What is the difference between distance and displacement?

Distance is how far you travel along the path, with no direction (a scalar). Displacement is the straight-line change in position from start to finish, with a direction (a vector). A runner who does one lap of a 400 m track covers 400 m of distance but has zero displacement.

What does the area under a velocity–time graph give you?

The displacement. Areas above the time axis are positive and areas below are negative. Add the sizes of the areas instead if the question asks for the total distance.

When can you use the SUVAT equations?

Only when the acceleration is constant and the motion is along a straight line. For a trip with several stages, use them on one stage at a time. They don't work while the acceleration is changing.

Is g positive or negative?

The size of g is 9.8 m s^−2. Its sign depends on your convention: if up is positive, the acceleration of a falling or thrown object is −9.8 m s^−2; if down is positive, it is +9.8 m s^−2. State your choice and keep it.

How do you measure g in Year 11?

By two graphical methods: time falls from rest from different heights and plot h against t^2 (gradient g ÷ 2), and measure a falling object's velocity at several times and plot v against t (gradient g). Then compare the two values with 9.8 m s^−2 and discuss their accuracy and reliability.

Year 11 Physics · Fundamentals of mechanics · Motion in a straight line · 12 min read

Motion graphs and SUVAT

Motion in a straight line is the first big skill of Year 11 Physics. You describe how something moves with three quantities (displacement, velocity and acceleration), show them on graphs, and solve problems with the three equations of uniform acceleration, often called SUVAT after the symbols s, u, v, a and t.

This guide covers what each graph's gradient and area mean, when the SUVAT equations work, vertical motion under gravity and how to measure g. Then a quick quiz, the mistakes that cost marks, and a full worked example. Air resistance is ignored throughout, as it is in Year 11 calculations.

What you need to know

Solve problems for displacement and for the total distance travelled.
Explain how to measure and calculate average and instantaneous speed and velocity.
Use v_av = Δs ÷ Δt to solve problems, and work out time, distance, displacement, speed and velocity.
Apply a = Δv ÷ Δt to straight-line problems (two-dimensional problems are in the vectors guide).
Construct displacement–time, velocity–time and acceleration–time graphs from data.
Analyse motion graphs to describe an object's acceleration, velocity and displacement, and to relate displacement, time, acceleration and initial and final velocity.
Use s = ut + ½at², v = u + at and v² = u² + 2as for one-dimensional uniform acceleration problems.
Find g, the acceleration due to gravity, in a laboratory experiment by two graphical methods; compare the two, and judge how accurate and reliable each result is.

The equations you will use:

v_av = Δs ÷ ΔtAverage velocity: displacement over time. Average speed uses total distance instead.

a = Δv ÷ Δt = (v − u) ÷ tAcceleration is the rate of change of velocity, in m s⁻².

v = u + atNo displacement in it: use it when s is neither given nor asked for.

s = ut + ½at²No final velocity in it: use it when v is not given and not asked for.

v² = u² + 2asNo time in it: use it for stopping distances and greatest heights.

Dot points: PY11-FM-05, PY11-FM-06, PY11-FM-07, PY11-FM-09, PY11-FM-10, PY11-FM-11, PY11-FM-12, PY11-FM-13, PY11-FM-14, PY11-FM-15 (2025 Physics syllabus, Year 11), in our own words.

Mini lesson

Distance or displacement, speed or velocity

Distance and speed are scalars: a size only. Displacement and velocity are vectors: a size and a direction. Walk 6 m east and then 2 m west and you have covered a distance of 8 m, but your displacement is only 4 m east.

Average velocity is displacement ÷ time for the whole trip. Instantaneous velocity is the velocity at one moment, which is what a speedometer shows (with a direction added). In the lab you measure an instantaneous velocity by timing the object over a very short distance, for example with a light gate, or from a video frame by frame.

In one dimension, a sign carries the direction. Choose which way is positive, write it down, and then a negative velocity simply means moving the other way.

Displacement–time graphs: the gradient is velocity

On a displacement–time (s–t) graph, the gradient is the velocity. A straight line means constant velocity; a flat line means the object is at rest; a line sloping down means it is moving in the negative direction.

A curve means the velocity is changing. The instantaneous velocity at a point is the gradient of the tangent there. The average velocity between two times is the gradient of the straight line (the chord) joining the two points.

Blue: the tangent's gradient is the instantaneous velocity at that moment. Orange: the chord's gradient is the average velocity from the start to that moment. The curve gets steeper, so the object is speeding up.

