Educerie · IB Diploma · Physics
Theme A Space, time and motion · A.1 Kinematics
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Describe motion in terms of position, displacement, velocity and acceleration, and say which are vectors | SL, HL | Paper 1A: "Which of these quantities is a scalar?" (1 mark) |
| Distinguish distance from displacement, and speed from velocity | SL, HL | "Calculate the average speed and the average velocity of…" (3 marks) |
| Use velocity as the rate of change of position and acceleration as the rate of change of velocity | SL, HL | Reading gradients off displacement–time and velocity–time graphs |
| Distinguish instantaneous from average values, and determine each from data or a graph | SL, HL | "Determine the instantaneous velocity at t = 5.0 s" (tangent, 2–3 marks) |
| Choose and use the four equations for uniformly accelerated motion | SL, HL | Paper 2: a braking, launching or falling calculation, 2–4 marks, working shown |
| Analyse uniform and non-uniform acceleration from graphs, using gradients and areas | SL, HL | Paper 1A and Paper 1B: "Estimate the distance travelled in the first 10 s" |
| Solve projectile problems with no fluid resistance, launched horizontally, above or below the horizontal | SL, HL | Paper 2, 4–8 marks across parts: time of flight, range, maximum height, landing velocity |
| Describe how fluid resistance changes a projectile's time of flight, path, velocity, acceleration and range, and explain terminal speed | SL, HL | "Explain why the ball reaches a terminal speed" (2–3 marks) |
| Use graphical analysis to find other quantities (NOS) | SL, HL | Paper 1B: plot d against t², find a from the gradient |
Before you start
You need to rearrange an equation, to use Pythagoras and the sine and cosine of an angle in a right-angled triangle, and to find the gradient of a straight line. That is all. The equations in this subtopic are in the data booklet, so the skill being tested is knowing which one fits and what each symbol means.
1The idea in one paragraph
Motion is described by three quantities that each come from the one before. Position says where a body is. Velocity is how fast position changes, and acceleration is how fast velocity changes. Draw position against time and the gradient is velocity; draw velocity against time and the gradient is acceleration, while the area underneath gives back the change in position. When the acceleration is constant, four equations connect everything and let you predict where a body will be and when. A projectile is two motions at once: steady sideways, uniformly accelerated up and down, joined only by the time. Air resistance spoils the neat version, and you must be able to say how.
2Where it is, and how far it has moved
Position is where a body is, measured from a chosen origin in a chosen direction. Displacement is the change in position: the straight-line distance from start to finish, with a direction. Distance is the total length of the path actually travelled, with no direction at all.
Figure 1 shows why the two are different. Walk 300 m east and then 400 m north.
The distance is 300 + 400 = 700 m. The displacement is the arrow from start to finish. By Pythagoras it is √(300² + 400²) = 500 m, and its direction is tan⁻¹(400 ÷ 300) = 53° north of east. A displacement without a direction is not finished.
This is the difference between a scalar, which has size only, and a vector, which has size and direction. Distance and speed are scalars. Displacement, velocity and acceleration are vectors. In one dimension the direction is carried by the sign: choose which way is positive, and anything the other way is negative.
Speed is distance divided by time; velocity is displacement divided by time. If the walk in Figure 1 took 350 s, the average speed is 700 ÷ 350 = 2.0 m s⁻¹, but the average velocity is 500 ÷ 350 = 1.4 m s⁻¹ at 53° north of east. A runner who finishes a 400 m lap back at the start line has run at, say, 5 m s⁻¹ on average, yet her displacement is zero and so is her average velocity.
3Rates of change, averages and instants
Velocity is the rate of change of position. Acceleration is the rate of change of velocity. "Rate of change" means "change divided by the time it took", so
average velocity = Δs ÷ Δt average acceleration = Δv ÷ Δt
where Δ ("delta") means "change in". Acceleration is measured in m s⁻²: metres per second, every second. An acceleration of −3 m s⁻² means the velocity loses 3 m s⁻¹ every second. If the body is moving in the positive direction, that is slowing down; if it is already moving in the negative direction, it is speeding up. Do not call every negative acceleration a "deceleration" until you know which way the body is going.
