Free Cheatsheet · Topic A.1 · SL/HL

IB Physics HL A.1 Kinematics — Complete Cheatsheet

Every SUVAT equation, motion-graph rule, projectile formula and air-resistance trap for IB Physics HL Topic A.1. Hand-built by an IBO-certified Singapore tutor with 15+ years of IB experience.

Topic: A.1 Kinematics Syllabus: Topic A.1 (SL/HL) · ~9 hours Read time: ~16 minutes Last updated: Apr 2026

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, temperaturedisplacement, 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

QuantityAverageInstantaneous (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 $v$–$t$
TrickAverage speed $\neq |\bar{v}|$ when the object reverses direction. Average speed $=$ total distance $\div$ time; average velocity $=$ displacement $\div$ time.
TrapIf an object reverses, $|\text{displacement}| < \text{distance}$. Never put distance into a SUVAT equation — SUVAT only takes the signed displacement $s$.
From the Photon question bank

What examiners actually test on this concept:

Classify scalar vs vectorDistance vs displacementAverage vs instantaneousAverage speed of a there-and-back trip
The "there and back" MCQ is a near-permanent fixture. frequent A runner returns to the start: displacement $=0$ and average velocity $=0$, but distance and average speed are not — the classic distractor swaps the two.
Sign is the whole point of a vector. Take one direction as positive and keep it for the entire question; a velocity of $-4$ m s⁻¹ means 4 m s⁻¹ in the negative direction, not "slowing down".
Know the standard lists cold. "Which of these is a vector?" is a guaranteed one-mark recall — acceleration and momentum are the ones students most often misfile as scalars.

§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.

st s–t : gradient = v vt gradient = a v–t : area = s at a a–t : constant

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 $\Delta v$.

GraphGradient givesArea 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).

NoteOn a curved $s$–$t$ graph, the instantaneous velocity at a point is the gradient of the tangent there — never of a chord.
TrickTo get displacement from a $v$–$t$ graph, split the area into triangles + rectangles or count squares. Areas below the time axis are negative displacement — but count as positive distance.
From the Photon question bank

What examiners actually test on this concept:

Gradient of a tangentArea under v–tDistance vs displacement from a graphMatch graph to motion
Area vs gradient is the deciding line between grades. very frequent "Displacement" → area under $v$–$t$; "acceleration" → gradient of $v$–$t$. Reading the wrong feature is the single most common lost mark in the topic.
Signed area for displacement, unsigned for distance. When the $v$–$t$ curve dips below the axis, subtract that area for displacement but add its magnitude for total distance.
"Which graph shows…" matching questions reward knowing that a straight sloped $v$–$t$ ⇔ a parabolic $s$–$t$ ⇔ a flat $a$–$t$. Curvature in $s$–$t$ always signals a changing velocity.

§3 — SUVAT Equations (Uniform Acceleration) A.1 SL+HL

Eq. 1 (no $s$):$v = u + at$
Eq. 2 (no $a$):$s = \tfrac{1}{2}(u + v)t$
Eq. 3 (no $v$):$s = ut + \tfrac{1}{2}at^2$
Eq. 4 (no $t$):$v^2 = u^2 + 2as$

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 askedUse 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$
TrapSUVAT requires constant acceleration. If $a$ varies (air resistance, variable thrust), you must use motion-graph areas or calculus — SUVAT will give a wrong answer.
TrickFree fall taking up as positive: $a = -9.81$ m s⁻². At maximum height $v = 0$ only momentarily — the acceleration is still $-g$.
From the Photon question bank

What examiners actually test on this concept:

Select the right SUVATFree fall a = gSign conventionsTwo-stage motion"Show that" derivations
The winning routine: list $u,v,a,s,t$ with a "?" for the unknown, then pick the equation that omits the one variable you neither know nor want. This is the fastest path to full marks and it prevents algebra errors.
Free-fall problems test sign discipline. frequent Up-as-positive means $a=-g$; a ball thrown up and caught returns with speed equal to launch but velocity of opposite sign.
Two-stage journeys (accelerate, then constant speed; or up then down) must be split at the point where $a$ changes — SUVAT is applied to each stage separately, never across the join.

§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.
track v gate 1 gate 2 d timer / data-logger tickertimer dots every 1/50 s — widening ⇒ accelerating (strobe photo: equal-Δt images)

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 (air resistance ignored) st vt gradient g at g s–tv–ta–t With air resistance straight: const. v st vₜ vt g a → 0 at s–tv–ta–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 $\tfrac{1}{2}g$.

Linearised:$h = \tfrac{1}{2}g\,t^2 \;\Rightarrow\; g = 2 \times (\text{gradient of } h\text{–}t^2)$
Δ(t²) Δh gradient = Δh / Δ(t²) = g/2 t² / s² h / m

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.

TrapThe gradient of the $h$ vs $t^2$ line is $\tfrac{1}{2}g$, not $g$ — remember to double it. Use a dense object (ball bearing) so air resistance is negligible; a light object gives a value of $g$ that is too low.
From the Photon question bank

What examiners actually test on this concept:

Light-gate / ℓ-over-t reasoningLinearise to find gGradient = g/2Random vs systematic errorRead free-fall graphs
The $h$-versus-$t^2$ graph is a Paper 3 / data-analysis favourite. frequent You'll draw the best-fit line, take a large gradient triangle, then double it — and comment that a non-zero intercept signals a systematic error.
Random vs systematic error is examined verbatim. Repeats and a best-fit line reduce random error; a timing delay or air resistance is systematic and shifts every point the same way.
Interpreting shape: a $v$–$t$ graph that curves and then flattens is the tell-tale of terminal velocity — the moment it flattens, $a=0$ and SUVAT is off the table.

