IB Math AI HL Kinematics & Projectiles — Complete Cheatsheet
Every motion skill for IB Mathematics Applications & Interpretation HL — the displacement–velocity–acceleration chain, particle at rest, total distance versus displacement, average velocity and projectile motion (range, maximum height and time of flight). Hand-built by an IBO-certified Singapore tutor, with a print-ready PDF to download.
Kinematics is where the IB Mathematics Applications & Interpretation HL calculus strand meets the physical world: it describes how an object moves along a line through its displacement $s(t)$, velocity $v(t)$ and acceleration $a(t)$. In AI these are never abstract — they model a train braking, a drone climbing or a ball in flight, and a GDC is always allowed. So the marks live not in grinding the calculus but in choosing the right operation (differentiate down the chain, integrate back up), knowing that $v=0$ means "at rest", and reading distance, direction and units correctly from the answer.
This cheatsheet condenses the whole range — from the $s\to v\to a$ chain and total distance in AHL 5.13 to projectile range and maximum height — onto one page, and flags the traps that quietly cost method marks: distance versus displacement, $v=0$ versus $a=0$, and adding velocity components instead of taking their magnitude. The print-ready PDF is at the bottom, free to download.
§1 — The kinematics chain: $s\to v\to a$ SL 5.4
Displacement, velocity and acceleration are tied together by calculus. Differentiate to step down the chain; integrate to climb back up.
Velocity:$v=\dfrac{ds}{dt}$ — the rate of change (and $s$–$t$ graph gradient) of displacement
Acceleration:$a=\dfrac{dv}{dt}=\dfrac{d^{2}s}{dt^{2}}$ — differentiate the displacement twice
Displacement:$s=\displaystyle\int v\,dt$ — integrate the velocity (add $+C$)
Velocity:$v=\displaystyle\int a\,dt$ — integrate the acceleration (add $+C$)
The two "downhill" steps are derivatives; the two "uphill" steps are integrals, and each integration introduces a constant that an initial condition must fix (§5).
NoteThe chain is the topic. Every question is really asking "which way am I moving along $s\to v\to a$?" — differentiate to reach a rate, integrate to reach a total. This is the motion version of differentiation and integration.
TrapDifferentiate first, substitute the time second. Plugging the number in before you differentiate collapses the function to a constant, whose derivative is $0$.
§2 — Velocity & acceleration by differentiation SL 5.4
Given a displacement model $s(t)$, the velocity and acceleration drop straight out of the power rule — differentiate term by term.
Power rule:$\dfrac{d}{dt}\left(kt^{n}\right)=kn\,t^{\,n-1}$ — differentiate each term
$s\to v$:differentiate once — $v(t)=s'(t)$
$v\to a$:differentiate again — $a(t)=v'(t)=s''(t)$
For instance, if $s(t)=t^{3}-6t^{2}+9t$ then $v=\dfrac{ds}{dt}=3t^{2}-12t+9$ and $a=\dfrac{dv}{dt}=6t-12$ — the same model that runs through the worked example below.
TrickOn Paper 2 you can skip the algebra: graph $s(t)$ and read the numerical derivative $\left.\tfrac{ds}{dt}\right|_{t=p}$ for the velocity, or graph $v(t)$ for the acceleration. Show one line of calculus to bank the method mark.
NoteVelocity is signed — its sign is the direction of travel. Speed is $|v|$, always $\ge 0$. Getting these two confused is the root of most distance-versus-displacement slips (§6).
§3 — When is the particle at rest? AHL 5.13
A particle is momentarily at rest at a turning point of its displacement, where the velocity is zero. Set $v=0$ and solve for $t$.
At rest:$v(t)=0$ — solve for the time(s)
Changes direction:$v$ changes sign at that $t$ — the motion reverses
With $v(t)=3t^{2}-12t+9=3(t-1)(t-3)$, the particle is at rest at $t=1\text{ s}$ and $t=3\text{ s}$, reversing direction at each.
Note"At rest", "stationary", "momentarily at rest", "instantaneously at rest" and "changes direction" are all the same cue: set $v=0$.
Trap"At rest" is not "not accelerating". At $t=1$ and $t=3$ above the velocity is zero but the acceleration $a=6t-12$ is not — the particle is still slowing or speeding up through the instant it is at rest.