An average value covers an interval. An instantaneous value is the value at one moment: the average over a time interval so short that nothing changes during it. A car's speedometer shows instantaneous speed; the journey time divided into the distance gives average speed.
On a graph, the difference is the difference between a chord and a tangent. Figure 2 shows the displacement of a body that is speeding up.
The average velocity between t₁ and t₂ is the gradient of the straight chord joining the two points. The instantaneous velocity at time t is the gradient of the tangent: a straight line that touches the curve at that one point and has the same slope there. To draw a tangent, lay a ruler against the curve so that the curve bends away from it equally on both sides, draw it long, and take the gradient over as big a triangle as the graph allows. A small triangle multiplies the effect of your reading errors.
The same pair of ideas applies to speed and to acceleration: average acceleration is the chord on a velocity–time graph, and instantaneous acceleration is the tangent.
4Three graphs of one journey
Every motion can be drawn three ways, and the exam moves between them freely. Figure 3 is one journey: a cyclist speeds up uniformly from rest to 8 m s⁻¹ in 4 s, rides steadily for 6 s, then brakes uniformly to rest in 4 s.
Read the three panels as a chain.
| Graph | Gradient gives | Area under the line gives |
|---|---|---|
| displacement–time (s–t) | velocity | nothing useful |
| velocity–time (v–t) | acceleration | displacement |
| acceleration–time (a–t) | nothing needed | change in velocity |
On the s–t graph, panel (a), the line curves upwards while the cyclist speeds up, is straight while the velocity is constant, and curves over to flat when she stops. A straight line means constant velocity; a horizontal line means at rest.
On the v–t graph, panel (b), the first section has gradient 8 ÷ 4 = 2 m s⁻², the middle is flat (zero acceleration) and the last has gradient −2 m s⁻². The area under the line is a trapezium: ½ × (14 + 6) × 8 = 80 m, which is exactly where panel (a) ends.
On the a–t graph, panel (c), the three sections are flat lines at +2, 0 and −2 m s⁻². The area of the first block, 2 × 4 = 8 m s⁻¹, is the velocity gained.
The area rule has one trap. Area below the time axis is negative displacement, motion in the negative direction. Figure 4 is a ball thrown straight up at 15 m s⁻¹ and caught again at the same height, with up taken as positive.
The velocity falls steadily through zero at the top, 15 ÷ 9.8 = 1.53 s, and carries on to −15 m s⁻¹. The gradient is −9.8 m s⁻² all the way, including at the top: the ball is momentarily at rest there, but its velocity is still changing, so its acceleration is not zero. The triangle above the axis is +½ × 1.53 × 15 = +11.5 m, the one below is −11.5 m. The displacement is their sum, 0 m, because the ball came back to the hand. The distance is the sum of the sizes, 23 m.
5Uniform acceleration and the four equations
When acceleration is uniform (constant), five quantities describe a stretch of motion, and the data booklet gives four equations that link them.
| Symbol | Meaning | Unit |
|---|---|---|
| s | displacement | m |
| u | initial velocity | m s⁻¹ |
| v | final velocity | m s⁻¹ |
| a | acceleration (constant) | m s⁻² |
| t | time taken | s |
v = u + at s = ut + ½at² v² = u² + 2as s = ((u + v) ÷ 2) t
They are not four facts to learn but one straight-line v–t graph read four ways. v = u + at is the definition of acceleration rearranged. s = ((u + v) ÷ 2) t is the area of the trapezium under the line. The other two follow by eliminating v or t.
Each equation leaves out exactly one of the five quantities: v = u + at has no s, s = ut + ½at² has no v, v² = u² + 2as has no t, and s = ((u + v) ÷ 2) t has no a. So the method is always the same. List the five symbols. Fill in the three you know and mark the one you want. The fifth is the one you neither know nor want, and the right equation is the one without it.
Worked example 1. A train travelling at 24 m s⁻¹ brakes uniformly and stops in 400 m. Find its acceleration and the time it takes to stop.
Known: u = 24 m s⁻¹, v = 0, s = 400 m. Wanted: a. Not needed: t. Use the equation without t.
