Describing motion: vectors, velocity and the three graphs that tell one story
By the end of this lesson you will be able to
- Distinguish scalar from vector quantities and use a sign convention for displacement, velocity and acceleration.
- Define velocity and acceleration as rates of change, with SI units, and calculate them from data.
- Interpret displacement–time graphs: the gradient is the velocity, a curve means a changing velocity and a negative gradient means returning.
- Interpret velocity–time graphs: gradient = acceleration, area = displacement, and calculate both with units.
- Convert between displacement–time, velocity–time and acceleration–time graphs of the same motion.
- Process light-gate data to the standard of the on-screen examination: table, graph, gradient with units, conclusion against a hypothesis.
You met this before
- Speed = distance ÷ time, in metres per second (m/s); 1 m/s = 3.6 km/h, so divide km/h by 3.6 to get m/s (Year 1, Year 3).
- On a distance–time graph the gradient is the speed: a steeper line is faster, a flat line is stopped, a curve is a changing speed (Year 1).
- Acceleration = change in speed ÷ time taken, in m/s²; a car going from 0 to 20 m/s in 5 s accelerates at 4 m/s² (Year 1, Year 3).
- On a speed–time graph the gradient is the acceleration and the area under the line is the distance travelled (Year 3).
- A resultant force changes motion: F = ma, so the same force gives a heavier object a smaller acceleration (Year 1 Lesson 4, Year 3).
A minibus travels 1.8 km along a straight road in 90 s. Calculate its average speed in m/s and in km/h, and state what the gradient of its distance–time graph would be.
Reveal
1.8 km = 1 800 m
speed = distance ÷ time
speed = 1 800 m ÷ 90 s = 20 m/s
in km/h: 20 × 3.6 = 72 km/h
The gradient of the distance–time graph is the speed, 20 m/s — the same number, because the gradient is Δdistance ÷ Δtime. The May 2023 Integrated Sciences paper gave one mark just for showing the gradient calculation and a third for the km/h conversion.
Revise it in full: Year 1 · Movement · Lesson 1 — How fast? Distance, time and speed · Year 1 · Movement · Lesson 3 — Speeding up: acceleration and speed–time graphs · Year 3 · Physics · Unit 9 · Lesson 1 — Describing motion: speed, velocity, acceleration and the graphs that show them
The puzzle
Letsile Tebogo won the Olympic 200 m in Paris on 8 August 2024 in 19.46 s. His average speed was 200 ÷ 19.46 = 10.3 m/s. A second runner — you — walks once round a 400 m track in 300 s and finishes exactly where you started.
Your average speed is 1.3 m/s. Your average velocity is zero. Tebogo’s is not zero, but it is only about 6 m/s, even though his speed was 10.3 m/s, because the 200 m race starts on a bend and his finish line is only roughly 120 m from his start in a straight line.
Same runner, same race, three different numbers. What is velocity measuring that speed is not — and why do examiners insist on the difference?
A ball is thrown straight up at 15 m/s, rises, and lands back in the thrower’s hand 3.1 s later. Predict its displacement, its distance travelled and its average velocity for the whole flight — then predict the sign of its acceleration on the way up and on the way down.
Compare your answer
Displacement 0 m (it ends where it began).
Distance ≈ 2 × 11.5 m = 23 m (it rises about 11.5 m and falls back).
Average velocity = displacement ÷ time = 0 m/s.
The acceleration is −9.8 m/s² throughout — downwards on the way up (slowing it) and downwards on the way down (speeding it up). One direction, one sign: that is why acceleration must be a vector. Section 1.2 makes the rule exact.
The science — structured notes
1.1 Scalars and vectors: a size, or a size and a direction
A scalar is fully described by a number and a unit: 20 m, 3.5 s, 60 kg, 500 J. A vector needs a direction as well: 20 m north, 3 m/s to the left, 40 N downwards.
In one dimension the direction is carried by a sign: choose one direction as positive (say, to the right or upwards), and every quantity pointing the other way is negative. The choice is yours; using it consistently is not optional.
| Scalar (size only) | Vector (size and direction) | The link |
|---|---|---|
| Distance, d (m) | Displacement, s (m) | Displacement is the straight-line change of position from start to finish; distance is the length of the path actually followed. Distance ≥ |displacement|. |
| Speed (m/s) | Velocity, v (m/s) | Velocity is the rate of change of displacement; speed is the rate at which distance grows. A car going round a roundabout at a steady 8 m/s has constant speed but a continually changing velocity. |
| Time, mass, energy, temperature, volume | Acceleration, force, weight, momentum | Acceleration is the rate of change of velocity, so it has the direction of the change, not of the motion: a car braking while moving right has an acceleration pointing left. |
A 400 m running track has two straights and two semicircular bends. An athlete runs one full lap in 50.0 s. Calculate her average speed and her average velocity, and explain the difference.
