Describing motion: speed, velocity, acceleration and the graphs that show them
By the end of this lesson you will be able to
- Distinguish distance from displacement and speed from velocity, and state the SI unit of each.
- Apply the steady-speed model (s = vt) and convert between km/h and m/s.
- Define acceleration as the rate of change of velocity, and calculate it from a change in velocity and a time.
- Apply the constant-acceleration model — v = u + at and s = ut + ½at² — to a car, a sprinter and a falling object.
- Interpret speed–time and velocity–time graphs: gradient as acceleration, area as distance.
- Process a set of ticker-tape data into a velocity–time graph and evaluate the method (Criterion C).
The puzzle
Letsile Tebogo runs 100 m in 9.86 s. Divide one by the other and you get 10.1 m/s — but at no moment in the race is he actually moving at 10.1 m/s. For the first second he is far slower, and in the middle of the race he is far faster. A single number cannot describe a race in which the speed is never the same twice. What would? And how would you get it from a stopwatch, a strip of paper tape, or a photo-finish camera?
Sketch, in your head, how the speed of a sprinter changes over 100 m: where is it rising fastest, where is it highest, and does it fall before the line? Then predict what a graph of speed against time would look like for a Gaborone combi that pulls away from a stop, cruises down Nelson Mandela Drive and brakes at the next stop.
Compare your answer
The sprinter’s speed rises steeply out of the blocks, climbs more slowly through 30–60 m, peaks at about 60 m and then falls slightly to the line — the fastest athletes in the world cannot hold top speed for more than three or four seconds. The combi’s graph is a trapezium: a sloping line up as it accelerates, a flat top while it cruises, a steeper line down as it brakes. In both cases the slope of the graph is the acceleration and the area under it is the distance covered. This lesson turns those two sentences into tools.
The science — structured notes
1.1 The kinematic variables — and why direction matters
Kinematics is the description of motion without asking what causes it (that is Lesson 9.2). It needs only five quantities, and the first job is to be precise about each.
| Quantity | Symbol | SI unit | What it is | Scalar or vector? |
|---|---|---|---|---|
| Distance | d | m | How far you have travelled along your path | Scalar (size only) |
| Displacement | s | m | How far you are from where you started, and in which direction | Vector (size and direction) |
| Speed | v | m/s | Distance covered per second | Scalar |
| Velocity | v | m/s | Displacement per second: speed in a stated direction | Vector |
| Acceleration | a | m/s² | Change in velocity per second | Vector |
| Time | t | s | The clock reading | Scalar |
A vector has a direction as well as a size; a scalar has only a size. The distinction is not fussiness. Drive from Gaborone to Francistown on the A1 and you cover about 430 km of road — the distance — but your displacement is about 400 km to the north-east, because the road bends. Walk once round the athletics track and your distance is 400 m while your displacement is zero: you are back where you began. A velocity of 10 m/s north and a velocity of 10 m/s south are the same speed but opposite velocities, which is why a ball thrown upwards and caught has an average speed but an average velocity of zero.
A netball player runs 12 m across the court to the right, then 5 m back to the left, in a total of 4.0 s.
(a) State her distance and her displacement.
(b) Calculate her average speed and her average velocity.
(c) Explain why the two answers to (b) differ.
Reveal answer
(a) Distance 17 m; displacement 7 m to the right (12 − 5).
(b) Average speed = 17 ÷ 4.0 = 4.3 m/s (to two significant figures); average velocity = 7 ÷ 4.0 = 1.8 m/s to the right.
(c) Speed counts every metre of path; velocity counts only the net change of position, and the 5 m back cancels 5 m of the run out. Velocity carries a direction, so the answer must say “to the right”.
1.2 The steady-speed model: s = vt
The simplest model of motion says the velocity does not change. Then distance is speed multiplied by time, and the three forms of the equation are one relationship written three ways:
s = v t v = st t = sv
The model is exact for light in a vacuum and for a satellite’s speed round a circular orbit; it is a good approximation for a combi on the open road; it is useless for the first two seconds of a sprint. Knowing when a model applies is as important as knowing the model.
