Why is "Using suvat equations when the acceleration is not constant" a common mistake in Kinematics?

Why it happens: Students reach for the familiar equations without checking whether the acceleration is uniform. How to avoid it: Before using suvat, confirm the acceleration is **constant** over the whole interval. If air resistance is significant, or the v–t graph is curved, suvat does **not** apply — use graphical methods instead. Source: IB Physics Theme A subject reports

Why is "Mixing horizontal and vertical quantities in one projectile equation" a common mistake in Kinematics?

Why it happens: The two motions feel like one problem, so students put a horizontal velocity into a vertical suvat equation. How to avoid it: Keep two separate columns (horizontal and vertical). The **only** quantity they share is the **time**. Solve one direction for t, then use t in the other. Source: IB Physics Paper 2 subject reports

Why is "Thinking acceleration is zero at the highest point of a projectile's flight" a common mistake in Kinematics?

Why it happens: Because the vertical velocity is zero there, students assume the acceleration is too. How to avoid it: At the top, **v_y = 0 but a = g** (still 9.81 m s⁻² downwards). Gravity acts throughout the flight; only the vertical *velocity* is momentarily zero. Source: IB Physics Theme A subject reports

Why is "Reading a value off a motion graph when a gradient or area is required" a common mistake in Kinematics?

Why it happens: Students see a number on the axis and quote it instead of processing the graph. How to avoid it: On a v–t graph, a point gives the **velocity**; the **gradient** gives acceleration and the **area** gives displacement. Always show the triangle (for a gradient) or the shapes (for an area). Source: IB Physics Paper 1B subject reports

Why is "Being inconsistent with the direction (sign) convention for vectors" a common mistake in Kinematics?

Why it happens: Up is taken as positive in one line and negative in the next, so signs of g, u and s get muddled. How to avoid it: State your convention at the start (e.g. 'up is positive') and apply it to **every** vector: velocities, accelerations and displacements. A downward g is then −9.81 m s⁻². Source: IB Physics Theme A subject reports

Why is "Omitting units or giving an inappropriate number of significant figures" a common mistake in Kinematics?

Why it happens: Under time pressure students write a bare number, or copy all the calculator digits. How to avoid it: Always attach the SI unit and round to a sensible number of significant figures (usually 2–3, matching the data). A numerically correct answer with no unit is not fully credited. Source: IB Physics general subject reports

IB DP Physics (100452)

Kinematics

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Detailed Study Notes

Detailed notes on Space, time and motion for IB DP Physics, covering key concepts, explanations, examples, and exam-focused revision points.

A.1 Kinematics — IB Physics SL Study Notes (Theme A: Space, Time and Motion)

Describing motion without asking what causes it: distance vs displacement, speed vs velocity, acceleration, the four suvat equations for uniform acceleration, reading and drawing motion graphs, and projectile motion by resolving into independent horizontal and vertical components.

At a glance

Distance & speed are scalars (size only); displacement, velocity & acceleration are vectors (size + direction).
Uniform acceleration only → use the four suvat equations. Check acceleration is constant before you use them.
Displacement–time graph: gradient = velocity. Velocity–time graph: gradient = acceleration, area under = displacement.
Acceleration = rate of change of velocity, so it changes when speed OR direction changes.
Projectiles: horizontal and vertical motion are independent. Horizontal: constant velocity ( $a_x=0$ ). Vertical: constant $a=g=9.81\,\text{m s}^{-2}$ down.
At the top of a projectile's flight the vertical velocity is zero, but the horizontal velocity and the acceleration $g$ are unchanged.
Air resistance reduces range and maximum height, makes the descent steeper than the ascent, and can lead to terminal velocity.
Always quote a final answer with correct units and sensible significant figures, and show full working for method marks.

What you’ll learn

Mapped to the 100452 subject guide (2025-onwards).

Distinguish scalar quantities (distance, speed) from vector quantities (displacement, velocity, acceleration) and use the correct one in a calculation.
Define average and instantaneous velocity and acceleration, and relate them to the gradients of motion graphs.
Select and apply the four equations of motion (suvat) for straight-line motion under uniform acceleration, including free fall.
Interpret and sketch displacement–time and velocity–time graphs, using gradient and area to extract velocity, acceleration and displacement.
Analyse projectile motion by resolving into independent horizontal (constant velocity) and vertical (constant acceleration) components, with and without air resistance.

Scalars vs vectors: getting the language right

Distance/speed have size only; displacement/velocity/acceleration have direction too.

Kinematics is the description of how things move — position, velocity and acceleration — without worrying about the forces that cause the motion (that is dynamics, topic A.2).

The first marks in almost every motion question come from using the right kind of quantity:

Quantity	Type	Meaning	SI unit
Distance	Scalar	Total path length travelled	m
Displacement	Vector	Straight-line change in position (with direction)	m
Speed	Scalar	Rate of change of distance	m s⁻¹
Velocity	Vector	Rate of change of displacement (with direction)	m s⁻¹
Acceleration	Vector	Rate of change of velocity	m s⁻²

A scalar has magnitude only (e.g. a speed of 20 m s⁻¹).
A vector has magnitude and direction (e.g. a velocity of 20 m s⁻¹ due east). We handle direction with + / − signs on a straight line, or by resolving into components in 2D.

Why the distinction earns marks: a runner who completes one lap of a 400 m track has travelled a distance of 400 m but has a displacement of 0 m (they finish where they started). Their average speed is not zero, but their average velocity is zero.

Instantaneous vs average:

Average velocity $= \dfrac{\Delta s}{\Delta t}$ over a whole interval.
Instantaneous velocity is the velocity at a single instant — the gradient of the tangent to a displacement–time graph at that point.

Scalars: distance, speed (magnitude only).
Vectors: displacement, velocity, acceleration (magnitude + direction).
Average = over an interval; instantaneous = at one instant (gradient of a tangent).