Average speed = distance ÷ time
Average speed = 400 m ÷ 50.0 s
Average speed = 8.00 m/s
Displacement = 0 m (she finishes where she started)
Average velocity = displacement ÷ time
Average velocity = 0 m ÷ 50.0 s = 0 m/s
Speed counts every metre of path; velocity counts only the straight line from start to finish. On the bends her velocity changes direction every instant even though her speed is steady — which is why a body moving in a circle at constant speed is accelerating (a resultant force, from her feet, acts towards the centre).
A delivery drone flies 300 m east, then 400 m north, in 100 s. Calculate its distance, its displacement (with direction), its average speed and its average velocity.
Reveal answer
Distance = 300 m + 400 m = 700 m
Displacement = √(300² + 400²)
Displacement = √250 000 = 500 m
Direction: tan θ = 400 ÷ 300, so θ = 53° north of east
Average speed = 700 m ÷ 100 s = 7.0 m/s
Average velocity = 500 m ÷ 100 s = 5.0 m/s at 53° N of E
A 3-4-5 triangle: the classic examiner set-up.
1.2 Velocity and acceleration as rates of change
Both definitions are “change ÷ time”, and both carry a sign:
v = ΔsΔt (m/s)
a = ΔvΔt = v − ut (m/s²)
Here u is the initial velocity, v the final velocity and t the time taken; Δ (“delta”) means “change in”. The unit of acceleration, m/s² (written m s⁻² in the exam’s formula sheet), means “metres per second, per second”: at 3 m/s² the velocity grows by 3 m/s every second.
A negative acceleration means the velocity is changing in the negative direction. If the object is moving in the positive direction that is a deceleration (slowing down); if it is moving in the negative direction, a negative acceleration makes it go faster.
Never translate “negative” as “slowing” without checking which way the object is moving.
An advert claims a car goes “0 to 100 km/h in 8.0 s”. Calculate its average acceleration in m/s².
Convert first: 100 km/h = 100 ÷ 3.6 = 27.8 m/s (100 000 m ÷ 3 600 s).
a = (v − u) ÷ t
a = (27.8 m/s − 0) ÷ 8.0 s
a = 3.5 m/s² (2 s.f., matching the data)
Substituting 100 straight in — 100 ÷ 8 = 12.5 “km/h per second” — is the kind of mistake the May 2023 physics report describes: values put straight into a formula without considering the units.
“An object with a negative acceleration is always slowing down.”
Reveal answer
False. Take upwards as positive: a ball thrown up has a = −9.8 m/s² and slows; the same ball on the way down still has a = −9.8 m/s² and speeds up, because its velocity is negative and becoming more negative. “Slowing down” means velocity and acceleration have opposite signs; “speeding up” means the same sign.
1.3 Displacement–time graphs: the gradient is the velocity
Plot displacement (vertical) against time (horizontal). The gradient of the line — Δs ÷ Δt — is the velocity. A straight line means constant velocity; the steeper it is, the faster.
A horizontal line means the object is at rest (displacement not changing). A line sloping downwards means the object is moving back towards the start: negative velocity. A curve means the velocity is changing — the object is accelerating — and the velocity at any instant is the gradient of the tangent there. Curving upwards more steeply: speeding up; flattening: slowing down.
A displacement–time graph for a hyena shows a straight line from (40 s, 120 m) to (50 s, 240 m) after a lion appears. Calculate the hyena’s velocity in m/s and in km/h.
Gradient = Δs ÷ Δt
Gradient = (240 − 120) m ÷ (50 − 40) s
Gradient = 120 m ÷ 10 s = 12 m/s
Show the two points and the division — the first mark is for “gradient calculation seen or implied”.
12 m/s × 3.6 = 43.2 km/h (12 m every second is 12 × 3 600 = 43 200 m every hour). The markscheme gave all three marks if 43 or 43.2 was seen.