A combi leaves Gaborone and holds a steady 100 km/h on the A1. How far does it travel in 12 minutes, and how long does it take to cover the 47 km to the Lobatse turn-off?
First convert to SI units. 100 km/h = 100 000 m in 3 600 s, so v = 100 0003 600 = 27.8 m/s. (The quick rule: divide km/h by 3.6.)
12 minutes = 720 s. s = v t = 27.8 × 720 = 20 000 m = 20 km.
47 km = 47 000 m. t = sv = 47 00027.8 = 1 690 s = 28 minutes.
Check the size: 100 km/h is 50 km in half an hour, so 47 km in 28 minutes is right. Always check.
Light travels at 300 000 km/s. Convert this to m/s in standard form, then find how long light takes to travel from the Sun to the Earth, 150 million km away.
Reveal answer
300 000 km/s = 300 000 000 m/s = 3.0 × 10⁸ m/s. Distance = 150 000 000 km = 1.5 × 10¹¹ m. t = 1.5 × 10¹¹3.0 × 10⁸ = 500 s = 8.3 minutes. When you look at the Sun you see it as it was eight minutes ago — a fact Lesson 9.4 turns into a way of stating distances.
1.3 Acceleration — how fast the velocity changes
Acceleration is the change in velocity divided by the time taken for it. If a combi pulls away from rest and reaches 12 m/s in 8 s, its velocity changes by 12 m/s in 8 s: 1.5 m/s every second, written 1.5 m/s².
a = Δvt = v − ut
Here u is the velocity at the start, v the velocity at the end, and Δ (“delta”) means “change in”. The unit m/s² reads “metres per second, per second”. A negative acceleration means the velocity is falling — a deceleration — or that it is growing in the negative direction; the sign convention decides which.
| Motion | Typical acceleration / m/s² | What it feels like |
|---|---|---|
| Combi pulling away from a stop | 1 to 2 | A gentle push back into the seat |
| Family car, hard acceleration | 3 to 4 | A firm push |
| Sprinter leaving the blocks (first 0.2 s) | 8 to 9 | The hardest push a human can produce horizontally |
| Any object in free fall near the Earth | 9.8 | Weightlessness — nothing pushes you at all |
| Car braking hard on a dry road | −8 to −9 | Thrown against the belt |
| Falcon 9 rocket just before engine cut-off | about 30 | Three times your weight pressing you into the seat |
A combi accelerates from rest to 12 m/s in 8.0 s, cruises, then brakes from 12 m/s to rest in 4.0 s. Find both accelerations.
Pulling away: a = v − ut = 12 − 08.0 = 1.5 m/s², forwards.
Braking: a = 0 − 124.0 = −3.0 m/s². The minus sign says the velocity is decreasing: a deceleration of 3.0 m/s², twice the size of the acceleration, which is why passengers lurch when it stops but not when it starts.
“A ball thrown straight up has zero acceleration at the top of its flight, because it is momentarily at rest.”
Reveal answer
False. Its velocity is zero for an instant, but its velocity is changing the whole time — upwards and slowing, then downwards and speeding up — at 9.8 m/s² downwards throughout. Zero velocity and zero acceleration are different things; if the acceleration were zero at the top, the ball would stay there.
1.4 The constant-acceleration model
If the acceleration stays the same — a falling stone, a trolley on a ramp, a car with the pedal held steady — two equations describe everything. The first is the acceleration equation rearranged; the second says how far you go.
A stone is dropped from rest and takes 2.0 s to reach the water (a = 9.8 m/s², air resistance ignored). Find its velocity as it hits, and the depth of the well.
v = u + at = 0 + 9.8 × 2.0 = 19.6 m/s downwards — about 70 km/h.
s = ut + ½at² = 0 + ½ × 9.8 × (2.0)² = ½ × 9.8 × 4.0 = 19.6 m.