Velocity–time graphs: gradient is acceleration, area is displacement

On a velocity–time (v–t) graph, the gradient is the acceleration and the area between the line and the time axis is the displacement. A straight sloping line means uniform acceleration, which is exactly when the SUVAT equations apply.

Area below the time axis counts as negative displacement (the object is moving backwards). To get the total distance, add the sizes of all the areas; to get the displacement, add them with their signs.

On an acceleration–time (a–t) graph, the area under the line is the change in velocity.

For uniform acceleration from u to v in time t: the gradient is a = (v − u) ÷ t, and the shaded area is s = ½(u + v)t. The SUVAT equations come from these two facts.

The SUVAT equations: choosing the right one

The three equations only work when the acceleration is uniform (constant) and the motion is in a straight line. If the motion has stages, such as speeding up and then braking, treat each stage separately and carry the velocity from the end of one stage into the next.

Each equation leaves out one of the five quantities, so pick the one without the quantity you don't know and don't need.

List the five: s, u, v, a, t. Mark the three you know and the one you want.
No s? Use v = u + at.
No v? Use s = ut + ½at².
No t? Use v² = u² + 2as.
Signs: put every vector in with its sign. A car slowing down while moving in the positive direction has a negative a.

Vertical motion and measuring g

Anything moving only under gravity has a uniform acceleration of 9.8 m s⁻² downward, whether it is going up, at the top or coming down. With up as positive, a = −9.8 m s⁻² for the whole flight. At the highest point the velocity is zero for an instant, but the acceleration is still 9.8 m s⁻² down.

You can find g in the lab by two graphical methods. Method 1: drop an object from several heights h, time each fall from rest, and plot h against t². Since h = ½gt², the gradient is g ÷ 2. Method 2: record the falling object's velocity at several times (light gates, a ticker timer or video analysis) and plot v against t. The gradient is g.

Compare the two values with the accepted 9.8 m s⁻² using percentage error (accuracy), and repeat each measurement to see how consistent the results are (reliability). Reaction time in hand timing is a common source of error, which is why electronic timing is better.

A ball thrown straight up (up positive). At the top (dashed): s is greatest, v crosses zero, and a is still −9.8 m s⁻². The v–t gradient equals the a–t value.

Common mistakes

Reading the gradient of a displacement–time graph as the acceleration.The s–t gradient is velocity. The v–t gradient is acceleration. Say the axes out loud before you read any graph.
Adding the area below the time axis of a v–t graph as positive displacement.Area below the axis is negative displacement. Add the signed areas for displacement, and the sizes for total distance.
Using one SUVAT equation across a whole trip that speeds up, cruises and then brakes.SUVAT needs one constant acceleration. Split the trip into stages and solve each, passing on the velocity between stages.
Throwing a ball up with u = +12 m s⁻¹ and also using a = +9.8 m s⁻².Up is positive, so gravity's acceleration is a = −9.8 m s⁻². Write the sign convention down before substituting.
Saying the acceleration is zero at the top of a vertical throw.Only the velocity is zero at the top. The acceleration is 9.8 m s⁻² down the whole time, or the ball would stay up there.
Finding average speed by averaging the speeds of two stages.Average speed = total distance ÷ total time. Stages that last different times don't count equally.

Worked example

A suburban train leaves a station from rest and accelerates uniformly at 0.80 m s⁻² for 25 s. It then travels at a constant velocity for 60 s. Finally it slows uniformly and stops at the next station 40 s later.

(a) Calculate the train's top speed. (b) Calculate the distance between the two stations. (c) Calculate the train's average velocity between the stations.

(5 marks)

I'll think out loud, the way you should in the exam, using KNOW-UD: Knowns, Not known, Organise, Work it out, Units, Direction.

Prediction first. The top speed should be about 0.8 × 25 = 20 m s⁻¹. The train is at top speed for about half the 125 s trip, and averages half its top speed while speeding up and slowing down, so it covers roughly ¾ of 20 × 125 = 2500 m. I predict a distance of about 1900 m and an average velocity a bit under ¾ of the top speed.

K · Knowns

I sketch the velocity–time graph first, because every part of this question is about it: the top speed is its height and the distance is its area. I put each value on the sketch and take the direction of travel as positive.

Stage 1: u = 0, a₁ = +0.80 m s⁻², t₁ = 25 s
Stage 2: constant velocity for t₂ = 60 s
Stage 3: uniform slowing to v = 0 in t₃ = 40 s

Knowns on the sketch: a = 0.80 m s⁻² for 25 s, constant velocity for 60 s, stopping in 40 s. Unknowns: the top speed v and the area (the distance).

N · Not known

I need three answers, so I write all three down now and tick them off at the end.