The suvat equations (uniform acceleration only)

Four equations link s, u, v, a, t — but only when acceleration is constant.

When the acceleration is constant (uniform), the five motion variables are linked by four equations. The IB data booklet provides them, but you must be able to select the right one instantly.

Symbol	Quantity
$s$	displacement
$u$	initial velocity
$v$	final velocity
$a$	acceleration
$t$	time

The four equations of motion:

$v = u + at$ $s = ut + \tfrac{1}{2}at^2$ $v^2 = u^2 + 2as$ $s = \tfrac{1}{2}(u+v)\,t$

How to choose the right equation — the "list the five" method:

Write down the three values you are given.
Write down the quantity you want.
Identify which quantity is missing / not needed — pick the equation that does not contain it.

For example, if you know $u$ , $a$ and $t$ and want $s$ (so $v$ is not involved), use $s = ut + \tfrac{1}{2}at^2$ .

Free fall is the most common special case: an object moving under gravity alone has $a = g = 9.81\,\text{m s}^{-2}$ directed downwards. If you take up as positive, then $g = -9.81\,\text{m s}^{-2}$ and any downward displacement is negative. Choose one convention and stick to it.

The single most important check: suvat works only for constant acceleration. It does not apply while air resistance is significant (acceleration changes), or to any curved-acceleration situation. If a graph of velocity against time is not a straight line, you cannot use suvat over that interval.

Four suvat equations, valid only for uniform (constant) acceleration.
Choose the equation that omits the variable you neither know nor want.
Free fall: a = g = 9.81 m s⁻² downwards; pick a sign convention and keep it.

See the full worked example for kinematics →

Where the suvat equations come from (derivation)

All four equations follow from the definition of acceleration and the area under a v–t graph.

You are not expected to reproduce these derivations under exam conditions, but understanding where the equations come from makes them impossible to misremember — and IB rewards students who can reason from first principles rather than pattern-match.

Start from a velocity–time graph for uniform acceleration: a straight line from initial velocity $u$ to final velocity $v$ over time $t$ .

1. The first equation comes straight from the definition of acceleration (gradient of the v–t graph): $a = \frac{\text{change in velocity}}{\text{time}} = \frac{v-u}{t} \quad\Rightarrow\quad \boxed{v = u + at}$

2. The fourth equation comes from the area under the graph = displacement. The area is a trapezium with parallel sides $u$ and $v$ and width $t$ : $s = \text{area} = \left(\frac{u+v}{2}\right)t \quad\Rightarrow\quad \boxed{s = \tfrac{1}{2}(u+v)\,t}$

3. The second equation comes from splitting that same area into a rectangle (height $u$ , area $ut$ ) plus a triangle (base $t$ , height $v-u = at$ , area $\tfrac{1}{2}t\cdot at$ ): $s = ut + \tfrac{1}{2}(at)(t) \quad\Rightarrow\quad \boxed{s = ut + \tfrac{1}{2}at^2}$

4. The third equation comes from eliminating $t$ . From equation 1, $t = \dfrac{v-u}{a}$ . Substitute into equation 4: $s = \tfrac{1}{2}(u+v)\cdot\frac{v-u}{a} = \frac{v^2-u^2}{2a} \quad\Rightarrow\quad \boxed{v^2 = u^2 + 2as}$

The displacement is the area under the v–t graph. Splitting it into a rectangle (ut) and a triangle (½at²) gives s = ut + ½at² directly — the geometry *is* the equation.

Why this matters for grade 9: if you ever blank on which equation to use, sketch the v–t graph and read off the gradient (for acceleration) or the area (for displacement). The graph never lets you down, even when your memory does.

v = u + at is the gradient (definition of acceleration) of the v–t line.
s = ½(u+v)t is the trapezium area; s = ut + ½at² is rectangle + triangle.
v² = u² + 2as comes from eliminating t between the other equations.

Motion graphs: gradient and area do the work

d–t gradient = velocity; v–t gradient = acceleration; v–t area = displacement.

Graphs are examined in every paper — you must be able to read them and sketch them.

Displacement–time (d–t) graph

Gradient = velocity.
A straight sloping line → constant velocity; a curve → changing velocity (acceleration).
A horizontal line → object is stationary.
The gradient of a tangent gives the instantaneous velocity.

Velocity–time (v–t) graph

Gradient = acceleration.
Area under the graph = displacement. (Split into triangles and rectangles, or find the area of a trapezium.)
A horizontal line → constant velocity (zero acceleration); a straight slope → constant acceleration.

Acceleration–time (a–t) graph

Area under the graph = change in velocity.

On a velocity–time graph the gradient of a section gives the acceleration (positive when speeding up, negative when slowing down) and the total area between the line and the time axis gives the displacement.

A recurring trap: a positive but decreasing gradient on a d–t graph still means the object is moving forwards — just more slowly. And on a v–t graph a line below the time axis means the velocity is negative (moving backwards), so that area counts as negative displacement.

d–t graph: gradient = velocity (tangent gives instantaneous velocity).
v–t graph: gradient = acceleration; area under = displacement.
a–t graph: area under = change in velocity.

See the full worked example for kinematics →

Projectile motion: split it into two independent problems

Horizontal = constant velocity; vertical = free fall. They share only the time.

A projectile is any object moving under gravity alone (we ignore air resistance unless told otherwise). The key idea — and the source of most marks — is that the horizontal and vertical motions are independent and are linked only by the common time $t$ .

Step 1 — resolve the launch velocity. If launched at speed $u$ at angle $\theta$ above the horizontal:

Horizontal component: $u_x = u\cos\theta$
Vertical component: $u_y = u\sin\theta$

Step 2 — treat each direction separately.