A student says: “The graph is a curve that gets steeper, so the object is decelerating, and at t = 6 s its velocity is 6 m/s because the displacement there is 36 m.” Identify two errors.
Reveal answer
(1) A displacement–time curve getting steeper has an increasing gradient, so the velocity is increasing — the object is accelerating, not decelerating. (2) The velocity at an instant is the gradient of the tangent at that point, not displacement ÷ time from the origin: 36 ÷ 6 = 6 m/s is the average velocity over the first 6 s.
If s = t² (which gives 36 m at 6 s) the tangent gradient at 6 s is 12 m/s — twice the average, as it always is for motion from rest with constant acceleration.
1.4 Velocity–time graphs: gradient is acceleration, area is displacement
Plot velocity (vertical) against time. Now the gradient, Δv ÷ Δt, is the acceleration: a straight sloping line means constant acceleration, horizontal means constant velocity (zero acceleration), a line sloping down means negative acceleration.
The area between the line and the time axis is the displacement, because velocity × time = displacement and the area of each strip is v × Δt. Area below the axis (negative velocity) counts as negative displacement.
Units check: the area has units of (m/s) × s = m, and the gradient (m/s) ÷ s = m/s².
A cage in a mine shaft descends 600 m. It accelerates uniformly from rest for 10 s, travels at a constant velocity for 50 s, then decelerates uniformly to rest in 10 s. Taking downwards as positive: (1) sketch the velocity–time graph; (2) calculate the constant velocity; (3) calculate the acceleration in each stage.
(1) A trapezium: a line rising from (0, 0) to (10 s, v), horizontal to (60 s, v), falling to (70 s, 0).
(2) The area under the graph is the displacement, 600 m.
Area of a trapezium = ½ × (sum of parallel sides) × height
Area = ½ × (50 s + 70 s) × v
Area = 60v
60v = 600 m
v = 10 m/s downwards
Check by pieces: triangle 5v + rectangle 50v + triangle 5v gives 60v.
(3) Stage 1: a = (10 − 0) m/s ÷ 10 s
a = +1.0 m/s² (downwards)
Stage 2: a = 0
Stage 3: a = (0 − 10) m/s ÷ 10 s
a = −1.0 m/s², i.e. 1.0 m/s² upwards
The upward acceleration in stage 3 is provided by the extra tension in the hoist rope. A person in the cage feels heavier during stage 3, not stage 1.
A cyclist’s velocity–time graph rises from 0 to 6 m/s in 5 s, stays at 6 m/s for 20 s, then falls to 0 in 3 s. Calculate the total displacement and the acceleration in the last stage.
Reveal answer
Triangle: ½ × 5 s × 6 m/s = 15 m
Rectangle: 20 s × 6 m/s = 120 m
Triangle: ½ × 3 s × 6 m/s = 9 m
Total displacement = 15 + 120 + 9 = 144 m
Last stage: a = (0 − 6) m/s ÷ 3 s
a = −2 m/s²
(Write “−2 m/s²” or “2 m/s² deceleration”; “−2 m/s² deceleration” says the same thing twice and confuses the sign.)
The cyclist’s velocity–time graph continues: after stopping, the velocity goes to −3 m/s for 10 s. What does this represent, and what is the cyclist’s displacement from the start now?
Reveal answer
A negative velocity means the cyclist is moving back the way she came at 3 m/s. The area below the axis is −3 m/s × 10 s = −30 m.
Displacement = 144 m − 30 m = 114 m from the start
Distance travelled = 144 m + 30 m = 174 m
Area below the axis subtracts from displacement but adds to distance.
1.5 Acceleration–time graphs, and moving between all three
An acceleration–time graph for uniformly accelerated motion is a set of horizontal steps: a positive block while speeding up, zero while steady, a negative block while slowing. Its area (a × Δt) is the change in velocity.
The three graphs are one story told three ways, and the exam expects you to translate freely: going “down” the chain (s → v → a) you take gradients; going “up” (a → v → s) you take areas.
| From | To | Do this | Check the units |
|---|---|---|---|
| displacement–time | velocity–time | gradient at each point (tangent on a curve) | m ÷ s = m/s |
| velocity–time | acceleration–time | gradient of each section | (m/s) ÷ s = m/s² |
| acceleration–time | velocity–time | area under each block = change in velocity; add to the starting velocity | (m/s²) × s = m/s |
| velocity–time | displacement–time | area up to each time = displacement so far (a straight v–t line gives a curved s–t line) | (m/s) × s = m |
Reading the picture. While the acceleration is +2 m/s² (0–4 s) the velocity line rises with gradient 2 and the displacement curve bends upwards — s = t², so 16 m at 4 s.