The two answers are the same number by coincidence of the 2.0 s; do not expect that in general. Note the square: a 4.0 s fall would be four times as deep, 78 m, not twice.
How far does the combi travel while accelerating from rest at 1.5 m/s² for 8.0 s, and while braking at 3.0 m/s² from 12 m/s to rest in 4.0 s?
Accelerating: s = ut + ½at² = 0 + ½ × 1.5 × 8.0² = ½ × 1.5 × 64 = 48 m.
Braking: u = 12 m/s, a = −3.0 m/s², t = 4.0 s. s = 12 × 4.0 + ½ × (−3.0) × 4.0² = 48 − 24 = 24 m. The negative term is the distance the braking “takes back” from the 48 m it would have covered at 12 m/s.
Same 12 m/s change, half the time, half the distance: 24 m. The next section shows both answers as areas on a graph.
A trolley is released from rest on a ramp and accelerates at 0.80 m/s².
(a) Calculate its velocity after 1.5 s.
(b) Calculate how far it has travelled in that time.
(c) State which model — steady speed or constant acceleration — a moving walkway at an airport fits, and why.
Reveal answer
(a) v = u + at = 0 + 0.80 × 1.5 = 1.2 m/s down the ramp.
(b) s = ut + ½at² = 0 + ½ × 0.80 × 1.5² = 0.40 × 2.25 = 0.90 m.
(c) Steady speed: the walkway moves at a constant velocity, so its acceleration is zero and s = vt describes it exactly — a passenger standing still on it covers equal distances in equal times.
1.5 Speed–time and velocity–time graphs
A graph of velocity against time shows a whole journey at once, and it does two calculations for you.
| Section of the combi journey | Time / s | Velocity / m/s | Shape on the graph | Gradient = a / m/s² | Area = s / m |
|---|---|---|---|---|---|
| Pulling away | 0 → 8 | 0 → 12 | Sloping line up | 12 ÷ 8 = 1.5 | ½ × 8 × 12 = 48 |
| Cruising | 8 → 28 | 12 | Flat line | 0 | 20 × 12 = 240 |
| Braking | 28 → 32 | 12 → 0 | Sloping line down | −12 ÷ 4 = −3.0 | ½ × 4 × 12 = 24 |
| Whole journey | 0 → 32 | Trapezium | 312 |
Use the table to find the combi’s average velocity for the whole 32 s, and explain why it is less than 12 m/s.
Total displacement = 48 + 240 + 24 = 312 m. Average velocity = 31232 = 9.8 m/s.
It spent 12 of its 32 seconds below cruising speed, so the average is pulled below 12 m/s. On the graph, the trapezium’s area is smaller than the 32 × 12 = 384 m rectangle by the two missing triangles at the ends.
Two students look at the same velocity–time graph of a lift. Amantle says the flat section at 2 m/s means the lift is stationary; Kagiso says it means the lift is moving. Explain who is right and how the area and gradient of that section support the answer.
Reveal answer
Kagiso. A flat line on a velocity–time graph means the velocity is constant, not zero; at 2 m/s the lift is moving steadily. The gradient of the flat section is zero, so the acceleration is zero — no change in velocity, which is exactly what steady motion means. The area under the section is 2 m/s × time, which is a real displacement: after 5 s of the flat section the lift has risen 10 m. Amantle has confused it with a distance–time graph, where a flat line would mean no change in position.
