Kinematics is the foundation of every IB Physics paper. Topic A.1 sets up the language — scalars vs vectors, displacement, velocity, acceleration — and the toolkit (SUVAT, motion graphs, projectiles) that you will reuse in every later mechanics topic, from forces and momentum to circular motion and rigid bodies. The new IB syllabus stays SL/HL-shared here, but examiners regularly hide multi-step traps inside what looks like a simple SUVAT or projectile problem: an air-resistance flag, a height mismatch, a "speed at the apex" question, or a graph where the gradient suddenly changes sign.
This cheatsheet condenses every formula, trick and trap from Topic A.1 Kinematics into one page you can revise from. It covers scalars and vectors, the four SUVAT equations, motion graphs (gradients and areas), 2-D projectile motion with and without air resistance, and the qualitative description of fluid resistance and terminal velocity. Scroll to the bottom for the printable PDF and the full Photon Academy library (notes, tutorials, marked solutions).
§1 — Scalars, Vectors & Definitions A.1 SL+HL
Kinematics describes how objects move without asking why. Every quantity is either a scalar (magnitude only) or a vector (magnitude and direction) — getting this split right is the first mark in almost every mechanics question.
| Scalars (magnitude only) | Vectors (magnitude + direction) |
|---|---|
| distance, speed, time, mass, energy, temperature | displacement, velocity, acceleration, force, momentum |
Distance vs displacement, speed vs velocity
- Displacement $\vec{s} = \vec{r}_f - \vec{r}_i$ — the straight-line change in position (a vector, can be negative).
- Distance — the total path length travelled (a scalar, always $\geq 0$).
- Velocity — rate of change of displacement (vector). Speed — rate of change of distance (scalar).
Average vs instantaneous
| Quantity | Average | Instantaneous (gradient) |
|---|---|---|
| Velocity | $\bar{v} = \dfrac{\Delta s}{\Delta t}$ | $v = \dfrac{ds}{dt}$ — gradient of $s$–$t$ |
| Acceleration | $\bar{a} = \dfrac{\Delta v}{\Delta t}$ | $a = \dfrac{dv}{dt}$ — gradient of $v$–$t$ |
What examiners actually test on this concept:
§2 — Motion Graphs A.1 SL+HL
The three motion graphs are linked by gradients (going down the chain) and areas (going up). Master this one picture and most non-numerical kinematics questions collapse to reading a slope or an area.
The gradient/area chain (uniform acceleration). Gradient of $s$–$t$ gives $v$; gradient of $v$–$t$ gives $a$; area under $v$–$t$ gives $s$; area under $a$–$t$ gives $\Delta v$.
| Graph | Gradient gives | Area under gives |
|---|---|---|
| $s$–$t$ | $v$ (velocity) | — |
| $v$–$t$ | $a$ (acceleration) | $s$ (displacement) |
| $a$–$t$ | — | $\Delta v$ (change in velocity) |
Shape clues: a curved $s$–$t$ graph means changing velocity; a straight $v$–$t$ line means uniform acceleration; a horizontal $a$–$t$ line means constant acceleration (SUVAT applies).
What examiners actually test on this concept:
§3 — SUVAT Equations (Uniform Acceleration) A.1 SL+HL
Variable list (SI units)
- $s$ — displacement (m) · $u$ — initial velocity (m s⁻¹) · $v$ — final velocity (m s⁻¹)
- $a$ — acceleration (m s⁻²) · $t$ — time (s)
Choosing the right SUVAT
| Variable NOT given / not asked | Use the equation |
|---|---|
| $s$ | $v = u + at$ |
| $v$ | $s = ut + \tfrac{1}{2}at^2$ |
| $u$ | $v^2 = u^2 + 2as$ |
| $a$ | $s = \tfrac{1}{2}(u + v)t$ |
| $t$ | $v^2 = u^2 + 2as$ |
What examiners actually test on this concept:
§4 — Measuring Motion & Free-Fall A.1 SL+HL
To study the motion of a real object you have to record it. The IB expects you to know three standard laboratory methods and to recognise the free-fall and terminal-velocity graphs they produce.
Practical methods to record motion
- Light gate(s). A gate times how long a beam is broken. A single gate with an interrupter card of length $\ell$ gives $\bar v = \ell/t$; two gates a distance $d$ apart give $\bar v = d/t$, and several gates on a data-logger give velocity and acceleration directly.
- Strobe photography. A strobe flashes at fixed intervals in a dark room while the shutter stays open, so one photo captures the object at equally-spaced instants — the image spacing maps out the motion.
- Ticker-timer tape. A timer prints dots on a tape at a fixed rate (typically every $\tfrac{1}{50}$ s). Widening dot spacing signals acceleration.
Recording motion in the lab. Two light gates a distance $d$ apart give $\bar v = d/t$; ticker-tape dots (every $\tfrac{1}{50}$ s) widen as the object speeds up.
Free-fall and terminal-velocity graphs
Free fall is motion under gravity alone (air resistance ignored). Every object then has the same acceleration $g \approx 9.81$ m s⁻², independent of mass. When drag is not negligible it grows with speed until it balances the weight — the acceleration falls to zero and the object descends at a constant terminal velocity $v_\text{t}$.
Free fall vs air resistance. Free fall: straight $v$–$t$ of gradient $g$, flat $a$–$t$. With drag: $v$ levels off at $v_\text{t}$, so $s$–$t$ straightens and $a$–$t$ falls to zero — and SUVAT no longer applies.