§4 — Maximum & minimum velocity AHL 5.13
The velocity is greatest or least not where $v=0$, but where its own rate of change vanishes — that is, where the acceleration is zero.
Extreme velocity:solve $a(t)=v'(t)=0$, then substitute that $t$ back into $v$
For $v=3t^{2}-12t+9$, $a=6t-12=0$ at $t=2\text{ s}$, and the velocity there is $v(2)=3(4)-24+9=-3\text{ m s}^{-1}$ — the minimum (most negative) velocity of the motion.
Trap$v=0$ means at rest; $a=0$ means fastest or slowest. Solving the wrong one is the classic mix-up. If a question wants an extreme velocity, set $a=0$, not $v=0$.
TrickOn a $v$–$t$ graph the extreme velocity is the graph's own turning point, so the GDC's maximum/minimum feature applied to $v(t)$ delivers it in a single step.
§5 — Integrating back up: velocity & displacement AHL 5.13
If you are handed the acceleration (or velocity) instead, integrate — and use an initial condition to fix the constant.
Velocity from $a$:$v(t)=\displaystyle\int a\,dt + C$, with $C$ set by $v(0)$
Displacement from $v$:$s(t)=\displaystyle\int v\,dt + C$, with $C$ set by $s(0)$
Change of position:$s(t_2)-s(t_1)=\displaystyle\int_{t_1}^{t_2} v\,dt$ — definite, no $+C$
Example: if $a=6t-12$ and the particle starts with $v(0)=9$, then $v=\int(6t-12)\,dt=3t^{2}-12t+C$, and $C=9$ gives $v=3t^{2}-12t+9$.
TrapThe $+C$ is not optional. Forgetting it — or evaluating your target time before applying $v(0)$ — is the single most common integration error in kinematics. Integrate, apply the initial condition, then substitute.
NoteA definite integral $\int_{t_1}^{t_2} v\,dt$ needs no constant and returns the displacement (net change of position) over the interval — which may be positive, negative or zero.
§6 — Displacement, distance & average velocity AHL 5.13
This is the section examiners love, because displacement and total distance are different whenever the velocity changes sign.
Displacement:$\displaystyle\int_{t_1}^{t_2} v\,dt$ — net change of position (signed)
Total distance:$\displaystyle\int_{t_1}^{t_2} |v|\,dt$ — every metre counted (unsigned)
Average velocity:$\dfrac{s(t_2)-s(t_1)}{t_2-t_1}=\dfrac{\text{displacement}}{\text{time}}$
Average speed:$\dfrac{\text{total distance}}{\text{time}}$ — generally not equal to average velocity
To get total distance by hand, find where $v=0$, split the journey there, and add the size of the displacement change on each leg: $\text{distance}=\sum|\Delta s|$. On the GDC, $\int|v|\,dt$ does the splitting for you.
TrapAverage velocity uses the two displacement endpoints: it is $\dfrac{s(t_2)-s(t_1)}{t_2-t_1}$, not the value of $v$ at either end, and not the average of the two end velocities.
TrickIf the question says "total distance", never integrate $v$ straight through — that returns only the displacement. Solve $v=0$ first and integrate $|v|$ leg by leg, or key $\int|v|\,dt$ into the GDC.
§7 — The velocity–time graph AHL 5.13
A velocity–time graph reads the whole motion at a glance: its gradient is the acceleration and the area between the curve and the $t$-axis is the displacement.
Gradient:$\dfrac{dv}{dt}=a$ — steeper means larger acceleration
Area:$\displaystyle\int v\,dt=$ displacement (area below the axis counts as negative)
A velocity–time graph: the area under it is the displacement and its gradient is the acceleration. For this speed-up · cruise · slow-down trip the distance is the trapezium area, $\tfrac12(\text{top}+\text{base})\times v_{\max}=\tfrac12(30+60)\times20=900$ m.
Reading the graph: the train speeds up for $10\text{ s}$, cruises at $20\text{ m s}^{-1}$ for $30\text{ s}$, then brakes for $20\text{ s}$. The distance is the trapezium's area, $\tfrac12(30+60)\times20=900\text{ m}$, and the braking acceleration is the final gradient, $\dfrac{0-20}{20}=-1\text{ m s}^{-2}$.
