IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 ~10 min read

Displacement, Velocity & Acceleration

Kinematics is motion-along-a-line with strict sign conventions. Displacement is signed (not distance), velocity is signed (not speed), and “accelerating” vs “decelerating” depends on whether v and a have the SAME or OPPOSITE sign. Most paper questions in this chapter sit in two flavours: interpreting velocity-time graphs, and turning English phrases (“dropped from a cliff”, “moving due east”) into signs.

📘 What you need to know

Displacement s is signed; distance is always positive. A bus returning to its depot has displacement 0 but distance = the full route length.
Velocity v is signed; speed = |v|. v = −6 means speed 6 m/s in the negative direction.
Acceleration a can be negative without the particle slowing down. The rule is: same sign as v ⟹ speeding up; opposite sign ⟹ slowing down.
“At rest” means v = 0. This can happen at the start, the end, or momentarily (e.g., at the top of a thrown ball’s flight).
v = 0 does NOT mean a = 0. A ball at the peak has v = 0 but a = −g still pulling it down.
Velocity-time graphs: gradient = acceleration; (signed) area = displacement; total |area| = distance.
Direction conventions: positive direction is usually right (horizontal motion) or up (vertical). Stick with one and apply it consistently.
Common phrases: “due east”/”right” → v > 0; “due west”/”left”/”dropped”/”down” → v < 0; “at rest” → v = 0.

The four quantities — and their signs

Each pair of words people use loosely in everyday speech has a STRICT meaning in kinematics. Get the signs right and the rest follows.

Quantity	Symbol & units	Sign convention
Displacement	s, m	SIGNED. Position relative to a fixed point. Negative = on the negative side.
Distance	d, m	ALWAYS POSITIVE. Total length travelled (regardless of direction).
Velocity	v, m s⁻¹	SIGNED. Rate of change of displacement. Negative = moving in negative direction.
Speed	\|v\|, m s⁻¹	ALWAYS POSITIVE. Magnitude of velocity (drop the sign).
Acceleration	a, m s⁻²	SIGNED. Rate of change of velocity. Sign by itself doesn’t tell you speeding up vs slowing down.

The trap: “negative acceleration” does NOT always mean the particle is slowing down. If the particle is moving in the negative direction (v < 0) and a < 0 too, it’s actually SPEEDING UP. The test is sign(v) × sign(a).

Speeding up

v and a have the SAME sign

both positive: moving forward, getting faster
both negative: moving backward, getting faster

Slowing down

v and a have OPPOSITE signs

v positive, a negative: forward but braking
v negative, a positive: backward but braking

Reading a velocity-time graph

Velocity-time graphs are the workhorse of this chapter. Two things about them you absolutely must know: the GRADIENT is the acceleration, and the AREA between the curve and the t-axis is the (signed) displacement.

Feature on graph	What it means physically
Above t-axis (v > 0)	Particle moving in POSITIVE direction
Below t-axis (v < 0)	Particle moving in NEGATIVE direction
Touches t-axis (v = 0)	Particle momentarily at rest (or changing direction)
Crosses t-axis	Particle CHANGES DIRECTION at that instant
Horizontal line	Constant velocity (acceleration = 0)
Straight line (non-horizontal)	Constant non-zero acceleration
Steeper line	Greater magnitude of acceleration
Sum of areas (with sign)	DISPLACEMENT over the time interval
Sum of \|areas\|	DISTANCE travelled over the interval

🧭 Recipe — interpret a velocity-time graph

State the direction convention — positive direction = right (or up, for vertical motion). Note this even if the question doesn’t.
Find where v = 0 — these are the times the particle is at rest. If the graph CROSSES the axis here (not just touches), the particle is changing direction.
Read the sign of v on each interval to determine direction of motion.
Use the gradient for acceleration; compare sign(v) and sign(a) to decide speeding up vs slowing down on each interval.
Sum areas with sign for displacement; sum |areas| for distance. Always state with units.