§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 $t$.

u u cos θ u sin θ θ v = u cos θ  (v_y = 0) H g R (range)

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)

Time to apex:$t_{\text{apex}} = \dfrac{u\sin\theta}{g}$
Max height:$H = \dfrac{u^2\sin^2\theta}{2g}$
Time of flight:$T = \dfrac{2u\sin\theta}{g}$
Range:$R = \dfrac{u^2\sin 2\theta}{g}$  — max at $\theta = 45°$; $\theta$ and $(90°-\theta)$ give equal range
TrapDo not use $R = u^2\sin 2\theta/g$ when launch height $\neq$ landing height (off a cliff, into a hoop, against a wall). Resolve into components and use SUVAT vertically and horizontally with a shared $t$.
TrickThe trajectory equation $y = x\tan\theta - \dfrac{gx^2}{2u^2\cos^2\theta}$ is not required in IB — don't waste exam time deriving it. Work with components instead.
From the Photon question bank

What examiners actually test on this concept:

Resolve u into componentsSpeed at the apexCliff / unequal-height launchesSymmetry of the pathRange vs angle
"Speed at the highest point" is the single most-set projectile trap. very frequent The answer is $u\cos\theta$, never zero — only the vertical component vanishes at the apex.
Unequal launch/landing heights break the neat formulas. Set up the vertical SUVAT with the correct signed displacement (e.g. $s_y = -h$ for a cliff), solve the quadratic for $t$, then feed that $t$ into the horizontal equation.
Symmetry and equal-range angles ($\theta$ and $90°-\theta$) are recurring MCQs; on flat ground the ascent and descent are mirror images, so launch speed equals landing speed.

§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.

vₜ start: drag ≈ 0, so a ≈ g drag = mg ⇒ a = 0 t v

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.
NoteIB only requires a qualitative description of air resistance in A.1. Never apply SUVAT when drag is present — the acceleration is no longer constant.
From the Photon question bank

What examiners actually test on this concept:

Force balance at terminal velocityShape of the v–t curveEffect of drag on a projectileWhy heavier ≠ faster in vacuum
"Explain terminal velocity" wants the force argument. frequent Drag rises with speed → net force falls → acceleration falls → at $v_\text{t}$ drag $= mg$ so $a=0$. State each link for the marks.
Drag questions are qualitative — resist the urge to calculate. Compare with the no-drag case in words: shorter range, lower peak, unsymmetrical path, slower landing.
Vacuum vs air. In a vacuum a feather and a coin fall together ($a=g$ for both); the everyday difference is entirely due to air resistance, a favourite conceptual MCQ.

§7 — Exam Attack Plan A.1 — all sections

When you see this in the question — reach for that:

Question triggerReach for
"Constant acceleration", numerical motion problemList $u,v,a,s,t$ first, then pick the SUVAT that omits the missing variable
Free fall, dropped object, ball thrown upSUVAT with $a = -g$ (up positive); split at the turning point if needed
Velocity from a position–time graphGradient of the tangent — not a chord
Displacement from a velocity–time graphSigned area under the curve
"Distance travelled" on a $v$–$t$ graphSum 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 groundRange / max-height formulas; check launch $=$ landing height
Projectile from a cliff or onto a platformResolve into components; SUVAT vertically (signed $s_y$) and horizontally
"Speed at the highest point"$u\cos\theta$ — not zero
Air resistance / drag mentionedQualitative 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

  1. 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)
  2. 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)
  3. 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)
  4. Take the positive root: $t = \dfrac{10.32 + 24.36}{9.81} = 3.54$ s.  (A1)  [part (a)]
  5. 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?
SUVAT is only valid when acceleration is constant. The classic exam trap is using SUVAT during free fall with air resistance — drag changes the acceleration, so SUVAT no longer applies. In that case use motion graphs (area under $v$–$t$ for displacement) or qualitative description. For free fall in vacuum near Earth, $a = g = 9.81$ m s⁻² is constant, so SUVAT is fine.
What is the difference between distance and displacement?
Distance is a scalar — the total path length, always positive. Displacement is a vector — the straight-line change in position from start to finish, with direction (sign). If an object reverses, distance $> |$displacement$|$. Average speed $=$ distance $\div$ time, while average velocity $=$ displacement $\div$ time. Confusing the two is the most-dropped mark on Paper 1 MCQs in this topic.
What is the speed of a projectile at its highest point?
Not zero. Only the vertical component $v_y$ is zero at the apex — the horizontal component is unchanged: $v_x = u\cos\theta$. So the speed at the apex equals $u\cos\theta$, not 0. This is one of the most common Paper 1 MCQ traps. The acceleration at the apex is also still $g$, pointing downward.
When can I use the range formula $R = u^2 \sin 2\theta / g$?
ONLY when launch height equals landing height. If a ball is launched from a cliff or hits a wall, you cannot use it — you must resolve the velocity into horizontal and vertical components and use SUVAT separately on each. Maximum range on flat ground is at $\theta = 45°$, and the angles $\theta$ and $(90° - \theta)$ give the same range.
How does air resistance change projectile motion?
With air resistance, 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 because energy is dissipated by drag. Drag opposes the velocity vector, so its direction changes throughout the flight. IB only requires a qualitative description — never apply SUVAT when air resistance is involved.

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