While the acceleration is zero (4–10 s) the velocity is constant at 8 m/s and the displacement is a straight line rising 48 m to 64 m. While the acceleration is −2 m/s² (10–14 s) the velocity falls to zero and the displacement curve flattens at 80 m — the whole area of the trapezium.
Cover one panel and reconstruct it from the other two: that is the exam skill.
A lift’s acceleration–time graph shows +1.5 m/s² for 2 s, 0 for 6 s, then −1.5 m/s² for 2 s. Explain, with calculations, what its velocity–time graph and its displacement look like.
Reveal answer
Area of the first block = 1.5 × 2 = 3 m/s, so the velocity rises in a straight line from 0 to 3 m/s in 2 s; it stays at 3 m/s for 6 s (zero acceleration); the last block’s area is −3 m/s, so the velocity falls in a straight line to 0 at 10 s — a trapezium.
Displacement = area of the trapezium
Displacement = ½ × (6 s + 10 s) × 3 m/s
Displacement = 24 m
That is about eight storeys. The displacement–time graph curves up, runs straight, then flattens.
1.6 Measuring motion without a stopwatch
A hand-held stopwatch adds about 0.2 s of reaction time to every reading — a random error that matters for times under a few seconds and that examiners expect you to name and remove. Light gates record the time a card of known length takes to pass a beam and divide to give the speed at that point; two gates give two speeds and, with the distance between them, an acceleration.
A motion sensor (ultrasound) or a phone’s video at 30 or 240 frames per second gives position every frame, from which software draws all three graphs. A ticker timer marks a tape 50 times a second: the spacing of the dots is the velocity, the change in spacing the acceleration.
Whichever you use, the exam wants the instrument named against the variable it measures (“time between gates, light gates and datalogger, to 0.001 s”).
A 5.0 cm card on a trolley blocks light gate 1 for 0.050 s and light gate 2 for 0.025 s; the gates are 0.60 m apart and the timer records 0.40 s between the two gates. Calculate the two speeds and the acceleration, and check the answer against the distance.
Reveal answer
v₁ = card length ÷ blocking time
v₁ = 0.050 m ÷ 0.050 s = 1.0 m/s
v₂ = 0.050 m ÷ 0.025 s = 2.0 m/s
a = (v₂ − v₁) ÷ t
a = (2.0 − 1.0) m/s ÷ 0.40 s
a = 2.5 m/s²
Check: distance = average velocity × time.
distance = ½(1.0 + 2.0) m/s × 0.40 s
distance = 0.60 m
That is the gate spacing, so the data are consistent. Lesson 2 gives a second route that needs no time at all:
v² = u² + 2as
a = (v² − u²) ÷ 2s
a = (2.0² − 1.0²) ÷ (2 × 0.60 m)
a = 2.5 m/s²
A displacement–time graph is the curve s = 2t² for 0 ≤ t ≤ 5 s. Without calculus, find the velocity at t = 3 s by taking the average velocity over shorter and shorter intervals centred on 3 s (from 2 to 4 s, then 2.5 to 3.5 s, then 2.9 to 3.1 s), and state the rule you discover.
Reveal answer
2–4 s: (32 − 8) m ÷ 2 s = 12 m/s
2.5–3.5 s: (24.5 − 12.5) m ÷ 1 s = 12 m/s
2.9–3.1 s: (19.22 − 16.82) m ÷ 0.2 s = 12 m/s
Every symmetric interval gives 12 m/s, so the tangent gradient at 3 s is 12 m/s. Rule: for s = kt² the velocity is 2kt — the gradient of the tangent grows in proportion to t, which is exactly “constant acceleration” (here a = 2k, which is 4 m/s²).
Also notice: for constant acceleration, the average velocity over any interval equals the instantaneous velocity at its midpoint — the fact behind s = ½(u + v)t in Lesson 2.
Try it — a journey drawn in real time: from the walk to the graph and back
The lab shows a walker going out along a 20 m path, pausing, and coming back, while the distance–time graph draws itself underneath. You set the walking speed (the independent variable), the time walked out and the pause.
The dependent variable the lab records is the total distance walked, but the real work is reading the graph: the gradient of each section is the velocity of that section, with the return leg negative if you treat “away from the start” as positive.