	Horizontal	Vertical
Acceleration	$a_x = 0$	$a_y = g = 9.81\,\text{m s}^{-2}$ (down)
Velocity	constant $= u\cos\theta$	changes: $v_y = u\sin\theta - gt$ (up positive)
Displacement	$x = u\cos\theta \cdot t$	$y = u\sin\theta \cdot t - \tfrac{1}{2}gt^2$

Key facts examiners test:

At the highest point, the vertical velocity is zero — but the object is still moving horizontally and its acceleration is still $g$ downwards.
The time to the top = time to fall back to launch height (for level ground), so total flight time is twice the time to the top.
Range is greatest for a launch angle of 45° (no air resistance).

Micro-example (do this in your head to lock the method in): a ball is thrown at $u = 20\,\text{m s}^{-1}$ at $\theta = 30°$ .

Components: $u_x = 20\cos30° = 17.3\,\text{m s}^{-1}$ , $u_y = 20\sin30° = 10.0\,\text{m s}^{-1}$ .
Time to the top (where $v_y = 0$ ): $t_{\text{up}} = u_y/g = 10.0/9.81 = 1.02\,\text{s}$ , so total flight time $= 2.04\,\text{s}$ .
Range $= u_x \times t = 17.3 \times 2.04 \approx 35\,\text{m}$ .

Notice the structure: resolve → use the vertical motion for the time → use the horizontal motion for the distance. Every angled-projectile question is that same three-move sequence.

A projectile's path is a parabola. The horizontal velocity stays constant throughout; the vertical velocity falls to zero at the peak and then grows downwards. The two motions share only the time.

With air resistance (a common "describe" question): the drag force opposes motion, so compared with the ideal parabola the projectile has a lower maximum height, a shorter range, and an asymmetric path — the descent is steeper than the ascent. A vertically falling object may reach terminal velocity when drag equals weight.

Resolve launch velocity: uₓ = u cos θ, u_y = u sin θ.
Horizontal: constant velocity. Vertical: free fall with a = g.
At the peak v_y = 0, but horizontal velocity and g are unchanged.

See the full worked example for kinematics →

Air resistance and terminal velocity

Drag grows with speed until it balances weight — then acceleration is zero.

Real motion is affected by air resistance (drag), a force that:

always acts opposite to the direction of motion, and
increases with speed (roughly with $v^2$ at higher speeds).

Falling with air resistance — the story to tell in a "describe/explain" answer:

At release, speed is zero, so drag is zero. The only force is weight, so the acceleration is $g$ (maximum).
As speed rises, drag increases, so the resultant force decreases, so the acceleration decreases (the object is still speeding up, but less quickly).
Eventually drag = weight, the resultant force is zero, so the acceleration is zero and the object falls at a constant terminal velocity.

Falling with air resistance: the initial gradient equals g, then the curve flattens as drag builds up, reaching a constant terminal velocity when drag balances weight and the resultant force is zero.

Terminal velocity does not mean the object stops — it means it moves at a constant maximum speed because the forces are balanced (Newton's first law, topic A.2). A parachutist reaches a high terminal velocity in free fall, then a much lower one once the parachute opens (larger area → more drag → new, lower balance).

Drag opposes motion and increases with speed.
Falling: acceleration decreases from g to zero as drag builds up.
Terminal velocity: drag = weight, resultant force = 0, constant speed.

Quick recap

Distance and speed are scalars; displacement, velocity and acceleration are vectors.
The four suvat equations apply only to uniform (constant) acceleration.
d–t gradient = velocity; v–t gradient = acceleration; area under v–t = displacement.
Projectiles: independent horizontal (constant v) and vertical (a = g) motions linked only by time; v_y = 0 at the peak.
Air resistance opposes motion and grows with speed; when drag = weight the object moves at constant terminal velocity (a = 0).
Always give the final answer with correct units and appropriate significant figures, and show working for method marks.

Memorise this

Verbatim phrases, formulae and definitions IB DP mark schemes credit (key for AO1 knowledge marks on Paper 1).

Scalars: distance, speed. Vectors: displacement, velocity, acceleration.
suvat (uniform a only): v = u + at; s = ut + ½at²; v² = u² + 2as; s = ½(u+v)t.
Free fall: a = g = 9.81 m s⁻² downwards.
d–t graph: gradient = velocity. v–t graph: gradient = acceleration, area = displacement.
Projectile: uₓ = u cos θ (constant); u_y = u sin θ (a = g). At the top v_y = 0.
Max range (no air resistance) at launch angle 45°.
Terminal velocity: drag = weight → resultant force = 0 → constant velocity (a = 0).

How it’s examined

Kinematics appears in every IB Physics exam and underpins all of Theme A. Paper 1A (MCQ): selecting the correct suvat equation, reading a value or gradient from a motion graph, and identifying which graph matches a described motion — no calculator, so numbers are chosen to be manageable. Paper 1B (data-based): determining velocity or acceleration from the gradient of a graph (showing the triangle and stating coordinates), and interpreting the shape of a curve. Paper 2: multi-step projectile problems (3–6 marks) that require resolving the launch velocity, using the vertical motion to find the time, and feeding that into the horizontal motion — plus 'describe/explain' questions on the effect of air resistance on a trajectory. Command terms: state, determine, calculate, sketch, describe, explain. Examiner reports repeatedly flag: applying suvat to non-uniform acceleration, mixing horizontal and vertical components, quoting a graph value instead of a gradient/area, and forgetting units or significant figures. Always show full working — method marks are awarded even when the final arithmetic slips.

Sources: IB Diploma Programme Physics Guide (first assessment 2025) — Theme A: Space, time and motion (A.1 Kinematics); IB Physics Data Booklet (2025); IB Physics subject reports and specimen papers (2023–2025). Last reviewed 2026-07-21.

Take this whole topic with you

Step-by-step worked examples — Kinematics

Step-by-step solutions to past-paper-style questions on kinematics, written exactly the way a tutor would explain them at the board.