Check the time another way: s = ((u + v) ÷ 2) t gives t = 400 ÷ 12 = 33 s. Two routes agreeing is the cheapest check there is.
Worked example 2. A ball is thrown vertically upwards at 12.0 m s⁻¹ from the edge of a flat roof 20.0 m above the ground, and misses the roof on the way down. Take g = 9.8 m s⁻² and ignore air resistance. Find the speed at which it hits the ground and the time it is in the air.
Choose up as positive and keep it. Then u = +12.0 m s⁻¹, a = −9.8 m s⁻², and the ground is 20.0 m below the start, so s = −20.0 m. The ball goes up, stops and comes down, but the acceleration is the same throughout, so one equation covers the whole flight.
The ball lands at 23.2 m s⁻¹ after 3.59 s. Nobody had to split the flight into "up" and "down". The sign convention did the work, and the wrong sign on s is the only thing that could have stopped it.
6Non-uniform acceleration
The four equations need a constant a. When the acceleration changes, they give wrong answers, and no amount of care fixes that. Use the graph instead: the definitions of velocity and acceleration as gradients, and displacement as area, are true for every motion.
Figure 5 is a car pulling away from traffic lights. It accelerates hard at first and less and less as it approaches its cruising speed, so its v–t graph is a curve whose gradient falls.
Instantaneous acceleration at t = 5 s is the gradient of the tangent at that point. The tangent in Figure 5 rises about 20 m s⁻¹ over 10 s, so a ≈ 2.0 m s⁻² at that instant; at 20 s the curve is much flatter and the acceleration is smaller.
Displacement is still the area under the curve. The area is no longer a triangle or a trapezium, so you estimate it: count the grid squares under the curve (counting a square that is more than half inside and ignoring one that is less), then multiply by the area one square represents, in metres. Or cut the area into thin vertical strips and treat each as a trapezium. Either method is accepted as long as the working shows what one square, or one strip, is worth.
A 1A question may show a curved v–t graph and ask which statement is true. The answer that uses v² = u² + 2as is always the wrong one.
7Projectiles with no fluid resistance
A projectile is a body moving under gravity alone once it has been launched. With no air resistance, the only force is its weight, which acts straight down. That single fact splits the motion into two independent parts.
- Horizontally there is no force, so no acceleration. The horizontal component of velocity stays the same for the whole flight.
- Vertically the acceleration is g downwards, constant, so the four equations apply.
Figure 6 shows both components at five points of a flight.
The horizontal arrow is the same length everywhere. The vertical arrow shrinks on the way up, is zero at the top, and grows downwards on the way down. The actual velocity at any point is the sum of the two arrows, along the tangent to the path, and its size is √(horizontal² + vertical²). At the top the velocity is not zero; it is the horizontal component alone. The path is a parabola. You will not be asked for its equation.
The two motions share only one quantity: time. So the method never changes.
- Resolve the launch velocity u at angle θ to the horizontal: uₓ = u cos θ horizontally, and u sin θ vertically (up positive).
- Use the vertical motion to find the time, because that is where the acceleration and the information about heights are.
- Use that time in the horizontal motion: x = uₓ t.
Figure 7 shows the three launches you must be able to handle.
Worked example 3 · horizontal launch. A marble rolls off a table 0.80 m high at 2.5 m s⁻¹. Where does it land, and how fast is it going? Take g = 9.8 m s⁻².
The vertical component of the launch velocity is zero, so the fall is a simple drop from rest. Down is positive here, which keeps every number positive.
Worked example 4 · above the horizontal. A football is kicked from level ground at 18 m s⁻¹, 35° above the horizontal. Find the time of flight, the range and the greatest height. Take g = 9.8 m s⁻².
Worked example 5 · below the horizontal. A stone is thrown from the top of a sea cliff 25 m high at 10 m s⁻¹, 20° below the horizontal. How far out from the foot of the cliff does it hit the water? Take g = 9.8 m s⁻².
Up is positive. The launch is downwards, so the vertical component is negative, and the water is 25 m below, so s = −25 m.