Common mistakes to avoid
- Velocity without a direction. “15 m/s” is a speed; “15 m/s north” or “+15 m/s” is a velocity.
- Distance ÷ time from the origin as the velocity at an instant. On a curve, the velocity at an instant is the gradient of the tangent there.
- Reading a displacement–time graph as a picture of the path. A rising line on an s–t graph is not “going uphill”; it is moving away from the start.
- Mixing up the two graphs. Gradient of s–t is velocity; gradient of v–t is acceleration; area under v–t is displacement; area under s–t means nothing.
- “Negative acceleration = slowing down”. Only if the velocity is positive. Compare the signs of v and a.
- km/h substituted into an SI formula. Convert to m/s first (÷ 3.6); the examiners’ reports name this every session.
- Units written as m/s^2. The markscheme wants a real superscript: m/s² or m s⁻².
- “Proportional” for any straight line. Proportional means straight and through the origin; test it with a ratio or a doubling.
Study smart: how this is examined
Study routine. Today: draw the three-graph diagram of Section 1.5 from memory for a different motion (a = +3 m/s² for 2 s, 0 for 4 s, −3 m/s² for 2 s) and check the numbers.
In two days: do the mine-hoist and drone EXAMPLES with the book closed, then explain aloud why average velocity round a lap is zero. In a week: run the lab as a full investigation, convert its distance–time graph into a velocity–time graph by hand, and take the quiz.
How we found out — graphs before Galileo, and a horse that settled a bet
The idea of drawing motion as a picture is older than the telescope. In the 1330s a group of scholars at Merton College, Oxford — William Heytesbury and his colleagues, the “Merton calculators” — proved the mean speed theorem: a body accelerating uniformly from rest covers the same distance as one moving steadily at half its final speed.
Around 1350 Nicole Oresme, in Paris, drew it: time along a line, speed as vertical bars, and the distance as the area under them — in effect a velocity–time graph, almost three hundred years before Descartes published his coordinate geometry (1637).
Galileo Galilei made the theorem physical. Unable to time free fall with the instruments of 1604, he rolled bronze balls down a grooved inclined plane and timed them with a water clock, weighing the water that flowed; the distances grew as 1, 4, 9, 16 — the squares of the times — which is exactly the curve of Section 1.5, and he published the result in Two New Sciences in 1638.
Isaac Newton, born the year Galileo died, turned “rate of change” into a mathematics — the calculus of the 1660s — in which the gradient of a tangent and the area under a curve are inverse operations, the fact your three graphs rely on.
Seeing motion frame by frame came much later. In 1878 the photographer Eadweard Muybridge, financed by Leland Stanford, set a row of cameras beside a track in California and had a galloping horse trip their shutters in turn; the sequence proved that all four hooves leave the ground at once — and gave one of the first displacement–time records at a fraction of a second.
Étienne-Jules Marey in Paris put many exposures on one plate in 1882 (chronophotography), and in the 1930s Harold Edgerton at MIT froze bullets and milk drops with a strobe flashing thousands of times a second.
The strobe photograph of a falling ball, with its widening gaps, is Galileo’s inclined plane made visible; the phone in your pocket, filming at 240 frames a second, is Muybridge’s row of cameras.
Galileo’s water clock measured time as a mass of water collected. Explain why he could still show that distance was proportional to time squared, and why he could not have measured g itself.
Reveal answer
Proportionality needs only a consistent scale: if a steady flow delivers the same mass per second, then mass ∝ time, and distance ∝ (mass)² shows distance ∝ t² without knowing how many seconds a gram represents.
To get g in m/s² he would have needed the time in seconds — a calibration his clock lacked — and a free fall too fast for it; the May 2021 physics paper asked exactly why a water clock made results hard to share (no standard unit, every clock different).
Where it is used now
- Athletics and football analysis. The GPS vests worn under football shirts record position ten or more times a second and turn it into velocity–time graphs for every player — total distance, number of sprints, top speed, decelerations that strain hamstrings. A 200 m sprinter’s velocity–time curve rises steeply for the first few seconds and then levels off; coaches read the gradient of those first seconds as closely as the finish time.
- Speed cameras. A fixed camera measures speed at one instant; an average-speed system reads a plate at two points and divides the distance by the time — a gradient on a distance–time graph over 2 km, which no burst of braking can fool.