(a) v = ? m s⁻¹ (b) s = ? m (c) v_av = ? m s⁻¹

O · Organise

Stage 1 is uniform acceleration from rest, and I know u, a and t, so v = u + at gives the top speed. For the distance I use the area under my graph: a triangle, a rectangle and another triangle. Average velocity is the total displacement over the total time.

v = u + a₁t₁
s = ½vt₁ + vt₂ + ½vt₃
v_av = Δs ÷ Δt, Δt = t₁ + t₂ + t₃

W · Work it out

I do the top speed first, because all three areas need it. Then each area on its own line, so a marker can follow me, and a quick check of stage 1 with s = ut + ½at².

(a) v = 0 + 0.80 × 25 = 20 m s⁻¹
Stage 1: ½ × 20 × 25 = 250 m (check: s = ½ × 0.80 × 25² = 250 m)
Stage 2: 20 × 60 = 1200 m
Stage 3: ½ × 20 × 40 = 400 m
(b) s = 250 + 1200 + 400 = 1850 m
(c) v_av = 1850 ÷ (25 + 60 + 40) = 1850 ÷ 125 = 14.8 m s⁻¹

U · Units

Speeds in m s⁻¹, distance in m. The data are given to two significant figures, so I round the final answers to match.

(a) 20 m s⁻¹ (b) 1.85 × 10³ m ≈ 1.9 km (c) ≈ 15 m s⁻¹

D · Direction

Average velocity is a vector, so it needs a direction: the train only moves one way, so it points along the track towards the second station. As a sense check, the average must lie between half the top speed and the top speed, because the train spends part of the trip cruising. 14.8 sits between 10 and 20.

v_av ≈ 15 m s⁻¹ along the track, from the first station towards the second

(a) 20 m s⁻¹. (b) About 1.9 km (1850 m). (c) About 15 m s⁻¹ along the track towards the second station.

Check against the prediction. I predicted about 1900 m and an average a bit under ¾ of the top speed. The distance is 1850 m and 14.8 ÷ 20 = 0.74, so both match.

Exam advice

What markers look for

When you find a gradient from a graph, draw a large triangle on your line of best fit, not between two data points. Show rise ÷ run with units, and say what the gradient means (for example, 'the gradient is g ÷ 2').NESA marking feedback, 2024, summarised
Draw a ruled line of best fit with the points scattered evenly on both sides. Don't join the dots, and don't force the line through the origin unless the physics says it must pass through it.NESA marking feedback, 2025, summarised
Read the axes before you read the graph. A displacement–time graph and a velocity–time graph can look identical, but the same slope means a different quantity on each.NESA marking feedback, 2024, summarised
Show the equation, the substitution and the answer with a unit. Correct working still earns marks when a number slips; a bare wrong answer earns none.General advice from NESA marking feedback, 2021 to 2025, summarised
Keep full calculator values between stages and round only the final answer. Rounding the velocity at the end of stage 1 makes every later stage drift.NESA marking feedback, 2025, summarised

Summaries in our own words. Read NESA's own feedback with the exam packs: 2024 · 2025.

Where this leads in Year 12

In Year 12 the same three equations come straight back in projectile motion: you use them twice at once, with constant velocity horizontally and a uniform 9.8 m s⁻² vertically, and time links the two. Vertical motion under gravity, signs and the 'v = 0 at the top' idea are exactly what you need there. Our projectile motion guide is written for the current HSC course (2017 syllabus), but the physics in it is the same physics you will use in Year 12.

FAQ

What is the difference between distance and displacement?: Distance is how far you travel along the path, with no direction (a scalar). Displacement is the straight-line change in position from start to finish, with a direction (a vector). A runner who does one lap of a 400 m track covers 400 m of distance but has zero displacement.
What does the area under a velocity–time graph give you?: The displacement. Areas above the time axis are positive and areas below are negative. Add the sizes of the areas instead if the question asks for the total distance.
When can you use the SUVAT equations?: Only when the acceleration is constant and the motion is along a straight line. For a trip with several stages, use them on one stage at a time. They don't work while the acceleration is changing.
Is g positive or negative?: The size of g is 9.8 m s⁻². Its sign depends on your convention: if up is positive, the acceleration of a falling or thrown object is −9.8 m s⁻²; if down is positive, it is +9.8 m s⁻². State your choice and keep it.
How do you measure g in Year 11?: By two graphical methods: time falls from rest from different heights and plot h against t² (gradient g ÷ 2), and measure a falling object's velocity at several times and plot v against t (gradient g). Then compare the two values with 9.8 m s⁻² and discuss their accuracy and reliability.

Related guides

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