Motion and reading graphs

6 visual steps · 3 worked examples · 5 exercises with solutions

First, let’s see the idea.

After walking out and back, does distance equal displacement?

Move forward one step at a time. Change the values and observe what happens.

Go and return0 m+xSpatial axis above; position–time graph below.
Drawing description

Choose zero and rightward. We start from the lilac zero. Being on the right gives positive position; being on the left gives negative position. In this model each leg lasts half the total time and direction changes instantly. Locate 0 m and the +x arrow on the upper line. Positive-direction path: 40 m. Return path: 15 m. Total time: 20 s.

Step 1 of 6

Set the origin and positive direction

Locate 0 m and the +x arrow on the upper line.

We start from the lilac zero. Being right gives positive position; being left gives negative position. In this model each leg lasts half the total time and direction changes instantly.

Positive-direction path
40 m
Return path
15 m
Total time
20 s

What if you change this?

20 m100 m
0 m100 m
10 s60 s

Before moving on

You walk 40 m and return 40 m in 20 s. What are distance, displacement and average velocity?

Compare with the explanation

Distance: 80 m. Displacement: 0 m. Average velocity: 0 m/s; average speed is 4 m/s.

You decide when to mark this lesson as studied.

Go and return

Slide the drawing to see the details.

Explore the full explanationThe foundations, principles, worked examples and all the exercises, at your own pace.

What you will understand

  • Choose a reference frame and calculate signed displacements.
  • Distinguish average speed from average velocity and instantaneous values.
  • Interpret slopes on position–time graphs and areas on velocity–time graphs.
  • Recognize acceleration as a change in velocity.

Before you start

You do not need to know everything by heart. The explanation revisits the necessary ideas and shows how they connect. If a foundation is still unclear, you can return to these lessons:

Position depends on a reference

A seated passenger is at rest relative to the bus and moving relative to the road. There is no contradiction: motion is a change of position over time relative to a reference frame. When describing a body, state what you consider fixed, the position origin, the axis direction and the positive sense.

On a straight road, one axis is enough. We can choose the school as x = 0 and the direction towards the market as positive. Being at x = −30 m means being thirty metres on the opposite side from the market; the sign does not mean negative size. Position is a coordinate and can be negative, zero or positive.

A body can be represented as a point particle if its size and rotation are irrelevant to the question. For studying a bus journey between cities, this approximation may work; for studying passage through a narrow doorway, dimensions matter. A model is an explained choice, not a complete photograph of the object.

Displacement and distance travelled

Displacement in one dimension is Δx = xf − xi, in metres. The Greek letter Δ indicates a change, always final value minus initial value. If you go from x = 10 m to x = −5 m, Δx = −15 m. The sign tells us the direction of the total change between the two positions.

Distance travelled adds the lengths of all path segments. It has no negative sign. Walking 40 m one way and returning 15 m produces distance 55 m but displacement only 25 m in the initial direction. If you return to the starting point, displacement is zero even after effort and considerable distance travelled.

Knowing only the initial and final points does not reveal total distance. We need information about the path. In straight-line motion without reversal, distance equals displacement magnitude. Equality no longer holds when direction changes. This care will be essential when interpreting signed areas.

Duration and averages

Duration is Δt = tf − ti and is measured in seconds. An instant says when something happens; duration says how much time passes between events. A stopwatch started at the first event makes ti = 0, but that choice is not compulsory.

Average velocity is vm = Δx/Δt, in m/s. Along a line, the sign expresses displacement direction. Average speed is total distance/Δt, also in m/s, but without direction or negative sign. The same journey can have zero average velocity and positive average speed. Do not use the word “velocity” to hide which quantity you calculated.

An average summarizes a whole interval and does not reveal each moment of motion. A trip with stops and accelerations can have the same average as uniform motion. When segment times differ, simply averaging their speeds arithmetically is not enough; first add all distances and durations, including stops belonging to the interval considered.

The position–time graph has a language

On a graph of x against t, the horizontal axis represents time in seconds and the vertical axis position in metres. Each point is a pair: the body's position at an instant. Connecting points with segments assumes what happened between measurements; isolated points do not guarantee that intermediate motion was uniform.

A straight line's slope is vertical change divided by horizontal change: Δx/Δt. Thus, on a position–time graph, a secant's slope gives average velocity between two instants. A rising line corresponds to positive velocity; a falling line to negative velocity; a horizontal line to rest in that frame.

A steeper line has greater velocity magnitude when axes keep the same scale. Graph height is not speed: a body far from the origin can be stationary. On a curved graph, slope changes; the local tangent represents instantaneous velocity, the value at one instant. You do not yet need differential calculus to recognize this interpretation.

x (m)t (s)forward motionstopreturn

Schematic diagram: slope changes from positive to zero and then negative. Segments do not represent the road's physical shape.

The velocity–time graph

The vertical axis now shows velocity, not position. A horizontal line above zero represents uniform motion in the positive direction. A horizontal line below zero represents uniform motion in the negative direction. It does not mean rest: rest requires v = 0. The meaning of the same shape depends on the quantities labelled on the axes.

For constant velocity, Δx = vΔt. On a velocity–time graph, the product is the rectangle's area between the line and time axis. Over varying sections, we can add rectangle and triangle areas. Areas below the axis count negatively for displacement; for distance travelled, add area magnitudes, separating direction changes.

Area units are (m/s) × s = m, confirming the interpretation. A velocity–time graph reveals displacements, but an initial position is needed to find positions. Two people can have identical velocity graphs and remain a constant distance apart.