When launch and landing are at the same height, the flight is symmetrical and the time up equals the time down. From a cliff it is not, and you must solve the quadratic. Know which situation you are in before you reach for a shortcut.
Some projectile questions give a height and ask only for a speed. Those can be done without the time at all, by conservation of energy, which is in A.3.
8What fluid resistance does
Everything in section 7 assumed no fluid resistance, the drag force a gas or a liquid exerts on a body moving through it. In reality the air pushes back, and the push grows as the body moves faster. You will only be asked about its effects in words, never to calculate a path with it. Figure 8 shows both effects the guide names.
Drag always acts opposite to the velocity. That one fact produces every effect in panel (a).
- Acceleration is no longer just g downwards. On the way up, drag points down and adds to the weight, so the body slows faster than g would make it. On the way down, drag points up and partly cancels the weight, so the body speeds up more slowly than g. Drag also has a horizontal component against the motion, so the horizontal velocity now falls throughout the flight.
- Velocity is smaller at every point after launch than it would have been in a vacuum.
- Maximum height is lower, because the upward speed is lost faster.
- Range is shorter, because the horizontal velocity keeps falling.
- Trajectory is no longer a parabola, nor symmetrical. The body comes down more steeply than it went up, because by then it has lost much of its horizontal velocity.
- Time of flight is usually shorter for a launch from level ground, because the body rises less high. For a fall from a great height, such as a high cliff, the slower descent can make it longer; say which effect you are describing.
Panel (b) is a body dropped from rest. At first it is slow, drag is tiny, and it accelerates at almost g. As it speeds up, drag grows, the resultant force (weight minus drag) shrinks, and so does the acceleration: the gradient of the v–t curve falls. Eventually drag equals weight. The resultant force is zero, the acceleration is zero, and the velocity stops changing. That constant speed is the terminal speed. A skydiver takes several seconds of falling to reach it; a light shuttlecock or a falling leaf reaches it almost at once.
Explaining terminal speed is a force argument, which is A.2's business, but the exam asks it here. The mark-earning sentence is always the same: drag increases with speed until it equals the weight, so the resultant force and therefore the acceleration become zero.
9Where marks are lost
Distance where displacement was asked. The ball in Figure 4 travelled 23 m but its displacement is 0 m. Read which one the question wants, and give a direction with every displacement and velocity.
Zero velocity means zero acceleration. At the top of a vertical throw v = 0 but a = −9.8 m s⁻². If the acceleration were zero there, the ball would hang in the air.
Using the equations of motion when a is not constant. A curved v–t graph means a changing acceleration. Use gradients and areas; never v = u + at.
Mixing signs. Choose a positive direction at the start, write it down, and give g, u and s the signs that go with it. A roof-to-ground drop with s = +20 m instead of −20 m gives a wrong answer that looks perfectly reasonable.
Taking the gradient from a tiny triangle, or from data points instead of the line. A gradient is read from the best-fit line or tangent, over as large a triangle as fits on the graph.
Forgetting that area below the axis is negative. Adding the sizes gives the distance, not the displacement.
Letting the horizontal velocity change in a projectile. With no air resistance nothing acts horizontally. uₓ at launch is uₓ at landing.
Saying the speed is zero at the top of a projectile's path. Only the vertical component is zero. The speed at the top is uₓ.
10Draw it right
Motion graphs are marked on precise features. Check these every time you sketch or plot one.
- Axes labelled with quantity and unit in the IB form: t / s, v / m s⁻¹, s / m. Time on the horizontal axis.
- Constant velocity: a straight s–t line. Constant acceleration: a straight v–t line and a curved s–t line. Mixing these up is the most common sketch error.
- Where one graph's gradient is zero, the next graph crosses or touches zero at the same time. Line the times up vertically when you sketch two graphs together.
- A body that returns in the opposite direction has a v–t line that crosses the time axis, not one that bounces off it.
- A tangent touches the curve at one point, is drawn with a ruler, and is long. Show the triangle you used and the two readings.
- A projectile's path is a symmetrical parabola without air resistance. With air resistance it is lower, shorter and steeper on the way down. Draw both on one set of axes when asked to compare.