Question type:

1Average speed vs average velocity
Getting startedWord problem• scalars & vectors, AO2
Question
A jogger runs 300 m due east in 100 s, then 100 m due west in 40 s. Calculate (a) her average speed and (b) the magnitude of her average velocity for the whole run. (3 marks)
Step-by-step solution
1. Step 1
  Average speed uses the total distance (a scalar): distance $= 300 + 100 = 400\,\text{m}$ ; total time $= 100 + 40 = 140\,\text{s}$ .
  $\text{average speed} = \frac{400}{140} = 2.86\,\text{m s}^{-1}$
2. Step 2
  Average velocity uses the displacement (a vector): taking east as positive, displacement $= 300 - 100 = 200\,\text{m}$ east.
  $\text{average velocity} = \frac{200}{140} = 1.43\,\text{m s}^{-1}\ \text{east}$
Answer
(a) average speed = 2.86 m s⁻¹; (b) average velocity = 1.43 m s⁻¹ due east.
Examiner tip
Mark scheme: (1) total distance/time for speed; (2) net displacement for velocity; (3) both values correct to 3 s.f. with units. The whole point of the question is that speed ≠ magnitude of velocity when the path is not straight.
2Choosing the right suvat equation
Getting startedDirect calculation• suvat, AO2
Question
A car accelerates uniformly from rest to 24 m s⁻¹ in 6.0 s. Calculate its acceleration and the distance travelled. (3 marks)
Step-by-step solution
1. Step 1
  List known values: $u = 0$ , $v = 24\,\text{m s}^{-1}$ , $t = 6.0\,\text{s}$ . For acceleration use the equation linking $v, u, a, t$ .
  $a = \frac{v-u}{t} = \frac{24-0}{6.0} = 4.0\,\text{m s}^{-2}$
2. Step 2
  For distance, use $s = \tfrac{1}{2}(u+v)t$ (no need to know $a$ ).
  $s = \tfrac{1}{2}(0+24)(6.0) = 72\,\text{m}$
Answer
a = 4.0 m s⁻²; s = 72 m.
Examiner tip
Mark scheme: (1) correct rearrangement for a; (2) correct equation for s; (3) both answers with units. Using s = ½(u+v)t avoids needing a first — either route earns the marks if working is shown.
3Free fall from a height
Getting startedDirect calculation• free fall, AO2
Question
A stone is dropped from rest from the top of a cliff and hits the sea 2.5 s later. Ignoring air resistance, calculate (a) the height of the cliff and (b) the speed of the stone as it hits the water. (g = 9.81 m s⁻²) (3 marks)
Step-by-step solution
1. Step 1
  Dropped from rest: $u = 0$ , $a = g = 9.81\,\text{m s}^{-2}$ , $t = 2.5\,\text{s}$ . Use $s = ut + \tfrac{1}{2}at^2$ for the height.
  $s = 0 + \tfrac{1}{2}(9.81)(2.5)^2 = 30.7\,\text{m}$
2. Step 2
  Use $v = u + at$ for the impact speed.
  $v = 0 + (9.81)(2.5) = 24.5\,\text{m s}^{-1}$
Answer
(a) height ≈ 31 m; (b) impact speed ≈ 25 m s⁻¹.
Examiner tip
Mark scheme: (1) correct equation and substitution for s; (2) height ≈ 30.7 m; (3) v ≈ 24.5 m s⁻¹. 'Dropped from rest' is the clue that u = 0 — students who carry a non-zero u lose marks.
4Reading a velocity–time graph
Building confidenceGraph or diagram• motion graphs, AO2, AO3
Question
A train starts from rest and accelerates uniformly to 20 m s⁻¹ over 10 s, travels at 20 m s⁻¹ for 30 s, then decelerates uniformly to rest in 20 s. (a) Find the acceleration in the first stage. (b) Find the total distance travelled. (4 marks)
Step-by-step solution
1. Step 1
  Acceleration in stage 1 = gradient of the v–t graph.
  $a = \frac{20-0}{10} = 2.0\,\text{m s}^{-2}$
2. Step 2
  Total distance = area under the v–t graph. Stage 1 is a triangle, stage 2 a rectangle, stage 3 a triangle.
  $s_1 = \tfrac{1}{2}(10)(20) = 100\,\text{m}$
3. Step 3
  Stage 2 (rectangle) and stage 3 (triangle):
  $s_2 = 20\times30 = 600\,\text{m};\quad s_3 = \tfrac{1}{2}(20)(20) = 200\,\text{m}$
4. Step 4
  Add the three areas for the total displacement.
  $s = 100 + 600 + 200 = 900\,\text{m}$
Answer
(a) a = 2.0 m s⁻²; (b) total distance = 900 m.
Examiner tip
Mark scheme: (1) gradient = 2.0 m s⁻²; (2) triangle areas; (3) rectangle area; (4) total 900 m. The distance is the AREA under the graph — a frequent error is to read the peak velocity (20) and stop there.
5Using v² = u² + 2as when time is unknown
Building confidenceDirect calculation• suvat, AO2
Question
A cyclist travelling at 12 m s⁻¹ brakes with a uniform deceleration of 3.0 m s⁻² until she stops. How far does she travel while braking? (3 marks)
Step-by-step solution
1. Step 1
  Known: $u = 12\,\text{m s}^{-1}$ , $v = 0$ , $a = -3.0\,\text{m s}^{-2}$ (deceleration is negative). Time is not given or wanted → use $v^2 = u^2 + 2as$ .
  $0 = 12^2 + 2(-3.0)s$
2. Step 2
  Rearrange for $s$ .
  $s = \frac{-12^2}{2(-3.0)} = \frac{-144}{-6.0} = 24\,\text{m}$
Answer
s = 24 m.
Examiner tip
Mark scheme: (1) select v² = u² + 2as; (2) correct signs (a negative); (3) s = 24 m with unit. Getting the sign of the deceleration right is essential — a positive a would give a negative distance, which is unphysical.
6Horizontal projectile off a table
Building confidenceMulti-step problem• projectile, AO2
Question
A ball rolls off a table 1.25 m high with a horizontal velocity of 3.0 m s⁻¹. Ignoring air resistance, calculate (a) the time to reach the floor and (b) the horizontal distance travelled. (g = 9.81 m s⁻²) (4 marks)
Step-by-step solution
1. Step 1
  Vertical: launched horizontally so $u_y = 0$ ; the fall is free fall through 1.25 m. Use $s = \tfrac{1}{2}gt^2$ to find $t$ .
  $1.25 = \tfrac{1}{2}(9.81)t^2$
2. Step 2
  Solve for the time of flight.
  $t = \sqrt{\frac{2(1.25)}{9.81}} = 0.505\,\text{s}$
3. Step 3
  Horizontal: velocity is constant at $3.0\,\text{m s}^{-1}$ . Use the same time.
  $x = u_x\,t = 3.0 \times 0.505 = 1.51\,\text{m}$
Answer
(a) t ≈ 0.51 s; (b) horizontal distance ≈ 1.5 m.
Examiner tip
Mark scheme: (1) recognise u_y = 0; (2) t from vertical free fall; (3) horizontal distance using constant velocity; (4) answer with unit. The vertical motion gives the time; the horizontal motion is independent and uses that time.
7Projectile launched at an angle
StretchMulti-step problem• projectile, AO2, resolving vectors
Question
A ball is kicked from level ground at 20 m s⁻¹ at 30° above the horizontal. Ignoring air resistance, calculate (a) the maximum height reached and (b) the horizontal range. (g = 9.81 m s⁻²) (5 marks)
Step-by-step solution
1. Step 1
  Resolve the launch velocity into components.
  $u_x = 20\cos30^\circ = 17.3\,\text{m s}^{-1};\quad u_y = 20\sin30^\circ = 10.0\,\text{m s}^{-1}$
2. Step 2
  Max height: at the top $v_y = 0$ . Use $v_y^2 = u_y^2 - 2g\,H$ vertically.
  $0 = 10.0^2 - 2(9.81)H \Rightarrow H = \frac{100}{19.62} = 5.10\,\text{m}$
3. Step 3
  Time of flight: time up = $u_y/g$ ; for level ground total flight time is twice this.
  $t = \frac{2u_y}{g} = \frac{2(10.0)}{9.81} = 2.04\,\text{s}$
4. Step 4
  Range: horizontal velocity is constant over the flight time.
  $R = u_x\,t = 17.3 \times 2.04 = 35.3\,\text{m}$
Answer
(a) maximum height ≈ 5.1 m; (b) range ≈ 35 m.
Examiner tip
Mark scheme: (1) both components; (2) v_y = 0 at top → H; (3) flight time = 2u_y/g; (4) range = u_x t; (5) answers with units. A* discipline: use v_y = 0 at the peak and remember the flight time is twice the time to the top for level ground.
8Instantaneous velocity from a curved graph
StretchGraph or diagram• motion graphs, AO3, tangent
Question
The displacement–time graph of an accelerating trolley is a curve. At t = 4.0 s a tangent to the curve passes through the points (2.0 s, 1.0 m) and (6.0 s, 9.0 m). Determine the instantaneous velocity of the trolley at t = 4.0 s, and explain why a tangent is used. (3 marks)
Step-by-step solution
1. Step 1
  Instantaneous velocity = gradient of the tangent to the d–t curve at that instant.
  $v = \frac{\Delta s}{\Delta t} = \frac{9.0 - 1.0}{6.0 - 2.0} = \frac{8.0}{4.0} = 2.0\,\text{m s}^{-1}$
2. Step 2
  A tangent is used because the graph is curved (the velocity is changing). The gradient of a chord would give an average velocity over an interval, whereas the tangent gives the velocity at the single instant $t = 4.0\,\text{s}$ .
Answer
v = 2.0 m s⁻¹; a tangent gives the instantaneous (not average) velocity because the velocity is changing on the curve.
Examiner tip
Mark scheme: (1) uses the two tangent points; (2) v = 2.0 m s⁻¹; (3) correct explanation (tangent = instantaneous, chord = average). Reading the coordinates off the stated tangent points, not the curve, is the key skill.