Worked examples

WE 1

Distance vs displacement on a single straight road

A cyclist rides 350 m east along a straight road, then turns around and rides 120 m west, stopping there. Taking east as the positive direction, find (a) the total distance travelled, and (b) the cyclist’s displacement from the starting point.

Step 1 — set up the sign convention east = positive direction (given) leg 1: +350 m (east) leg 2: −120 m (west, so signed as negative) Step 2 — distance ignores direction d = 350 + 120 = 470 m Step 3 — displacement is the SIGNED sum s = (+350) + (−120) = 230 m Distance = 470 m; Displacement = 230 m east distance = how far the wheels turn; displacement = where the cyclist actually ends up. Two different numbers, two different meanings.

WE 2

Speed and direction from given velocities

A particle moves along a horizontal line. Its velocity, in m s⁻¹, is recorded at three instants: at t = 1 the velocity is −5; at t = 4 the velocity is 3; at t = 7 the velocity is −8. For each instant, state the speed of the particle and the direction in which it is moving.

Step 1 — speed = |v| (drop the sign) at t = 1: v = −5 → speed = 5 m s⁻¹ at t = 4: v = 3 → speed = 3 m s⁻¹ at t = 7: v = −8 → speed = 8 m s⁻¹ Step 2 — direction from the sign of v v < 0 → moving in negative direction v > 0 → moving in positive direction t=1: 5 m/s, negative direction; t=4: 3 m/s, positive direction; t=7: 8 m/s, negative direction speed alone tells you “how fast” but NOT “which way”. The sign of v gives the direction; |v| gives the speed.

WE 3

Speeding up or slowing down — four cases

At a particular instant, a particle has the velocity v (in m s⁻¹) and acceleration a (in m s⁻²) given in each row below. State, with reason, whether the particle is speeding up or slowing down at that instant.

(a) v = 4, a = 2 (b) v = 6, a = −3 (c) v = −5, a = −2 (d) v = −7, a = 4

Rule: same sign for v and a ⇒ speeding up; opposite signs ⇒ slowing down (a) v = 4 (positive), a = 2 (positive) same sign → SPEEDING UP (b) v = 6 (positive), a = −3 (negative) opposite signs → SLOWING DOWN (c) v = −5 (negative), a = −2 (negative) same sign → SPEEDING UP (← “negative a” but speeding up!) (d) v = −7 (negative), a = 4 (positive) opposite signs → SLOWING DOWN (a) speeding up (b) slowing down (c) speeding up (d) slowing down case (c) is the classic trap: “negative acceleration” sounds like braking, but if the particle is already moving in the negative direction, negative acceleration pushes it FASTER in that direction.

Piecewise-linear velocity-time graph: rises from 0 to 4 over t ∈ [0,3], constant at 4 on [3,7], then drops linearly to −2 on [7,10], crossing zero at t = 9. Green = positive velocity (forward); red = negative velocity (backward).

WE 4

Reading a velocity-time graph — features of motion

The diagram above shows the velocity (in m s⁻¹) of a particle moving along a horizontal line, between t = 0 and t = 10 seconds. The positive direction is to the right.

(a) State the times at which the particle is instantaneously at rest. (b) State the time at which the particle changes direction. (c) State the interval(s) during which the particle is decelerating.

(a) at rest means v = 0 (graph touches t-axis) v = 0 at t = 0 and at t = 9 (b) changes direction means v crosses zero (sign flips) at t = 9, v goes from positive (just before) to negative (just after) → changes direction note: at t = 0 the particle is starting from rest in positive direction — not a “change” (c) decelerating means |v| is decreasing (sign(v) and sign(a) differ) on [0, 3]: v > 0, gradient > 0 → speeding up ✗ on [3, 7]: gradient = 0 → constant velocity (neither) ✗ on [7, 9]: v > 0, gradient < 0 → SLOWING DOWN ✓ on [9, 10]: v < 0, gradient < 0 → speeding up (in negative direction) ✗ (a) t = 0 and t = 9 (b) t = 9 (c) decelerating on 7 < t < 9 on [9, 10] the velocity is becoming more negative — that’s speeding up, not slowing down. The gradient alone is misleading; always compare with the sign of v.