1.6 A race as data — Letsile Tebogo over 100 m
Tebogo’s 100 m best is 9.86 s. Timing companies split a race into 10 m sections with cameras at each line; the split times below are modelled on a race of that time — they are typical of an elite 9.8 s race rather than an official record of one — and the section speed is 10 m divided by the section time.
| Section / m | Time for the section / s | Clock at end of section / s | Section speed / m/s |
|---|---|---|---|
| 0–10 | 1.83 | 1.83 | 5.5 |
| 10–20 | 1.03 | 2.86 | 9.7 |
| 20–30 | 0.92 | 3.78 | 10.9 |
| 30–40 | 0.87 | 4.65 | 11.5 |
| 40–50 | 0.85 | 5.50 | 11.8 |
| 50–60 | 0.84 | 6.34 | 11.9 |
| 60–70 | 0.85 | 7.19 | 11.8 |
| 70–80 | 0.86 | 8.05 | 11.6 |
| 80–90 | 0.89 | 8.94 | 11.2 |
| 90–100 | 0.92 | 9.86 | 10.9 |
Read the pattern, quoting the numbers. The speed rises from 5.5 m/s in the first section to 11.9 m/s in the sixth — most of the gain, 5.5 to 10.9 m/s, in the first 30 m — and then falls to 10.9 m/s in the last. The first section includes about 0.15 s of reaction time before he moves at all. Top speed, 11.9 m/s, is 43 km/h — faster than the 40 km/h limit past a school gate. The average of 10.1 m/s is reached at no point in the first 20 m and is exceeded everywhere after it.
A student writes: “Tebogo’s acceleration in the first 10 m is 5.5 ÷ 1.83 = 3.0 m/s². His acceleration between 50 and 60 m is 11.9 m/s².”
Reveal answer
Two errors. First, 5.5 m/s is the average speed over the first section, not the velocity at its end; the velocity at 10 m is higher (about 8 m/s), so the acceleration is larger than 3.0 m/s² — closer to 4.5 m/s² averaged over the section, and higher still in the first 0.2 s. Second, 11.9 is a speed in m/s, not an acceleration: between 50 and 60 m the speed barely changes (11.8 → 11.9 → 11.8 m/s), so the acceleration there is close to zero. Speed and acceleration have different units, and the unit is the giveaway.
Try it — two labs: the graph that draws itself, and the speed trap
First, the walker: watch a distance–time graph draw itself as a walker goes out, pauses and returns. Read the gradient of the outward section and check it equals the speed you set; notice the flat section. Then the speed trap: a battery car passes two light gates, and you change the distance between them. If the car keeps a steady speed, distance ÷ time gives the same number every row — the steady-speed model, tested. Change the motor setting and find the new speed from the gradient of a distance–time plot.
Suppose your speed-trap results at motor setting 3 were: 20 cm → 0.44 s, 40 → 0.89, 60 → 1.34, 80 → 1.78, 100 → 2.22 s.
(a) Show that the car moved at a steady speed and find that speed.
(b) Explain why light gates, not a stopwatch, were used, naming the error a stopwatch would add and whether it is random or systematic.
(c) Sketch what a velocity–time graph of the whole run would look like, and state its gradient during the timed section.
Reveal answer
(a) Distance ÷ time: 0.20 ÷ 0.44 = 0.45 m/s; 0.40 ÷ 0.89 = 0.45; 0.60 ÷ 1.34 = 0.45; 0.80 ÷ 1.78 = 0.45; 1.00 ÷ 2.22 = 0.45 m/s. Every ratio is the same to two significant figures, so the speed was steady at 0.45 m/s; on a distance–time graph the points lie on a straight line through the origin with gradient 0.45 m/s.
(b) A hand-held stopwatch adds the operator’s reaction time, about 0.2 s, at the start and stop; the two do not cancel and vary from trial to trial, so it is mainly a random error that scatters times by ± 0.1 s or more — a fifth of the shortest time. A light gate starts and stops the timer when the car’s flag breaks the beam, at the same point every time, so the reaction error is removed.
(c) A short sloping line as the car starts, then a long flat line at 0.45 m/s through both gates; the gradient during the timed section is zero — zero acceleration, steady velocity.
Common mistakes to avoid
- Speed for velocity — velocity needs a direction; a change of direction at constant speed is a change of velocity.
- Acceleration as “going fast” — it is the rate of change of velocity, unit m/s²; a car at a steady 120 km/h has zero acceleration.
- Mixing units — km/h into an equation with seconds. Convert to m, s and m/s first; divide km/h by 3.6.