Model Answers — Kinematics

High-scoring sample answers for kinematics on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
2 marks
Q (2 marks). Define (a) displacement and (b) acceleration, stating whether each is a scalar or a vector.
Model answer
(a) Displacement is the change in position of an object measured in a straight line from its starting point, together with its direction. It is a vector. (SI unit: metre.)

(b) Acceleration is the rate of change of velocity. It is a vector. (SI unit: m s⁻².)
Why this scores
Why this scores 2/2. (1) displacement = change in position with direction (vector); (2) acceleration = rate of change of velocity (vector). Saying acceleration is 'rate of change of speed' is not enough — it must be velocity, because a change of direction is also an acceleration.
Question 2
3 marks
Q (3 marks). An aeroplane accelerates uniformly from rest along a runway at 3.2 m s⁻² and needs to reach a take-off speed of 64 m s⁻². Calculate the minimum length of runway required.
Model answer
Known values: $u = 0$ , $v = 64\,\text{m s}^{-1}$ , $a = 3.2\,\text{m s}^{-2}$ ; find $s$ (time not given → use $v^2 = u^2 + 2as$ ).

$v^2 = u^2 + 2as \Rightarrow 64^2 = 0 + 2(3.2)s$

$s = \frac{4096}{6.4} = 640\,\text{m}$

The runway must be at least 640 m long.
Why this scores
Why this scores 3/3. (1) correct equation selected (no time); (2) correct substitution; (3) s = 640 m with unit. Choosing v² = u² + 2as because time is neither given nor asked for is the mark-earning decision.
Question 3
3 marks
Q (3 marks). A velocity–time graph for a car is a straight line rising from 5 m s⁻¹ at t = 0 to 25 m s⁻¹ at t = 8.0 s. (a) State what the gradient represents and calculate it. (b) State what the area under the graph represents.
Model answer
(a) The gradient of a velocity–time graph represents the acceleration.