WE 5

Displacement and distance from areas on a v-t graph

Using the same velocity-time graph as WE 4, find (a) the displacement of the particle from its starting position after 10 seconds, and (b) the total distance travelled by the particle in those 10 seconds.

Step 1 — split into four areas (3 above, 1 below) A₁ (triangle, 0–3): (1/2)(3)(4) = 6 A₂ (rectangle, 3–7): (4)(4) = 16 A₃ (triangle, 7–9): (1/2)(2)(4) = 4 A₄ (triangle, 9–10): (1/2)(1)(2) = 1 ← below axis Step 2 — displacement = SIGNED sum of areas s = A₁ + A₂ + A₃ − A₄ = 6 + 16 + 4 − 1 = 25 Step 3 — distance = sum of |areas| d = A₁ + A₂ + A₃ + A₄ = 6 + 16 + 4 + 1 = 27 Displacement = 25 m; Distance = 27 m the particle goes 26 m forward (A₁+A₂+A₃) then 1 m back, ending up 25 m from start — but its odometer reads 27 m.

WE 6

Words to signs — assigning v and a from descriptions

For each scenario, state the signs of v and a at the moment described. Assume the positive direction is to the east (horizontal cases) or upward (vertical case).

(a) A car is moving due west and accelerating (its speed is increasing). (b) A train is travelling due east and the brakes are applied (it slows down). (c) A stone is in free-fall, having been dropped from a cliff. (d) A car, initially at rest, starts moving eastward.

(a) moving west = negative direction; speeding up = v, a same sign v < 0 and a < 0 (both negative) (b) moving east = positive; brakes = slowing down = opposite signs v > 0 and a < 0 (c) dropped = falling downward; up is positive, so v points down v < 0 (moving downward, against positive direction) speeding up (gravity pulls it faster down) → a same sign as v a < 0 (the famous a = −g) (d) initially at rest, then starts moving east at the moment it starts: v = 0, but becoming positive to gain positive velocity from rest: a > 0 (a) v<0, a<0 (b) v>0, a<0 (c) v<0, a<0 (d) v=0, a>0 case (c) is a clean illustration of the trap: a negative acceleration with negative velocity means SPEEDING UP — exactly what gravity does to a falling stone.

💡 Top tips

Declare your convention before assigning any sign. “Taking east as positive…” or “taking upward as positive…” — even when not asked, this prevents sign errors.
Speed and distance are the magnitudes of velocity and displacement. If you ever compute a negative speed or negative distance, you’ve made a sign error somewhere.
Test “speeding up” with the product v · a: positive product = speeding up; negative product = slowing down. One number, one decision.
For v-t graph problems, mark every key point first: zero crossings, peaks, troughs, kinks. These are where your areas split.
Units: m for displacement/distance, m s⁻¹ for velocity/speed, m s⁻² for acceleration, s for time. Lose them and lose marks.

⚠ Common mistakes

Treating distance and displacement as the same thing — they only agree if the particle never reverses direction.
Assuming “negative acceleration” = “slowing down”. It only means slowing down if velocity is positive. If v < 0 and a < 0, the particle is speeding up in the negative direction.
Assuming v = 0 implies a = 0. At the top of a ball’s flight, v = 0 but a = −g (the ball is about to fall).
Adding signed areas for distance — that gives you DISPLACEMENT. For distance you must add the |areas|. Forgetting this turns a 27 m answer into a 25 m answer.
Treating “instantaneously at rest” as the end of motion — at v = 0 the particle is momentarily still, but it’s about to move again (often the other way).

Up next: Calculus for Kinematics. The next note connects these three quantities by differentiation and integration: v = ds/dt, a = dv/dt, and conversely s = ∫v dt, v = ∫a dt. You’ll handle problems where v(t) is a FORMULA (not a graph), and you find max speed, times of rest, total distance travelled, and where the particle changes direction — using the calculus you already know.

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