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.
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
- The first segment is 60 m; the return is 60 − 20 = 40 m. Total distance is 100 m.
- Displacement uses only endpoints: Δx = 20 − 0 = +20 m.
- Average velocity is 20/50 = +0,40 m/s. Average speed is 100/50 = 2,0 m/s.
Example 2
A straight line with an initial position
- The duration between the data points is 6 s and displacement 18 m. Slope is v = +3 m/s.
- From t = 2 s to t = 5 s, 3 s pass. Intermediate displacement is 3 × 3 = 9 m.
- Add to the reference position: x = 8 + 9 = 17 m.
Example 3
Area and slope give different results
- During the start, area is triangular: Δx₁ = 4 × 6/2 = 12 m.
- During the uniform segment, area is rectangular: Δx₂ = 3 × 6 = 18 m.
- Add areas: Δx = 30 m. Initial acceleration, obtained from slope, is 6/4 = 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
A hint
See the worked solution
Exercise 2
A hint
See the worked solution
Exercise 3
A hint
See the worked solution
Exercise 4
A hint
See the worked solution
Exercise 5
A hint
See the worked solution
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:
- OpenStax, College Physics 2e — 2.1 Displacement
Reference frame, position, distance and displacement.
- OpenStax, College Physics 2e — 2.3 Time, Velocity, and Speed
Duration and the distinction between averages and instantaneous values.
- OpenStax, College Physics 2e — 2.4 Acceleration
Acceleration and interpreting signs.
- OpenStax, College Physics 2e — 2.8 Graphical Analysis of One-Dimensional Motion
Slopes and graphical representations of motion.
The organisation by school year is a study proposal. Pedagogical review by teachers remains a stage of the project.