- Aircraft flight recorders. The “black box” stores velocity, acceleration and position many times a second; investigators reconstruct the three graphs of Section 1.5 for the last minutes of a flight.
- Earthquake early warning. Sensors near the epicentre time the fast P-wave and the slower, damaging S-wave; the time gap and the two speeds give the distance and the seconds of warning — the May 2021 Integrated paper built a question on it.
- Self-driving cars. Radar measures the velocity of every object around the car directly from the Doppler shift of the reflected wave; the software then integrates to predict where each object will be in the next two seconds — area under a v–t graph, computed fifty times a second.
- Your phone. Its accelerometer reads acceleration many times a second; the step counter counts the peaks in that graph, one per footfall, and estimates distance from the number of steps.
Science now — and the scientist you need to be
What is happening in 2026
Motion is now measured everywhere, all the time, and the skill in demand is reading the graphs the sensors produce. Satellite positioning receivers in phones and cars fix position to a few metres several times a second, and the newer dual-frequency receivers bring that to well under a metre — good enough that the velocity–time graphs of delivery bikes and buses are used to plan cities.
In elite sport, computer-vision systems track every athlete from broadcast video without any sensor on the body, producing displacement and velocity data for every second of a match; the same techniques, applied to a phone video of a school sprint, give a student a velocity–time graph within minutes.
Vehicle safety has gone the same way: the automatic emergency braking systems fitted to new cars sold in the European Union since July 2024 continuously estimate the velocity and deceleration of the car ahead from radar and camera data, and act when the predicted stopping distance is too short — Lesson 4 does that calculation by hand.
Careers built on these ideas run from sports science and biomechanics through traffic engineering and crash investigation to the software teams that turn raw accelerometer data into a graph a coach or an engineer can trust. The physics in every one of them is Section 1.5.
The mindshift
What you need to be able to do
Given any one of the three graphs, sketch the other two with numbers on the axes. Calculate a gradient with two labelled points and give it a unit. Find an area in pieces and give it a unit. Convert km/h to m/s before touching a formula. And when a question says “describe the motion”, write times, values and the words accelerating, constant velocity, decelerating and at rest — not “it goes up then down”.
See an example
A 10 m walk out in 8 s (gradient +1.25 m/s), a 3 s pause (0), and a 10 m return in 6 s (gradient −1.7 m/s); the assistant gave the lap example correctly but described a car “accelerating at −2 m/s” — the wrong unit.
Key terms
| Scalar | A quantity with size only: distance, speed, time, mass, energy. |
| Vector | A quantity with size and direction: displacement, velocity, acceleration, force; in one dimension the direction is a sign. |
| Displacement, s | Straight-line change of position from start to finish, in m, with direction. |
| Velocity, v | Rate of change of displacement, v = Δs ÷ Δt, in m/s; the gradient of a displacement–time graph. |
| Acceleration, a | Rate of change of velocity, a = (v − u) ÷ t, in m/s²; the gradient of a velocity–time graph. |
| Deceleration | An acceleration opposite to the velocity, so the object slows. |
| Tangent | The straight line touching a curve at a point; its gradient is the instantaneous rate of change there. |
| Area under a v–t graph | The displacement; area below the time axis counts as negative. |
| Uniform acceleration | Constant acceleration: a straight v–t line, a curved s–t line, a flat a–t line. |
| Light gate | A beam and sensor that times how long a card takes to pass, giving speed without reaction time. |
Think like a scientist
Criterion C A group placed a motion sensor at the top of a straight ramp and let a trolley run from rest. The sensor logged the trolley’s velocity every 0.5 s. Their hypothesis was: “If the time since release doubles, the velocity will double, because the trolley has a constant acceleration down the ramp.” Three runs were logged:
| Time / s | Velocity run 1 / m/s | Velocity run 2 / m/s | Velocity run 3 / m/s | Mean velocity / m/s |
|---|---|---|---|---|
| 0.5 | 0.31 | 0.29 | 0.30 | |
| 1.0 | 0.60 | 0.62 | 0.58 | |
| 1.5 | 0.89 | 0.91 | 0.90 | |
| 2.0 | 1.21 | 1.19 | 1.52 | |
| 2.5 | 1.50 | 1.48 | 1.51 |
(a) Identify the anomalous reading and suggest a specific cause. [2]
(b) Complete the mean column, excluding the anomaly and rounding appropriately. [2]
(c) Describe the graph you would draw: axes, labels, scale, points and line. [3]
(d) Calculate the gradient of the line and state what it represents, with its unit. [3]
(e) Explain, with a calculation, whether the data support the hypothesis, and whether the relationship is proportional or merely linear. [3]
(f) Use the graph to determine the displacement of the trolley in the first 2.0 s. [2]
(g) State one source of random error and one of systematic error, the effect of each on the graph, and an improvement for each. [4]
Model answer
(a) Run 3 at 2.0 s, 1.52 m/s — far above the other two (1.21, 1.19) and above the 2.5 s readings, breaking the trend [1]. Cause: the sensor picked up a reflection from a hand or the trolley’s edge at that instant, or the trolley was nudged at release in run 3 — a specific event, not “a mistake” [1].