$a = \frac{\Delta v}{\Delta t} = \frac{25 - 5}{8.0} = 2.5\,\text{m s}^{-2}$

(b) The area between the line and the time axis represents the displacement (distance travelled) of the car.
Why this scores
Why this scores 3/3. (1) gradient = acceleration; (2) a = 2.5 m s⁻² with working; (3) area = displacement. These two facts (gradient = acceleration, area = displacement) are examined constantly — state them explicitly.
Question 4
4 marks
Q (4 marks). A stone is thrown horizontally at 8.0 m s⁻¹ from the top of a vertical cliff and lands in the sea 3.0 s later. Ignoring air resistance, calculate (a) the height of the cliff and (b) the horizontal distance from the base of the cliff at which it lands. (g = 9.81 m s⁻²)
Model answer
Vertical motion (thrown horizontally, so initial vertical velocity $u_y = 0$ ; free fall):

$h = u_y t + \tfrac{1}{2}gt^2 = 0 + \tfrac{1}{2}(9.81)(3.0)^2 = 44.1\,\text{m}$

So the cliff is about 44 m high.

Horizontal motion (constant velocity, same time of flight):

$x = u_x t = 8.0 \times 3.0 = 24\,\text{m}$

The stone lands 24 m from the base of the cliff.
Why this scores
Why this scores 4/4. (1) recognise u_y = 0; (2) height ≈ 44 m; (3) horizontal velocity constant; (4) distance = 24 m. The independence of the two directions — with time as the only shared quantity — is the assessed idea.
Question 5
5 marks
Q (5 marks). A car travelling at 30 m s⁻¹ sees a hazard. The driver's reaction time is 0.70 s, during which the car continues at constant speed, and then the brakes provide a uniform deceleration of 6.0 m s⁻². Calculate the total stopping distance (thinking distance + braking distance).
Model answer
Thinking distance (constant velocity during reaction time):

$s_{\text{think}} = v\,t = 30 \times 0.70 = 21\,\text{m}$

Braking distance (uniform deceleration from 30 m s⁻¹ to rest; time unknown → use $v^2 = u^2 + 2as$ ):

$0 = 30^2 + 2(-6.0)s_{\text{brake}}$

$s_{\text{brake}} = \frac{-900}{-12} = 75\,\text{m}$

Total stopping distance:

$s = s_{\text{think}} + s_{\text{brake}} = 21 + 75 = 96\,\text{m}$
Why this scores
Why this scores 5/5. (1) thinking distance = vt; (2) 21 m; (3) select v² = u² + 2as with a negative; (4) braking distance 75 m; (5) total 96 m. The reaction phase is constant velocity (no acceleration) — treating it as part of the braking is a common, costly error.
Question 6
4 marks
Q (4 marks). A skydiver jumps from a stationary balloon and falls without opening her parachute. Explain, in terms of the forces acting, how her acceleration changes from the moment she jumps until she reaches terminal velocity.
Model answer
Just after jumping her speed is low, so the air resistance (drag) is small. Her weight is much greater than the drag, so there is a large resultant downward force and her acceleration is at its maximum (equal to $g$ ).

As she speeds up, the drag increases (drag grows with speed). The resultant downward force (weight − drag) therefore decreases, so her acceleration decreases — she is still speeding up, but less and less quickly.

Eventually the drag becomes equal to her weight. The resultant force is now zero, so by Newton's first law her acceleration is zero and she falls at a constant terminal velocity.
Why this scores
Why this scores 4/4. (1) initially drag small, resultant ≈ weight, a = g; (2) as speed rises drag increases; (3) resultant force decreases so acceleration decreases; (4) drag = weight → resultant zero → constant terminal velocity. Answers must talk about the resultant force changing, not just 'gravity pulls her down'.
Question 7
6 marks
Q (6 marks). A projectile is launched from level ground at 25 m s⁻¹ at an angle of 40° above the horizontal. Ignoring air resistance, calculate (a) the time of flight, (b) the maximum height reached and (c) the horizontal range. (g = 9.81 m s⁻²)
Model answer
Resolve the launch velocity: $u_x = 25\cos40^\circ = 19.2\,\text{m s}^{-1}, \qquad u_y = 25\sin40^\circ = 16.1\,\text{m s}^{-1}$

(a) Time of flight — for level ground the projectile returns to launch height, so the total time is twice the time to the top (where $v_y = 0$ ): $t = \frac{2u_y}{g} = \frac{2(16.1)}{9.81} = 3.28\,\text{s}$

(b) Maximum height — using $v_y^2 = u_y^2 - 2gH$ with $v_y = 0$ at the top: $H = \frac{u_y^2}{2g} = \frac{16.1^2}{2(9.81)} = 13.2\,\text{m}$

(c) Horizontal range — horizontal velocity is constant over the flight time: $R = u_x\,t = 19.2 \times 3.28 = 63.0\,\text{m}$
Why this scores
Why this scores 6/6. (1) both components resolved; (2) flight time = 2u_y/g; (3) t = 3.28 s; (4) max height using v_y = 0; (5) H = 13.2 m; (6) range = u_x t = 63 m. The discriminators are v_y = 0 at the top and the symmetric flight time — mixing components loses multiple marks.
Question 8
6 marks
Q (6 marks). A ball is thrown vertically upward and returns to the thrower's hand. Taking upward as positive and ignoring air resistance: (a) sketch the velocity–time graph for the whole flight, and (b) explain the shape of the graph, including the sign of the velocity and the value of the acceleration at the highest point.
Model answer
(a) The velocity–time graph is a straight line with a constant negative gradient. It starts at a positive value (the launch speed, e.g. $+u$ ) at $t = 0$ , crosses the time axis at the highest point (where $v = 0$ ), and continues down to $-u$ when the ball returns to the hand — a single straight line sloping downwards.

