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: ~12 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

Scalars vs vectors

  • Scalar quantities: distance, speed, time, mass, energy.
  • Vector quantities: displacement, velocity, acceleration, force.

Distance vs displacement

  • Displacement $\vec{s} = \vec{r}_f - \vec{r}_i$ — change in position (vector, can be negative).
  • Distance — total path length, always $\geq 0$.

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 $\neq |\bar{v}|$ if the object reverses direction. Average speed $=$ total distance $\div$ time.
TrapIf an object reverses, $|\text{displacement}| < \text{distance}$. Never use distance inside a SUVAT equation — SUVAT only takes signed displacement $s$.

§2 — SUVAT Equations (Uniform Acceleration) A.1

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 / askedUse
$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 graphs (areas) or calculus — SUVAT will give a wrong answer.
TrickFor free fall taking upward as positive: $a = -9.81$ m s⁻². At maximum height $v = 0$ only momentarily — the acceleration is still $-g$.

§3 — Motion Graphs A.1

Displacement-time, velocity-time and acceleration-time graphs for motion with constant acceleration
Motion graphs: gradient of $s$–$t$ gives $v$, gradient of $v$–$t$ gives $a$, area under $v$–$t$ gives displacement.
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 non-uniform velocity; a straight $v$–$t$ line means uniform acceleration; a horizontal $a$–$t$ line means constant acceleration (SUVAT applies).

NoteFor a curved $s$–$t$ graph, the instantaneous velocity at a point is the gradient of the tangent at that point — not of any chord.
TrickTo find displacement from a $v$–$t$ graph, split the area into triangles + rectangles, or count squares. Areas below the time axis count as negative displacement.

§4 — Projectile Motion (No Air Resistance) A.1

Projectile motion trajectory diagram showing parabolic path with horizontal and vertical velocity components
Projectile path: parabolic shape, $u_x = u\cos\theta$ constant, $u_y = u\sin\theta$ subject to gravity.

Resolve into independent components. Horizontal motion is uniform; vertical motion has constant acceleration $g$ downward.

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$

Useful results (launch height = landing height only)

Time to apex:$t_{\text{apex}} = \dfrac{u\sin\theta}{g}$
Max height:$H = \dfrac{u^2\sin^2\theta}{2g}$
Total flight time:$T = \dfrac{2u\sin\theta}{g}$
Range:$R = \dfrac{u^2\sin 2\theta}{g}$
Max range:$\theta = 45°$; equal ranges for $\theta$ and $(90° - \theta)$.
TrapDo NOT use $R = u^2\sin 2\theta / g$ if launch height $\neq$ landing height (e.g. a ball off a cliff, into a hoop, against a wall). Resolve into components and use SUVAT vertically and horizontally.
TrickAt the apex: $v_x = u\cos\theta$ (unchanged), $v_y = 0$. Speed at the apex $= u\cos\theta$, not zero. The acceleration is still $g$ downward.
TrapThe full equation of trajectory $y = x\tan\theta - \dfrac{gx^2}{2u^2\cos^2\theta}$ is not required in IB. Don't waste exam time deriving it.

§5 — Fluid Resistance & Terminal Velocity A.1

Velocity-time graph for an object falling through a fluid, showing approach to terminal velocity
Terminal velocity: drag grows with $v$ until it equals weight; net force $\to 0$, so $v$ levels off.

Projectile WITH air resistance — qualitative

Compared to the ideal (no-drag) case:

  • Range decreases.
  • Max height decreases.
  • Time of flight decreases.
  • Trajectory is not symmetric — the descent is steeper than the ascent.
  • Landing speed is less than launch speed.
  • Drag opposes the velocity vector, so its direction changes throughout the flight.

Terminal velocity (free fall)

Drag $= mg$  ⇒  net force $= 0$  ⇒  acceleration $= 0$. The object falls at constant terminal speed $v_T$.

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

§6 — Exam Attack Plan A.1 — All sections

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

Question triggerReach for
"Constant acceleration", numerical motion problemPick the right SUVAT (table above) — list $u, v, a, s, t$ first
Free fall, dropped object, ball thrown upSUVAT with $a = -g$ (taking up as positive)
Velocity from a position-time graphGradient (tangent) — not chord
Displacement from a velocity-time graphArea under curve (signed)
"Distance travelled" on a $v$–$t$ graphSum of $|$areas$|$ — count negative areas as positive
2-D projectile, flat groundRange / max-height formulas; check launch = landing height
Projectile from a cliff or onto a higher platformResolve into components; SUVAT vertically 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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