- Reading a velocity–time graph as a map — flat means steady speed, not stopped; the gradient is acceleration, the area is distance.
- Forgetting the ½ — from rest, s = ½at², not at². The average velocity is half the final velocity.
- Forgetting the square — twice the time gives four times the distance under constant acceleration.
How we found out — a ramp, a water clock and a galloping horse
For nearly two thousand years the accepted account of motion was Aristotle’s (fourth century BC): heavy things fall faster than light things, in proportion to their weight, and a moving object needs a continuous push to keep moving. Nobody had measured a fall, because falls were over too quickly to time. Galileo Galilei solved the timing problem around 1604 by slowing the fall down: he let a bronze ball roll along a groove in a tilted beam and timed it with a water clock, weighing the water that flowed out while the ball rolled. He found that the distance grew with the square of the time — 1, 4, 9, 16 units in successive equal times — and that the rule was the same whatever the tilt and whatever the ball. Published in Two New Sciences in 1638, four years before he died under house arrest, the result was the first quantitative law of motion and the origin of s = ½at². He could not measure a straight fall, but he argued that a ramp is a fall spread out, and that the acceleration of a free fall was the same rule at its steepest.
The velocity of a moving thing at an instant remained hard to catch until the camera. In 1878 the photographer Eadweard Muybridge, hired by Leland Stanford to settle the question whether a galloping horse ever has all four feet off the ground, lined a track at Palo Alto with twelve cameras triggered by threads the horse broke as it ran. The photographs, taken a fraction of a second apart, showed the horse in the air — and, more importantly, gave the first frame-by-frame record of motion, from which speed at each instant could be read. Electronic timing arrived in athletics in the 1930s, fully automatic photo-finish timing to 0.01 s at the 1968 Olympics, and the 10 m split cameras behind the Tebogo table in the 1990s. Every improvement in the description of motion came from a better way of stopping time: a water clock, a thread across a track, a light beam broken by a runner.
Galileo could not time a stone falling 2 m — it takes 0.64 s, too short for a water clock. Explain how the ramp solved the problem and why his conclusion about free fall was a prediction rather than a measurement.
Reveal answer
Tilting the beam reduced the acceleration — at a gentle slope the ball took several seconds to roll a metre, long enough for the water clock to give a repeatable reading — while keeping the same kind of motion, a start from rest with a steady acceleration. He found the distance–time-squared rule held at every tilt and reasoned that a vertical drop was the limiting case of the steepest possible ramp, so the same rule must apply. That last step is an extrapolation beyond his data, a prediction from a model; it was confirmed only when later instruments (Atwood’s machine in 1784, the electromagnet-and-trapdoor timer in the freefall lab) could time a straight fall directly and found the 9.8 m/s² his rule implied.
Where it is used now
- Road safety in Botswana. Speed cameras on the A1 and in Gaborone measure the time a vehicle takes between two marks on the road or two radar pulses — the speed trap lab at full scale. Stopping distance rises with the square of the speed (s = ½at² read backwards), which is why the limit falls to 60 km/h in town and 40 km/h outside schools.
- Athletics. Botswana’s sprinters and their coaches train from split-time tables like the one above; the reaction time is read from a sensor in the blocks, and a start under 0.100 s is ruled a false start because no human can react faster.
- Your phone. The accelerometer inside it measures acceleration thousands of times a second; the step counter, the screen rotation and the crash-detection feature are all kinematics.
- Aircraft and spacecraft. A pilot’s take-off is a constant-acceleration problem — will the runway be long enough for v = u + at to reach flying speed? — and a spacecraft’s navigation integrates acceleration to velocity to position, exactly as the area under the graph does.
- Wildlife. GPS collars on elephants in the Okavango give position every few minutes; the differences give velocity, and the pattern of speed against time tells researchers when an animal is feeding, walking to water or fleeing.