- A terminal-speed v–t graph starts with gradient g, curves over, and approaches a horizontal line without crossing it.
- In Paper 1B, draw a single best-fit line that follows the trend of the points, not a dot-to-dot.
11Try it
Marks in brackets. Take g = 9.8 m s⁻². Answers and marker's notes are at the end. Show your working every time: the method earns marks even when the arithmetic slips.
Q1. A ball is thrown vertically upwards. Air resistance is negligible. What is the ball's acceleration at the highest point of its flight? 1 mark
A. zero · B. 9.8 m s⁻² upwards · C. 9.8 m s⁻² downwards · D. it depends on the mass of the ball
Q2. A delivery robot moves 60 m north along a pavement, then 80 m west, taking 70 s in total. Calculate its average speed and its average velocity. 3 marks
Q3. A sprinter accelerates uniformly from rest to 9.0 m s⁻¹ in 3.0 s, then runs at constant speed for the rest of a 100 m race.
(a) Calculate her acceleration. 1 mark
(b) Calculate the distance she covers while accelerating. 2 marks
(c) Determine her total time for the 100 m. 2 marks
Q4. (data-based, Paper 1B style) A trolley is released from rest on a ramp. A student measures the time t it takes to travel distances d down the ramp.
| d / m | 0.20 | 0.40 | 0.60 | 0.80 | 1.00 |
|---|---|---|---|---|---|
| t / s | 0.90 | 1.27 | 1.54 | 1.80 | 2.01 |
| t² / s² | 0.81 | 1.61 | 2.37 | 3.24 | 4.04 |
(a) Show that, if the acceleration a is uniform, a graph of d against t² is a straight line through the origin with gradient a ÷ 2. 2 marks
(b) Using the first and last rows of the table, determine the acceleration of the trolley. 2 marks
(c) Another student's graph of d against t² is straight but crosses the t² axis at a positive value. Suggest one cause. 1 mark
Q5. A ball is thrown from the top of a cliff 12.0 m above the sea, at 15.0 m s⁻¹ and 40° above the horizontal. Air resistance is negligible.
(a) Determine the greatest height of the ball above the sea. 2 marks
(b) Calculate the time the ball takes to reach the sea. 3 marks
(c) Calculate the horizontal distance from the cliff to where it lands. 2 marks
Q6. A shuttlecock is hit high into the air. Describe and explain how air resistance affects its acceleration during its fall, and explain why it reaches a terminal speed. 4 marks
12In one breath
Position is where you are; displacement is how far you ended up from the start, with a direction; distance is the length of the path, with none. Velocity is the rate of change of position and acceleration the rate of change of velocity, both vectors, both averages over an interval (a chord on a graph) or values at an instant (a tangent). On an s–t graph the gradient is velocity; on a v–t graph the gradient is acceleration and the area is displacement, negative below the axis; on an a–t graph the area is the change in velocity. When a is constant, v = u + at, s = ut + ½at², v² = u² + 2as and s = ((u + v) ÷ 2) t apply: pick the one without the quantity you neither know nor want, fix a positive direction first, and keep the signs. When a changes, only gradients and areas work. A projectile with no air resistance moves at constant velocity horizontally and with constant acceleration g vertically, sharing only the time, so find t from the vertical motion and then x = uₓt, whether launched horizontally, upwards or downwards. Air resistance opposes the velocity, so it lowers and shortens the path, makes it steeper on the way down, and on a falling body grows until it equals the weight, when the acceleration is zero and the body falls at terminal speed.
Answers
Q1. C. The only force on the ball is its weight, at every point of the flight, so its acceleration is g downwards throughout, including the instant it is at rest at the top. C only. A is the classic error, confusing zero velocity with zero acceleration; D contradicts the fact that all bodies fall with the same acceleration without air resistance.
Q2. Distance = 60 + 80 = 140 m, so average speed = 140 ÷ 70 = 2.0 m s⁻¹. Displacement = √(60² + 80²) = 100 m, at tan⁻¹(80 ÷ 60) = 53° west of north. Average velocity = 100 ÷ 70 = 1.4 m s⁻¹, 53° west of north (equivalently 37° north of west). 1 for the average speed, 1 for the 100 m displacement, 1 for the average velocity with a direction. A velocity with no direction loses the last mark.