(b) Means: 0.30, 0.60, 0.90, 1.20 (mean of 1.21 and 1.19, anomaly excluded), 1.50 m/s — 2 d.p. to match the sensor [1 for excluding the anomaly, 1 for the rest].
(c) Time (s) on the horizontal axis as the independent variable, mean velocity (m/s) on the vertical [1].
Both axes labelled with quantity and unit, even scales starting at zero (e.g. 0.5 s and 0.2 m/s per large square) filling more than half the grid, not the data values used as the scale [1]; five points as crosses with a ruler line of best fit through the origin — the points lie on a straight line [1].
(d) Using two far-apart points on the line, e.g. (0, 0) and (2.5, 1.50): gradient = (1.50 − 0) ÷ (2.5 − 0) = 0.60 [1]; unit m/s² [1]; it is the acceleration of the trolley down the ramp — the gradient of a velocity–time graph is the acceleration [1].
(e) Doubling the time from 1.0 to 2.0 s doubles the velocity from 0.60 to 1.20 m/s, and v ÷ t = 0.60 m/s² at every row [1]; the graph is a straight line through the origin, so velocity is directly proportional to time — not merely linear — and the hypothesis is supported [1]; the constant of proportionality, 0.60 m/s², is the constant acceleration the hypothesis predicted [1].
(f) Displacement = area under the line from 0 to 2.0 s
Displacement = ½ × 2.0 s × 1.20 m/s [1]
Displacement = 1.2 m [1]
(g) Random: the sensor’s readings fluctuate by about ±0.02 m/s from run to run (0.31, 0.29, 0.30), scattering points either side of the line; reduced by the mean of three and by a higher sampling rate [2].
Systematic: if the trolley was already moving when the log started (released a fraction before t = 0), every velocity is high by the same amount and the line has a positive intercept instead of passing through the origin — the gradient is unaffected.
Improvement: start the log with the same trigger that releases the trolley, or fit the line and read the intercept [2]. (Vague answers such as “mistakes were made” or “do more repeats” score 0.)
Probe deeper — exam-style questions
Attempt every question before opening the answer. Marks in brackets show what a full answer is worth.
Select the vector quantities from the list — distance · displacement · speed · velocity · mass · acceleration · time · force — and state what distinguishes a vector from a scalar.
Reveal answer
Displacement, velocity, acceleration, force [1 — all four and no others]; a vector has a direction as well as a size, a scalar has size only [1].
A train travelling at 108 km/h slows uniformly to 36 km/h in 40 s. Calculate its acceleration in m/s².
Reveal answer
108 km/h ÷ 3.6 = 30 m/s and 36 km/h ÷ 3.6 = 10 m/s [1]
a = (v − u) ÷ t
a = (10 − 30) m/s ÷ 40 s [1]
a = −0.50 m/s² [1]
That is a deceleration of 0.50 m/s²; the unit is required.
A displacement–time graph shows a straight line from (0 s, 0 m) to (12 s, 84 m), then a horizontal line to 20 s, then a straight line down to (30 s, 0 m). Describe the motion with values and calculate the velocity in the final stage.
Reveal answer
0–12 s: constant velocity of 84 ÷ 12 = 7.0 m/s away from the start [1]; 12–20 s: at rest, 84 m from the start [1]; 20–30 s: returning to the start with velocity (0 − 84) ÷ 10 = −8.4 m/s, i.e. 8.4 m/s back towards the start [1].
A velocity–time graph for a motorbike shows a straight line from (0, 0) to (6 s, 18 m/s), a horizontal line to 16 s, then a straight line to (20 s, 0). Calculate the acceleration in the first stage, the displacement for the whole journey and the average velocity.