Changing velocity and choosing the model

Average acceleration is am = Δv/Δt and is measured in m/s². Acceleration 2 m/s² means velocity increases by 2 m/s each second if acceleration is constant. On a velocity–time graph, slope gives average acceleration. This differs from area, which gives displacement.

Along a line, positive acceleration does not necessarily mean increasing speed. If v = −6 m/s and a = +2 m/s², velocity approaches zero and speed decreases until stopping. To decide whether speed increases or decreases, compare velocity and acceleration directions.

Start every problem with a verbal description and a simple table. Identify intervals, choose signs, convert units and only then calculate. A constant-velocity model does not describe a vehicle's start or braking. It can be a useful approximation for a section, but its validity must be recognized.

Worked examples, without skipping steps

Follow the reasoning first. Then cover the solution and try to reconstruct it in your own words.

Example 1

Outward trip and short return

A person starts at x = 0, walks to x = 60 m and returns to x = 20 m. The journey lasts 50 s. Calculate distance, displacement and both averages.
  1. The first segment is 60 m; the return is 60 − 20 = 40 m. Total distance is 100 m.
  2. Displacement uses only endpoints: Δx = 20 − 0 = +20 m.
  3. Average velocity is 20/50 = +0,40 m/s. Average speed is 100/50 = 2,0 m/s.
Result
Distance 100 m; displacement +20 m; average v +0,40 m/s; average speed 2,0 m/s.
Check the result
Average speed exceeds the magnitude of average velocity because there was a return trip. Both units are m/s.

Example 2

A straight line with an initial position

A moving body has x = 8 m at t = 2 s and x = 26 m at t = 8 s. The graph is a straight line between those points. Where is it at t = 5 s?
  1. The duration between the data points is 6 s and displacement 18 m. Slope is v = +3 m/s.
  2. From t = 2 s to t = 5 s, 3 s pass. Intermediate displacement is 3 × 3 = 9 m.
  3. Add to the reference position: x = 8 + 9 = 17 m.
Result
x(5 s) = 17 m.
Check the result
Five seconds is the midpoint of the time interval and 17 m is halfway between 8 m and 26 m, as the straight line requires.

Example 3

Area and slope give different results

Velocity increases uniformly from 0 to 6 m/s over 4 s and stays at 6 m/s for another 3 s. Calculate total displacement.
  1. During the start, area is triangular: Δx₁ = 4 × 6/2 = 12 m.
  2. During the uniform segment, area is rectangular: Δx₂ = 3 × 6 = 18 m.
  3. Add areas: Δx = 30 m. Initial acceleration, obtained from slope, is 6/4 = 1,5 m/s².
Result
Δx = +30 m; initial a = +1,5 m/s².
Check the result
Velocity is never negative, so distance is also 30 m. Do not confuse the area's 30 m with the slope's 1,5 m/s².

Now try it yourself

There is no timer. Write what you know, choose a relationship and justify each step. Hints and solutions are available whenever you need them.

Exercise 1

On an x(t) graph, a horizontal segment runs from t = 4 s to t = 9 s at x = 15 m. Interpret it.
A hint
Ask whether position changes.
See the worked solution
For 5 s, position remains 15 m. Slope is (15 − 15)/(9 − 4) = 0 m/s: the body is at rest in the chosen frame. The 15 m represents position, not distance travelled in that interval.

Exercise 2

A body moves from x = −7 m to x = 5 m in 3 s without reversing direction. Calculate its average velocity and average speed.
A hint
First calculate displacement.
See the worked solution
Δx = 5 − (−7) = +12 m. Average velocity is +12/3 = +4 m/s. Without reversal, distance is 12 m and average speed 4 m/s. Negative initial position does not require negative velocity.

Exercise 3

A round trip lasts 80 s and covers a total distance of 240 m. Calculate the average velocity and average speed.
A hint
The endpoint coincides with the starting point.
See the worked solution
Displacement is zero, so average velocity is 0/80 = 0 m/s. Average speed is 240/80 = 3 m/s. Zero average velocity does not mean rest during the journey.

Exercise 4

A v(t) graph has v = −3 m/s for 6 s. Calculate displacement and distance.
A hint
Keep the sign for displacement.
See the worked solution
Signed area is −3 × 6 = −18 m: displacement in the negative direction. Distance is 18 m because there is no reversal. The horizontal line shows zero acceleration, even with negative velocity.

Exercise 5

Velocity changes from −8 to −2 m/s in 3 s. Calculate average acceleration and say whether speed increases.
A hint
The sign of change is not the sign of velocity.
See the worked solution
Δv = −2 − (−8) = +6 m/s; average a = +6/3 = +2 m/s². Speed decreases from 8 to 2 m/s. Positive acceleration points opposite to motion.

Mistakes worth avoiding

The position–time graph draws the road.
The graph relates two quantities; a curve can correspond to motion on a straight road.
Negative velocity means a body is slowing down.
The sign indicates direction; slowing depends on the relationship between velocity and acceleration.
Average speed is always the magnitude of average velocity.
They are equal only when distance travelled equals displacement magnitude.

The essentials to take with you

  • State reference frame, origin and positive direction.
  • Displacement connects endpoints; distance adds the path.
  • Averages belong to intervals; instantaneous values belong to instants.
  • In x(t), slope is velocity; in v(t), area is displacement.
  • Acceleration measures velocity change.

If you still cannot explain one of these ideas, return to the corresponding section and work through an example again. Recognising the difficulty is part of studying.

References for further study

Text and examples prepared for this learning path. These sources help check the principles and continue studying:

The organisation by school year is a study proposal. Pedagogical review by teachers remains a stage of the project.