(b) Explanation:

The only force is gravity, so the acceleration is constant at $g = 9.81\,\text{m s}^{-2}$ downwards throughout — hence the graph is a straight line with constant gradient $-g$ .

On the way up the velocity is positive but decreasing (the ball slows as gravity opposes its motion).

At the highest point the velocity is zero, but the acceleration is still $g$ downwards (not zero) — this is why the graph crosses the axis but its gradient does not change.

On the way down the velocity is negative and its magnitude increases as the ball speeds up. By symmetry it returns to the hand with the same speed it left, so $v = -u$ .
Why this scores
Why this scores 6/6. (a) 2 marks: straight line, negative gradient, starts +, ends −, passes through zero. (b) 4 marks: constant a = g (1); velocity positive but decreasing going up (1); at the top v = 0 but a = g still (1); velocity negative and increasing in magnitude on the way down (1). The examiner's favourite check is the highest point: v = 0 does NOT mean a = 0.
Question 9
10 marks
Q (10 marks — extended response). A stone is thrown from the edge of a cliff with a speed of 18 m s⁻¹ at an angle of 35° above the horizontal. The base of the cliff is 25 m below the launch point, and the stone lands on level ground at the base. Ignoring air resistance (g = 9.81 m s⁻²): (a) show that the vertical component of the launch velocity is about 10 m s⁻¹; (b) calculate the maximum height reached above the launch point; (c) calculate the total time of flight until the stone lands; (d) calculate the horizontal distance from the base of the cliff at which the stone lands; (e) determine the speed and direction of the stone as it lands.
Model answer
Resolve the launch velocity (take up as positive): $u_x = 18\cos35^\circ = 14.7\,\text{m s}^{-1}, \qquad u_y = 18\sin35^\circ = 10.3\,\text{m s}^{-1}$

(a) $u_y = 18\sin35^\circ = 18 \times 0.574 = 10.3 \approx 10\,\text{m s}^{-1}$ ✓ (shown).

(b) Maximum height above launch point (at the top $v_y = 0$ ): $H = \frac{u_y^2}{2g} = \frac{10.3^2}{2(9.81)} = 5.41\,\text{m}$

(c) Total time of flight. The stone lands 25 m below launch, so the vertical displacement is $y = -25\,\text{m}$ . Using $y = u_y t - \tfrac{1}{2}gt^2$ : $-25 = 10.3\,t - 4.905\,t^2$ $4.905\,t^2 - 10.3\,t - 25 = 0$ Solving the quadratic (taking the positive root): $t = \frac{10.3 + \sqrt{10.3^2 + 4(4.905)(25)}}{2(4.905)} = \frac{10.3 + \sqrt{106 + 490.5}}{9.81} = \frac{10.3 + 24.4}{9.81} = 3.54\,\text{s}$

(d) Horizontal range (constant horizontal velocity over the whole flight): $x = u_x\,t = 14.7 \times 3.54 = 52.0\,\text{m}$

(e) Landing velocity. Horizontal component is unchanged: $v_x = 14.7\,\text{m s}^{-1}$ . Vertical component at landing: $v_y = u_y - gt = 10.3 - 9.81(3.54) = -24.4\,\text{m s}^{-1}\ (\text{i.e. } 24.4\,\text{m s}^{-1}\ \text{downward})$ Resultant speed (Pythagoras): $v = \sqrt{v_x^2 + v_y^2} = \sqrt{14.7^2 + 24.4^2} = 28.5\,\text{m s}^{-1}$ Direction below the horizontal: $\phi = \tan^{-1}\!\left(\frac{24.4}{14.7}\right) = 58.9^\circ\ \text{below the horizontal}$
Why this scores
Why this scores 10/10. (a) both components, u_y ≈ 10 m s⁻¹ (2); (b) H = u_y²/2g = 5.4 m (2); (c) set y = −25 m, form and solve the quadratic, t = 3.54 s (2); (d) range = u_x t = 52 m (2); (e) combine v_x and v_y by Pythagoras → 28.5 m s⁻¹ at 59° below horizontal (2). This is a model IB Paper 2 extended-response: the grade-9 discriminators are (i) recognising the landing point is below launch so the vertical displacement is negative and a quadratic is unavoidable, and (ii) recombining the components at landing rather than assuming symmetry.

Key Formulae — Kinematics

The formulae you need to memorise for kinematics on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

First equation of motion
$v = u + at$
$v$
final velocity (m s⁻¹)
$u$
initial velocity (m s⁻¹)
$a$
acceleration (m s⁻²)
$t$
time (s)
When to use
When you know three of v, u, a, t and want the fourth — and displacement is not involved.
Example
Speed after 6.0 s at a = 4.0 m s⁻² from rest: v = 0 + 4.0×6.0 = 24 m s⁻¹.
Second equation of motion
$s = ut + \tfrac{1}{2}at^2$
$s$
displacement (m)
$u$
initial velocity (m s⁻¹)
$a$
acceleration (m s⁻²)
$t$
time (s)
When to use
When you know u, a and t and want the displacement (final velocity not needed).
Example
Fall from rest for 2.5 s: s = ½(9.81)(2.5)² = 30.7 m.
Third equation of motion
$v^2 = u^2 + 2as$
$v$
final velocity (m s⁻¹)
$u$
initial velocity (m s⁻¹)
$a$
acceleration (m s⁻²)
$s$
displacement (m)
When to use
When time is neither given nor wanted — links velocities, acceleration and displacement.
Example
Braking from 12 m s⁻¹ at −3.0 m s⁻²: 0 = 12² + 2(−3.0)s → s = 24 m.
Fourth equation of motion
$s = \tfrac{1}{2}(u+v)\,t$
$s$
displacement (m)
$u$
initial velocity (m s⁻¹)
$v$
final velocity (m s⁻¹)
$t$
time (s)
When to use
When acceleration is not needed — uses the average of the initial and final velocities.
Example
0 to 24 m s⁻¹ in 6.0 s: s = ½(0+24)(6.0) = 72 m.
Resolving a launch velocity
$u_x = u\cos\theta, \qquad u_y = u\sin\theta$
$u$
launch speed (m s⁻¹)
$\theta$
launch angle above the horizontal
$u_x$
horizontal component of velocity (m s⁻¹)
$u_y$
vertical component of velocity (m s⁻¹)
When to use
At the start of any angled-projectile problem — split the velocity into independent horizontal and vertical parts.
Example
20 m s⁻¹ at 30°: uₓ = 20cos30° = 17.3 m s⁻¹; u_y = 20sin30° = 10.0 m s⁻¹.