Experiment: measuring the free-fall acceleration $g$
Release an object from rest and let it fall a height $h$ in time $t$. With $u = 0$, $s = ut + \tfrac{1}{2}at^2$ becomes $h = \tfrac{1}{2}gt^2$, so $g = 2h/t^2$. Rather than trust one reading, measure $t$ for a range of heights and plot $h$ against $t^2$: a straight line through the origin of gradient $\tfrac{1}{2}g$.
Determining $g$. The line passes through the origin; its gradient is $\tfrac{1}{2}g$, so double it. A non-zero intercept would reveal a systematic (e.g. timing) error.
What examiners actually test on this concept:
§5 — Projectile Motion (No Air Resistance) A.1 SL+HL
Resolve into independent components. The horizontal motion is uniform ($a_x = 0$); the vertical motion has constant acceleration $g$ downward. The two are linked only by a shared time $t$.
Projectile with equal launch and landing height. $u_x = u\cos\theta$ stays constant; at the apex $v_y = 0$ so the speed is $u\cos\theta$, not zero — and the acceleration is still $g$ downward throughout.
| Horizontal (constant $v$) | Vertical (constant $a = g$ down) |
|---|---|
| $v_x = u\cos\theta$ | $v_y = u\sin\theta - gt$ |
| $x = u\cos\theta \cdot t$ | $y = u\sin\theta \cdot t - \tfrac{1}{2}gt^2$ |
Standard results (launch height = landing height only)
What examiners actually test on this concept:
§6 — Fluid Resistance & Terminal Velocity A.1 SL+HL
A drag force always opposes the velocity vector and grows with speed. In free fall this makes the acceleration fall from $g$ towards zero; the object then travels at a constant terminal velocity.
Approach to terminal velocity. As $v$ rises, drag grows until it equals the weight; the net force — and the acceleration — falls to zero and $v$ levels off at $v_\text{t}$.
Projectile with air resistance — qualitative only
Compared with the ideal (no-drag) path:
- Range, max height and time of flight all decrease.
- The trajectory is no longer symmetric — the descent is steeper than the ascent.
- Landing speed is less than launch speed (energy is dissipated by drag).
- Drag opposes the velocity vector, so its direction changes throughout the flight.
What examiners actually test on this concept:
§7 — Exam Attack Plan A.1 — all sections
When you see this in the question — reach for that:
| Question trigger | Reach for |
|---|---|
| "Constant acceleration", numerical motion problem | List $u,v,a,s,t$ first, then pick the SUVAT that omits the missing variable |
| Free fall, dropped object, ball thrown up | SUVAT with $a = -g$ (up positive); split at the turning point if needed |
| Velocity from a position–time graph | Gradient of the tangent — not a chord |
| Displacement from a velocity–time graph | Signed area under the curve |
| "Distance travelled" on a $v$–$t$ graph | Sum of $|$areas$|$ — count areas below the axis as positive |
| Light gates / ticker tape / find $g$ | $\bar v = d/t$; plot $h$ vs $t^2$, gradient $= \tfrac{1}{2}g$ (double it) |
| 2-D projectile, flat ground | Range / max-height formulas; check launch $=$ landing height |
| Projectile from a cliff or onto a platform | Resolve into components; SUVAT vertically (signed $s_y$) and horizontally |
| "Speed at the highest point" | $u\cos\theta$ — not zero |
| Air resistance / drag mentioned | Qualitative description only — do NOT apply SUVAT |
| "Terminal velocity" | Drag $= mg$ ⇒ $a = 0$ ⇒ constant speed |
Worked Example — IB-Style Projectile from a Cliff
Question (HL Paper 2 style — 6 marks)
A stone is launched from the top of a 25 m vertical cliff with initial speed 18 m s⁻¹ at an angle of 35° above the horizontal. Air resistance is negligible. Take $g = 9.81$ m s⁻². Calculate (a) the time the stone takes to reach the sea below, and (b) its horizontal range from the base of the cliff.
Solution
- Resolve the launch velocity. Take up as positive.
$u_x = 18\cos 35° = 14.74$ m s⁻¹; $\;u_y = 18\sin 35° = 10.32$ m s⁻¹. (M1) - For vertical motion use $s_y = u_y t - \tfrac{1}{2}g t^2$ with $s_y = -25$ m (sea is 25 m below launch):
$-25 = 10.32\,t - 4.905\,t^2$. (M1) - Rearrange: $4.905\,t^2 - 10.32\,t - 25 = 0$. Apply the quadratic formula:
$t = \dfrac{10.32 \pm \sqrt{10.32^2 + 4(4.905)(25)}}{2(4.905)} = \dfrac{10.32 \pm 24.36}{9.81}$. (A1) - Take the positive root: $t = \dfrac{10.32 + 24.36}{9.81} = 3.54$ s. (A1) [part (a)]
- Horizontal range: $x = u_x \cdot t = 14.74 \times 3.54 = 52.1$ m. (M1)(A1) [part (b)]
Examiner's note: Using the flat-ground range formula here would give $R = 18^2 \sin 70° / 9.81 = 31.0$ m — completely wrong, because launch height $\neq$ landing height. Whenever the projectile starts and finishes at different heights, you must split into components and use SUVAT vertically. Sign convention matters: $s_y = -25$, not $+25$.
Common Student Questions
When can I use SUVAT in IB Physics?
What is the difference between distance and displacement?
What is the speed of a projectile at its highest point?
When can I use the range formula $R = u^2 \sin 2\theta / g$?
How does air resistance change projectile motion?
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