TrapFor a speed-up/cruise/slow-down journey the area is the whole trapezium, not a triangle or a rectangle. The short parallel side (the "top") is the length of the constant-speed phase — not the maximum speed.
NoteArea below the $t$-axis (negative velocity) subtracts from the displacement but adds to the total distance — the graph version of the distance-versus-displacement split in §6.
§8 — Projectile motion AHL 5.13
A projectile launched with speed $u$ at angle $\theta$ is two motions at once: constant velocity across, and free-fall (acceleration $g\approx9.81\text{ m s}^{-2}$) down. Resolve into components first.
Horizontal:velocity $u\cos\theta$ stays constant (no horizontal force)
Vertical:velocity $u\sin\theta$ initially, changing at $-g$
Time of flight:$T=\dfrac{2u\sin\theta}{g}$ — to return to launch height
Projectile motion splits into a constant horizontal velocity $u\cos\theta$ and a vertical velocity $u\sin\theta$ under gravity. The vertical component alone fixes the time of flight and the maximum height $H$; the horizontal component carries the range $R$.
At the top of the flight the vertical velocity is momentarily zero — an "at rest" condition, vertically — which is exactly what sets the maximum height; the horizontal velocity never changes, so it plays no part in $H$.
TrickThe range is largest at $\theta=45^\circ$, where $\sin2\theta=1$. Keep the calculator in degrees for projectile angles.
TrapOnly the vertical component decides the time of flight and the maximum height; only the horizontal component carries the range. Never mix the two — resolve the launch velocity before you substitute.
Mixed topic — Kinematics × Vectors. In two dimensions the position is a vector $\mathbf{r}(t)=x(t)\,\mathbf{i}+y(t)\,\mathbf{j}$. Differentiate each component for the velocity vector, then take its magnitude for the speed.
Speed:$|\mathbf{v}(t)|=\sqrt{\left(\dfrac{dx}{dt}\right)^{2}+\left(\dfrac{dy}{dt}\right)^{2}}$ — a Pythagoras step
For an object moving with constant velocity, $\mathbf{r}(t)=\mathbf{r}_0+t\mathbf{v}$, so the squared distance to a point (or between two such objects) is a quadratic in $t$: $D^{2}=at^{2}+bt+c$. Its minimum — the closest approach — is at $t=-\dfrac{b}{2a}$; substitute back and square-root for the least distance.
TrickMinimise $D^{2}$, not $D$ — the vertex $t=-\tfrac{b}{2a}$ gives the time of closest approach. Put that $t$ back and take the square root just once for the distance.
TrapSpeed is the magnitude of the velocity vector, $\sqrt{\dot{x}^{2}+\dot{y}^{2}}$ — differentiate the components first, then apply Pythagoras. Never add $v_x+v_y$, and never take the size of the position vector $\mathbf{r}$.
§10 — Exam attack plan All sections
Question cue
What to do
Watch for
"Find velocity / acceleration"
Differentiate $s$ once for $v$, twice for $a$
Differentiate first, substitute the time second
"When is it at rest / changes direction?"
Solve $v(t)=0$
$v=0$, not $a=0$; check $v$ changes sign
"Maximum / minimum velocity"
Solve $a(t)=0$, then sub into $v$
$a=0$ (not $v=0$); it is the $v$–$t$ turning point
"Given acceleration, find $v$ or $s$"
Integrate and add $+C$
Fix $C$ with $v(0)$ or $s(0)$ before substituting
"Total distance travelled"
Split at $v=0$ and sum $|\Delta s|$, or $\int|v|\,dt$ on the GDC
Vertical sets height & time, horizontal sets range; degrees
Worked Example — Distance, Direction & Average Velocity (AI HL Kinematics)
Question (AI HL Paper 2 style — 9 marks)
A particle $P$ moves along a straight line. Its velocity, in $\text{m s}^{-1}$, at time $t$ seconds ($t\ge0$) is $v(t)=3t^{2}-12t+9$. When $t=0$ the particle is at the origin, so $s(0)=0$. (a) Find the acceleration of $P$ at time $t$. (b) Find the two times at which $P$ is momentarily at rest. (c) Find the total distance travelled by $P$ during the first $4$ seconds. (d) Find the average velocity of $P$ over the first $4$ seconds.
