours. Determine whether the boats collide; if so, when and where.
Set r_A = r_B and solve componentwise
x: 2 + 3t = 10 β t β 4t = 8 β t = 2
y: β1 + 4t = 19 β 6t β 10t = 20 β t = 2 β
Same t in both components β collision
Find collision point: substitute t = 2
r_A = (2, β1) + 2(3, 4) = (8, 7)
Collide at (8, 7) at t = 2 hours
“collide” means same place at the SAME time β both components must give equal t
WE 6Find the initial position from a later observation
A car moves in a straight line with constant velocity v = (4, β3) m/s. After 5 seconds, the car is at the point (15, β8). Find the initial position of the car.
Step 1: Rearrange r = rβ + vt β rβ = r β vt
Step 2: Substitute t = 5, r = (15, β8), v = (4, β3)
rβ = (15, β8) β 5(4, β3)
= (15, β8) β (20, β15)
= (15 β 20, β8 + 15)
Initial position: (β5, 7)
working “backwards in time” β subtract velocity Γ time from the known later position
π‘ Top tips
- Velocity is a vector; speed is its magnitude β don’t confuse them in word problems.
- Distance travelled = speed Γ time only when velocity is constant.
- For collisions, the same t must satisfy all components β different t‘s mean the paths cross but the objects miss each other.
- Watch the units: if velocity is m/s and time is in minutes, convert.
- Visualise r0 as the launchpad and v as the arrow showing where one second of motion takes you.
β Common mistakes
- Computing displacement as v Γ t and calling it distance β distance is |v| Γ t, not the vector itself.
- Mixing up “paths cross” with “objects collide”. Paths can intersect at different times β not a collision.
- Forgetting to verify all components when solving for time β one component agreeing is not enough.
- Using r instead of r β vt when working backwards to find r0.
- Treating speed as a vector (writing it with components) β speed is a single non-negative number.
Next: Coincident, Parallel, Intersecting & Skew Lines. In 3D, two lines can do four very different things β sit on top of each other, run in parallel, cross at a point, or pass without ever meeting. The dot/scalar product and parameter-matching tell you which.
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