Key Definitions and Keywords — Kinematics

Definitions to memorise and the exact keywords mark schemes credit for kinematics answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Distance
Examiner keyword
The total length of the path travelled by an object. A scalar quantity (magnitude only). SI unit: metre (m).
Displacement
Examiner keyword
The change in position of an object measured in a straight line from start to finish, including direction. A vector. SI unit: metre (m).
Speed
Examiner keyword
The rate of change of distance with time. A scalar. SI unit: m s⁻¹.
Velocity
Examiner keyword
The rate of change of displacement with time, including direction. A vector. SI unit: m s⁻¹.
Acceleration
Examiner keyword
The rate of change of velocity with time. A vector — an object accelerates if its speed OR its direction changes. SI unit: m s⁻².
Average velocity
Total displacement divided by the total time taken for that displacement: v_avg = Δs/Δt.
Instantaneous velocity
Examiner keyword
The velocity at a single instant, found from the gradient of the tangent to a displacement–time graph at that point.
Uniform (constant) acceleration
Examiner keyword
Acceleration that does not change in size or direction. Only under uniform acceleration are the four suvat equations valid.
Free fall
Examiner keyword
Motion under the influence of gravity alone (no air resistance), giving a constant downward acceleration g = 9.81 m s⁻² near Earth's surface.
Projectile
Examiner keyword
An object moving under gravity alone after launch, whose horizontal (constant velocity) and vertical (constant acceleration g) motions are independent.
Terminal velocity
Examiner keyword
The constant maximum velocity reached by a falling object when the drag force equals its weight, so the resultant force and acceleration are zero.
Air resistance (drag)
A resistive force that opposes an object's motion through air and increases with speed. Not present in idealised (free-fall / no-drag) projectile problems.
Component of a vector
Examiner keyword
The projection of a vector onto a chosen axis, e.g. the horizontal (u cos θ) and vertical (u sin θ) components of a launch velocity.

Common Mistakes and Misconceptions — Kinematics

The traps other students keep falling into on kinematics questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Using suvat equations when the acceleration is not constant
IB Physics Theme A subject reports
Why it happens
Students reach for the familiar equations without checking whether the acceleration is uniform.
How to avoid it
Before using suvat, confirm the acceleration is constant over the whole interval. If air resistance is significant, or the v–t graph is curved, suvat does not apply — use graphical methods instead.
✕Mixing horizontal and vertical quantities in one projectile equation
IB Physics Paper 2 subject reports
Why it happens
The two motions feel like one problem, so students put a horizontal velocity into a vertical suvat equation.
How to avoid it
Keep two separate columns (horizontal and vertical). The only quantity they share is the time. Solve one direction for t, then use t in the other.
✕Thinking acceleration is zero at the highest point of a projectile's flight
IB Physics Theme A subject reports
Why it happens
Because the vertical velocity is zero there, students assume the acceleration is too.
How to avoid it
At the top, v_y = 0 but a = g (still 9.81 m s⁻² downwards). Gravity acts throughout the flight; only the vertical velocity is momentarily zero.
✕Reading a value off a motion graph when a gradient or area is required
IB Physics Paper 1B subject reports
Why it happens
Students see a number on the axis and quote it instead of processing the graph.
How to avoid it
On a v–t graph, a point gives the velocity; the gradient gives acceleration and the area gives displacement. Always show the triangle (for a gradient) or the shapes (for an area).
✕Being inconsistent with the direction (sign) convention for vectors
IB Physics Theme A subject reports
Why it happens
Up is taken as positive in one line and negative in the next, so signs of g, u and s get muddled.
How to avoid it
State your convention at the start (e.g. 'up is positive') and apply it to every vector: velocities, accelerations and displacements. A downward g is then −9.81 m s⁻².
✕Omitting units or giving an inappropriate number of significant figures
IB Physics general subject reports
Why it happens
Under time pressure students write a bare number, or copy all the calculator digits.
How to avoid it
Always attach the SI unit and round to a sensible number of significant figures (usually 2–3, matching the data). A numerically correct answer with no unit is not fully credited.

Kinematics

Detailed Study Notes

At a glance

What you’ll learn

Scalars vs vectors: getting the language right

The suvat equations (uniform acceleration only)

Where the suvat equations come from (derivation)

Motion graphs: gradient and area do the work

Projectile motion: split it into two independent problems

Air resistance and terminal velocity

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Average speed vs average velocity

2Choosing the right suvat equation

3Free fall from a height

4Reading a velocity–time graph

5Using v² = u² + 2as when time is unknown

6Horizontal projectile off a table

7Projectile launched at an angle

8Instantaneous velocity from a curved graph

First equation of motion

Second equation of motion

Third equation of motion

Fourth equation of motion

Resolving a launch velocity

Distance

Displacement

Speed

Velocity

Acceleration

Average velocity

Instantaneous velocity

Uniform (constant) acceleration

Free fall

Projectile

Terminal velocity

Air resistance (drag)

Component of a vector

✕Using suvat equations when the acceleration is not constant

✕Mixing horizontal and vertical quantities in one projectile equation

✕Thinking acceleration is zero at the highest point of a projectile's flight

✕Reading a value off a motion graph when a gradient or area is required

✕Being inconsistent with the direction (sign) convention for vectors

✕Omitting units or giving an inappropriate number of significant figures