Q3. (a) a = (v − u) ÷ t = (9.0 − 0) ÷ 3.0 = 3.0 m s⁻². A1 with unit.
(b) s = ((u + v) ÷ 2) t = ((0 + 9.0) ÷ 2) × 3.0 = 13.5 m (or ½ × 3.0 × 3.0² = 13.5 m). M1 for a correct equation with substitution, A1 for 13.5 m.
(c) Remaining distance = 100 − 13.5 = 86.5 m at 9.0 m s⁻¹ takes 86.5 ÷ 9.0 = 9.61 s. Total = 3.0 + 9.61 = 12.6 s. M1 for the constant-speed time for the remaining 86.5 m, A1 for 12.6 s. Using s = ut + ½at² with a = 3.0 m s⁻² over the whole 100 m scores 0: the acceleration was not uniform for the whole race.
Q4. (a) Starting from rest, u = 0, so s = ut + ½at² becomes d = ½at². This has the form y = mx with y = d and x = t², so the graph is a straight line through the origin with gradient ½a. M1 for d = ½at² from u = 0, R1 for matching it to y = mx and identifying gradient a ÷ 2. AG, so the reasoning must be shown.
(b) Gradient = (1.00 − 0.20) ÷ (4.04 − 0.81) = 0.80 ÷ 3.23 = 0.248 m s⁻². a = 2 × gradient = 0.50 m s⁻². M1 for the gradient from the two rows, A1 for a = 0.49 to 0.50 m s⁻². Forgetting to double the gradient scores M1 only.
(c) A systematic error in the timing: for example, the timer was started slightly before the trolley was released, so every time is too long, or the distances were measured from the wrong point on the trolley. any sensible systematic error. "Human error" or "random error" scores 0; a random error scatters points, it does not shift the whole line.
Q5. Components: uₓ = 15.0 cos 40° = 11.5 m s⁻¹ and the vertical component 15.0 sin 40° = 9.64 m s⁻¹ upwards.
(a) At the top the vertical velocity is zero: 0 = 9.64² − 2 × 9.8 × h, so h = 9.64² ÷ 19.6 = 4.74 m above the cliff. Height above the sea = 12.0 + 4.74 = 16.7 m. M1 for v² = u² + 2as with the vertical component, A1 for 16.7 m. Using 15.0 instead of the vertical component scores 0.
(b) Up positive, s = −12.0 m: −12.0 = 9.64t − 4.9t², so 4.9t² − 9.64t − 12.0 = 0, giving t = (9.64 + √(9.64² + 4 × 4.9 × 12.0)) ÷ 9.8 = 2.83 s. M1 for the vertical equation with s = −12.0 m and u = +9.64 m s⁻¹, M1 for solving the quadratic, A1 for 2.83 s. A student who finds the time to the top and the time to fall 16.7 from rest (0.98 + 1.85 s) gains full marks too.
(c) x = uₓ t = 11.5 × 2.83 = 32.5 m (32 to 33 m accepted). M1 for using the horizontal component with the time from (b), A1 for the answer. Allow error carried forward from (b).
Q6. Air resistance acts opposite to the velocity, so on the way down it acts upwards, against the weight. The resultant force is weight minus air resistance, so the acceleration is less than g. As the shuttlecock speeds up, the air resistance increases, so the resultant force and the acceleration decrease. When the air resistance has grown to equal the weight, the resultant force is zero, so the acceleration is zero and the velocity stays constant: that is the terminal speed. 1 for drag opposing the motion and so acting upwards, 1 for acceleration less than g because the resultant force is reduced, 1 for drag increasing with speed so acceleration decreasing, 1 for drag = weight so zero resultant force and zero acceleration at terminal speed. "The forces balance so it stops accelerating" with no mention of drag growing with speed earns 1 of the last 2.
Educerie · written from the published IB Diploma Programme Physics guide, first assessment 2025, section A.1 Kinematics. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.