Solution
(a) Differentiate the velocity: $a(t)=\dfrac{dv}{dt}=6t-12\ \text{m s}^{-2}$. (M1)(A1)
(b) At rest when $v=0$: $\;3t^{2}-12t+9=3(t-1)(t-3)=0$, so $t=1\text{ s}$ and $t=3\text{ s}$. (M1)(A1)
(c) Integrate for displacement: $s(t)=\displaystyle\int(3t^{2}-12t+9)\,dt=t^{3}-6t^{2}+9t$ (with $s(0)=0$, so $C=0$). Since $v$ changes sign at $t=1$ and $t=3$, split the journey there: $s(0)=0,\ s(1)=4,\ s(3)=0,\ s(4)=4$. (M1) The leg distances are $|4-0|=4$, $|0-4|=4$ and $|4-0|=4$, so the total distance $=4+4+4=12\text{ m}$. (A1)(A1)(GDC check: $\int_0^4|v|\,dt=12$.)
(d) Average velocity $=\dfrac{s(4)-s(0)}{4-0}=\dfrac{4-0}{4}=1\text{ m s}^{-1}$. (M1)(A1)
Examiner's note: part (c) is the discriminator. Integrating $v$ straight through gives $\int_0^4 v\,dt=4$, which is the displacement, not the distance. Because $P$ reverses at $t=1$ and $t=3$ you must split at those times (or key $\int_0^4|v|\,dt$ into the GDC) to collect every metre. Keep the units throughout ($\text{m}$, $\text{m s}^{-1}$, $\text{m s}^{-2}$), and notice how (c) $=12$ m and (d) $=1\ \text{m s}^{-1}$ describe the same motion two ways: plenty of back-and-forth, little net progress.
Common Student Questions
What is the difference between distance and displacement?
Displacement is the net (signed) change of position, $\int v\,dt$ — it can be positive, negative or zero. Total distance counts every metre travelled regardless of direction, $\int|v|\,dt$. They agree only while the velocity keeps one sign; the moment $v$ changes sign the object doubles back, so you must split the journey where $v=0$ and add the size of each leg.
How do I know whether to differentiate or integrate?
Locate what you are given on the chain $s\to v\to a$. To move down it (displacement to velocity to acceleration) you differentiate — that gives a rate. To move up it (acceleration to velocity to displacement) you integrate — that gives a total, and you must add $+C$ and fix it from an initial condition. "Rate" or "how fast" points down; "how far" or "total" points up.
Is "at rest" the same as "zero acceleration"?
No — they are different equations. At rest means the particle has momentarily stopped, so $v=0$. Zero acceleration, $a=0$, is where the velocity is greatest or least (fastest or slowest), because $a$ is the rate of change of $v$. A particle can be at rest while still accelerating, and can be accelerating hardest while moving fast. Match the equation to the word: at rest $\Rightarrow v=0$; fastest/slowest $\Rightarrow a=0$.
Why does a projectile's maximum height only use the vertical component?
Height is a purely vertical measurement, and at the top of the flight the vertical velocity is momentarily zero — that is what caps the height at $H=\dfrac{(u\sin\theta)^{2}}{2g}$. The horizontal velocity $u\cos\theta$ is unchanged throughout the flight (no horizontal force), so it carries the range along the ground but contributes nothing to how high the projectile climbs. Resolve first, then treat the two directions separately.
Do I have to show the algebra if my GDC gives the answer?
On Paper 2 the GDC is ideal — the numerical derivative, $\int|v|\,dt$ for total distance, and the graph maximum/minimum for extreme velocity. But method marks reward a line of working: show $v=0$ or $a=0$ being solved, or the integral being set up, then quote the GDC value as the check. On Paper 1 you differentiate and integrate by hand throughout. Best practice everywhere: show the calculus, confirm with the GDC.
What's NOT in this cheatsheet
This page gives you the formulas and traps for free. The full Photon Academy IB Math AI HL library (enrolled students, or the lifetime resource library) adds:
Notes PDF — the full displacement–velocity–acceleration chain, the total-distance method and every projectile formula, with worked derivations and edge cases.
Tutorial booklet — 25+ AI HL-style questions from chain fluency to total distance, velocity–time graphs and full projectile applications.
Tutorial Solutions — mark-scheme-style solutions with M1/A1 annotations and GDC screenshots for the total-distance and projectile questions.
Predicted-paper questions — the exact kinematics and projectile question types most likely in the next session.