Reveal answer
a = Δv ÷ Δt = 18 m/s ÷ 6 s
a = 3.0 m/s² [1]
Area 1: ½ × 6 s × 18 m/s = 54 m
Area 2: 10 s × 18 m/s = 180 m
Area 3: ½ × 4 s × 18 m/s = 36 m [1]
Total displacement = 54 + 180 + 36 = 270 m [1]
Average velocity = 270 m ÷ 20 s
Average velocity = 13.5 m/s [1]
True or false, with a reason: “If the area under a velocity–time graph is zero, the object has not moved.”
Reveal answer
False [1]: zero area means zero displacement — equal areas above and below the axis — so the object may have travelled out and back, covering a large distance and finishing where it started [1].
A ball is dropped and the velocity–time graph is drawn with downwards positive: a straight line of gradient 9.8 m/s². Sketch and describe how the graph changes if upwards is chosen as positive instead, and state what stays the same.
Reveal answer
The line now slopes downwards from the origin with gradient −9.8 m/s² and the velocities are negative [1]; the acceleration is still 9.8 m/s² in size, directed downwards — only the sign convention changed [1]; the area (displacement) has the same size and the distance fallen is unchanged; the physics does not depend on the choice of positive direction, only the signs do [1].
7. Two cars on a straight road: car A passes a marker at a constant 20 m/s; at that instant car B starts from rest at the same marker with a constant acceleration of 2.5 m/s². Using velocity–time graphs on the same axes, find when B catches A and how far from the marker they are — then explain why it is not when their velocities are equal.
Reveal answer
A: horizontal line at 20 m/s, so the area up to time t is 20t
B: line of gradient 2.5 m/s², so the area is ½ × 2.5 × t²
Area for B = 1.25t² [1]
Catch-up when the areas are equal: 20t = 1.25t²
t = 20 ÷ 1.25 = 16 s [1]
distance = 20 m/s × 16 s
distance = 320 m [1]
The velocities are equal at t = 8 s (B reaches 20 m/s), but at that moment B has covered only 80 m to A’s 160 m — B has been slower for the whole of the first 8 s, and needs a further 8 s of being faster to make up the 80 m gap; equal velocity is the moment the gap stops growing, not the moment it closes [1].
8. A phone’s accelerometer, taped to a lift, logs: +1.2 m/s² for 2.0 s, 0 for 5.0 s, −1.2 m/s² for 2.0 s, then 0. Construct the velocity–time graph, calculate the height the lift rose, and explain why the same log could not tell you whether the lift went up or down without knowing which direction the phone called positive.
Reveal answer
Change in velocity = area of first block = 1.2 m/s² × 2.0 s
Top velocity = 2.4 m/s
So the velocity rises to 2.4 m/s in 2.0 s, stays at 2.4 m/s for 5.0 s and falls to 0 in 2.0 s [1].
height = area of the trapezium
height = ½ × (5.0 s + 9.0 s) × 2.4 m/s
height = 16.8 m [1] (about five storeys)
The accelerometer reports a sign relative to its own axis; a lift going down with the same log would show exactly the same pattern if the phone’s positive axis pointed down [1] — the log fixes the size of every change but the direction is a convention, and a real phone resolves it by also measuring the constant 9.8 m/s² of gravity to find which way is down [1].
Summary
- Scalars have size; vectors have size and direction, carried by a sign in one dimension. Displacement is the straight-line change of position; velocity is its rate of change; acceleration is the rate of change of velocity, with the direction of the change.
- v = Δs ÷ Δt (m/s); a = (v − u) ÷ t (m/s²). Convert km/h to m/s (÷ 3.6) before using either. “Slowing down” means v and a have opposite signs.
- Displacement–time: the gradient is the velocity; flat means at rest; a negative gradient means returning; a curve means accelerating, with the velocity from the tangent.
- Velocity–time: gradient = acceleration; area = displacement (negative below the axis); trapezium areas in pieces, with units.
- Acceleration–time: blocks whose areas are changes in velocity. Down the chain take gradients, up the chain take areas; the three graphs are one motion.
- Measure short times with light gates, sensors or video, not a stopwatch; name the instrument against its variable in every plan.
Next time → The equations of motion: four equations for uniform acceleration, how to choose one and rearrange it, and free fall at g = 9.8 m/s² — from a dropped stone to a measurement of g in the lab.