Science now — and the scientist you need to be
What is happening in 2026
Motion is now measured continuously and everywhere. World Athletics events track every runner with radar and camera systems that report position and velocity many times a second, and AI tools trained on those data can pick out which part of a stride is costing time — coaches then redesign training and technique from the pattern, which is the statement of inquiry in practice. Cars sold in 2026 measure their own velocity from wheel sensors, GPS and cameras and compare them constantly; automatic emergency braking is a velocity–time calculation done by software faster than a driver can react. Spacecraft do the same with no GPS at all: the Artemis II crew capsule that flew in April 2026 navigated by measuring its acceleration and adding it up over time, checked against star positions. The equations in this lesson are three hundred years old; what has changed is how many times a second they are used.
The mindshift
What you need to be able to do
Three examinable skills: convert units before calculating and write the unit on every answer; use v = u + at and s = ut + ½at² with the ½ and the square in place; and read a velocity–time graph — gradient with its unit, m/s², and area with its unit, m. Criterion C tasks in this unit hand you ticker-tape or light-gate readings and ask for the graph, the gradient and a conclusion that quotes values; the Think task below is the rehearsal.
See an example
Bicycle on a flat road: 20 m in 4.1, 4.3, 4.0 s (mean 4.1 s → 4.9 m/s); 40 m in 8.3, 8.5, 8.2 s (mean 8.3 s → 4.8 m/s). The two speeds agree within the scatter of my timing (reaction time about 0.2 s, a random error), so the steady-speed model fits. The AI gave a car going round a roundabout at a constant 30 km/h as an example of changing velocity at constant speed — correct, because the direction is changing.
Key terms
| Kinematics | The description of motion — position, velocity, acceleration and time — without reference to its causes. |
| Distance / displacement | Length of path travelled (scalar) / change of position with direction (vector); both in metres. |
| Speed / velocity | Distance per second (scalar) / displacement per second, with direction (vector); both in m/s. |
| Acceleration | Rate of change of velocity, a = Δv ÷ t, in m/s²; negative for slowing down (deceleration). |
| Steady-speed model | Velocity constant: s = vt. Applies when the resultant force is zero. |
| Constant-acceleration model | Acceleration constant: v = u + at and s = ut + ½at², with u the initial velocity and v the final velocity. |
| Velocity–time graph | A graph whose gradient is the acceleration and whose area under the line is the displacement. |
| Gradient | Rise ÷ run of a graph, with the unit y-unit ÷ x-unit; on a v–t graph, m/s ÷ s = m/s². |
| Ticker-tape timer | A device that marks a paper tape 50 times a second; the spacing of the dots gives the velocity every 0.02 s. |
| Light gate | A beam and sensor that start or stop an electronic timer when an object breaks the beam, removing reaction time. |
Think like a scientist
Criterion C A trolley was released from rest on a ramp, pulling a paper tape through a ticker-timer that makes 50 dots a second. Each tape was cut into strips of 5 dot-spaces (0.10 s each) and the length of each strip measured with a rule to 0.1 cm. Three tapes were taken. The distance from the release point to the end of the sixth strip, measured along the ramp with a rule, was 19.5 cm.
| Strip (time interval / s) | Mid-time / s | Tape 1 length / cm | Tape 2 / cm | Tape 3 / cm | Mean length / cm | Velocity / m/s |
|---|---|---|---|---|---|---|
| 0.0–0.1 | 0.05 | 1.2 | 1.1 | 1.3 | ||
| 0.1–0.2 | 0.15 | 2.0 | 2.1 | 1.9 | ||
| 0.2–0.3 | 0.25 | 2.8 | 2.9 | 2.7 | ||
| 0.3–0.4 | 0.35 | 3.6 | 3.5 | 4.8 | ||
| 0.4–0.5 | 0.45 | 4.4 | 4.5 | 4.3 | ||
| 0.5–0.6 | 0.55 | 5.2 | 5.3 | 5.1 |
(a) Identify the anomalous reading and suggest a specific cause. [2]
(b) Calculate the mean strip lengths (excluding the anomaly), rounded to match the rule, and convert each to a velocity in m/s. [4]
(c) Describe how you would plot the velocity–time graph, and state the shape you expect. [3]
(d) Calculate the gradient of the graph with its unit and state what it represents. [3]
(e) Use the area under the graph, from 0 to 0.60 s, to find the distance travelled, and compare it with the 19.5 cm measured along the ramp. [3]
(f) Evaluate the method: one random and one systematic error, the effect of each on the results, and an improvement for each. [4]
(g) Write a conclusion that states whether the trolley’s motion fits the constant-acceleration model, using values from your processing. [3]
Model answer
(a) Tape 3, strip 0.3–0.4 s: 4.8 cm, well above the other two (3.6, 3.5) and larger than the following strip on the same tape (4.3 cm), which is impossible for a trolley speeding up. Specific cause: the strip was cut at the wrong dot — six spaces instead of five — or two strips were confused; alternatively the tape snagged and then jerked free during that interval.
(b) Means to 0.1 cm: 1.2, 2.0, 2.8, 3.6 (from 3.6 and 3.5 only: 3.55 rounds to 3.6), 4.4, 5.2 cm. Velocity = strip length ÷ 0.10 s, converted to metres: 0.012 ÷ 0.10 = 0.12 m/s; then 0.20, 0.28, 0.36, 0.44, 0.52 m/s. The velocity belongs to the middle of each interval, so it is paired with the mid-time.
(c) Mid-time on the x-axis (0 to 0.6 s, 0.1 s per large square), velocity on the y-axis (0 to 0.6 m/s, 0.1 m/s per large square), both labelled with units; six crosses; a straight line of best fit. Expected shape: a straight line rising at a constant slope, because the velocity increases by the same 0.08 m/s in every 0.1 s. Extended back to t = 0 it does not pass through the origin: it meets the velocity axis at about 0.08 m/s, which tells you the tape was already moving when the first dot was printed — the timer was switched on before the trolley was released, or the first, crowded dots could not be cut into a strip. That intercept is information, not an error.
(d) Using two points far apart on the line: gradient = 0.52 − 0.120.55 − 0.05 = 0.400.50 = 0.80 m/s². The gradient of a velocity–time graph is the acceleration: the trolley gained 0.80 m/s of velocity every second, and because the line is straight the acceleration was constant.
(e) Because the line starts at 0.08 m/s rather than at the origin, the area under it from 0 to 0.60 s is a trapezium, not a triangle: the velocity at 0.60 s read from the line is 0.08 + 0.80 × 0.60 = 0.56 m/s, so the area is ½ × (0.08 + 0.56) m/s × 0.60 s = 0.192 m. That is exactly the sum of the six mean strip lengths, 1.2 + 2.0 + 2.8 + 3.6 + 4.4 + 5.2 = 19.2 cm = 0.192 m, which is the check that the graph has been read correctly. The measured 19.5 cm agrees with 19.2 cm to within 0.3 cm (about 2 %); the difference is the rounding on six strips plus the short distance the trolley had already covered before the first dot, which the tape cannot show but the rule includes. (Using s = ½at² here would give 0.144 m and be wrong, because that form assumes the trolley was at rest at t = 0, and the intercept shows it was not.)
(f) Random: reading the strip length with a rule to 0.1 cm when the dots are 0.5 mm wide — the spread of 1.1 to 1.3 cm on the first strip — scatters the points about the line; improvement: measure the length of 10 dot-spaces and halve it, or measure between dot centres with a magnifier. Systematic: the tape drags through the timer with a friction force that opposes the motion on every run, so every velocity is slightly low and the gradient underestimates the trolley’s free acceleration; improvement: use light gates with a card on the trolley, which touch nothing, or lubricate the timer’s tape guide. (Also acceptable as systematic: the timer’s frequency not exactly 50 Hz, which scales every velocity by the same factor.)
(g) The motion fits the constant-acceleration model: the velocity rose from 0.12 m/s at 0.05 s to 0.52 m/s at 0.55 s in equal steps of 0.08 m/s per 0.1 s, giving a straight velocity–time line of gradient 0.80 m/s², and the area under the line, 19.2 cm, matches the 19.5 cm measured along the ramp to within 2 %. The acceleration was constant to within the scatter of the readings.
Probe deeper — exam-style questions
Attempt every question before opening the answer. Marks in brackets show what a full answer is worth.
A bus travels at 90 km/h. Convert this to m/s and calculate how far it travels in 45 s.
Reveal answer
90 km/h = 90 000 m ÷ 3 600 s = 25 m/s [1]; s = vt = 25 × 45 [1] = 1 125 m, about 1.1 km [1].
A motorbike accelerates from 8.0 m/s to 20.0 m/s in 6.0 s. Calculate its acceleration and the distance it covers while accelerating.
Reveal answer
a = (20.0 − 8.0) ÷ 6.0 = 2.0 m/s² [1]; s = ut + ½at² = 8.0 × 6.0 + ½ × 2.0 × 6.0² = 48 + 36 [1] = 84 m [1].
A student writes: “The velocity–time graph is flat at 15 m/s from 10 s to 30 s, so the car has stopped and travelled 15 m.” Identify three errors.
Reveal answer
A flat line on a velocity–time graph means constant velocity, not stopped — the car is moving at 15 m/s [1]; the distance is the area under the line, 15 m/s × 20 s = 300 m, not 15 m [1]; the acceleration in that section is zero (gradient zero), which is what “flat” actually tells you [1].
A ball is dropped from a bridge and hits the water after 1.8 s (a = 9.8 m/s², ignore air). Calculate its velocity as it hits and the height of the bridge, and explain why doubling the height would not double the time.
Reveal answer
v = u + at = 0 + 9.8 × 1.8 = 17.6 m/s ≈ 18 m/s [1]; s = ½at² = ½ × 9.8 × 1.8² = 15.9 m ≈ 16 m [1]; distance grows with the square of the time under constant acceleration [1], so doubling the height multiplies the time by √2 ≈ 1.4, not 2 — the ball is moving faster in the second half of the fall and covers it in less time [1].
True or false, with a reason: “An object moving at a constant speed cannot be accelerating.”
Reveal answer
False [1]: acceleration is a change in velocity, and velocity includes direction, so an object moving in a circle at constant speed — a satellite, a car on a roundabout — is accelerating towards the centre because its direction changes continuously [1].
A velocity–time graph of a lift shows: 0 to 2 s, a straight line from 0 to 3 m/s; 2 to 8 s, flat at 3 m/s; 8 to 11 s, a straight line down to 0. Calculate the acceleration in each sloping section and the total distance the lift rises.
Reveal answer
First section: a = 3 ÷ 2 = 1.5 m/s² [1]; last section: a = (0 − 3) ÷ 3 = −1.0 m/s², a deceleration [1]; distance = areas: ½ × 2 × 3 = 3 m, plus 6 × 3 = 18 m, plus ½ × 3 × 3 = 4.5 m [1]; total 25.5 m, about seven floors [1].
Summary
- Distance and speed are scalars; displacement, velocity and acceleration are vectors — direction matters, and in a straight line a sign carries it.
- Steady-speed model: s = vt. Convert km/h to m/s by dividing by 3.6 before any calculation.
- Acceleration a = Δv ÷ t, in m/s²; constant-acceleration model: v = u + at and s = ut + ½at².
- On a velocity–time graph the gradient is the acceleration and the area under the line is the displacement; flat means steady, not stopped.
- Galileo timed a ramp because a fall was too fast; Muybridge photographed a horse; every better description of motion came from a better way of stopping time.
- A race or a journey as a data table shows a pattern, and the pattern is what a coach or a designer changes.
Next time → Forces that change motion: why velocity changes at all — resultant force, F = ma, friction and air resistance, and how data redesign a bicycle.