Displacement-Time Graphs: How to Read Motion in HSC Physics

Learn how to read displacement, slope, direction, rest, and changing velocity from displacement-time graphs, with worked examples and common misconceptions explained.

September 17, 2026 12 min read

A displacement-time graph can look simple, but it is easy to read it completely backwards.

Suppose a person walks away from a starting point, stops for five seconds, then walks back past where they started. What should the graph look like?

A common first guess is that the line on the graph should somehow trace the person’s path. It does not. The graph is not a picture of the journey. It records where the person is relative to an origin at each moment in time.

That one idea makes almost everything else on a displacement-time graph easier to understand.

01What the axes are actually telling you

On a displacement-time graph:

the horizontal axis is time, usually measured in seconds (s)
the vertical axis is displacement, usually measured in metres (m)

Displacement tells you the object’s position relative to a chosen origin, including direction.

For example:

\(x = +6\text{ m}\) means the object is 6 m on the positive side of the origin
\(x = 0\text{ m}\) means the object is at the origin
\(x = -4\text{ m}\) means the object is 4 m on the negative side of the origin

Here, \(x\) represents displacement.

If the difference between distance and displacement is still a little fuzzy, it is worth fixing that first in Distance and Displacement in HSC Physics Explained Clearly.

Prediction: what does a point on the graph mean?

Imagine the graph contains the point \((3\text{ s}, 8\text{ m})\).

Does this mean the object travelled 8 m during the third second?

No. It means that at \(t = 3\text{ s}\), the object’s displacement is \(+8\text{ m}\).

A single point tells you position at one instant. To work out how the object is moving, you need to look at how the graph changes over time.

02Slope tells you velocity

Consider an object that moves from \(x = 2\text{ m}\) at \(t = 1\text{ s}\) to \(x = 10\text{ m}\) at \(t = 5\text{ s}\).

Its displacement has changed by:

\[
\Delta x = 10 – 2 = 8\text{ m}
\]

over a time interval of:

\[
\Delta t = 5 – 1 = 4\text{ s}
\]

So its average velocity is:

\[
v = \frac{\Delta x}{\Delta t}
= \frac{8}{4}
= 2\text{ m s}^{-1}
\]

That calculation is exactly the same as finding the slope, or gradient, of the line.

For a displacement-time graph:

\[
v = \frac{\Delta x}{\Delta t}
\]

where:

\(v\) is velocity in metres per second (\(\text{m s}^{-1}\))
\(\Delta x\) is the change in displacement in metres (m)
\(\Delta t\) is the change in time in seconds (s)

So the key rule is:

The slope of a displacement-time graph represents velocity.

This is not just a mathematical trick. If displacement changes rapidly, the object is moving quickly. If displacement changes slowly, the object is moving slowly.

Question: which object is moving faster?

Object A’s graph rises by 12 m in 3 s.

Object B’s graph rises by 10 m in 5 s.

Which object has the greater speed?

Answer

For A:

\[
v_A = \frac{12}{3} = 4\text{ m s}^{-1}
\]

For B:

\[
v_B = \frac{10}{5} = 2\text{ m s}^{-1}
\]

Object A is moving faster because its graph has the steeper slope.

The important comparison is not which line is higher on the page. It is which line changes displacement more rapidly with time.

03Positive slope, negative slope, and direction

Velocity is a vector quantity, so its sign tells you direction.

A line sloping upwards from left to right has a positive slope. That means positive velocity.

A line sloping downwards from left to right has a negative slope. That means negative velocity.

This is one of the most useful things a displacement-time graph can show.

Shape of graph	Velocity	Motion
Straight line sloping upwards	Positive and constant	Moving in the positive direction
Horizontal line	Zero	Stationary
Straight line sloping downwards	Negative and constant	Moving in the negative direction
Curve getting steeper	Velocity magnitude is increasing	Speed is increasing
Curve getting flatter	Velocity magnitude is decreasing	Speed is decreasing

The tempting mistake: “downwards” means the object is moving down

A student sees a line sloping downwards and thinks, “The object must be travelling downhill.”

That feels reasonable because the graph looks like a hill.

But remember, the graph is not a map.

If displacement falls from \(+8\text{ m}\) to \(+3\text{ m}\), the object is moving in the negative direction along whatever one-dimensional axis we chose. That could mean west, left, backwards, or towards the origin.

Nothing about the graph tells us the object is physically moving downwards.

Think of the graph as a diary rather than a photograph. Each point says, “At this time, the object was here.” The line joins those position records together.

That analogy has a limit. A real graph can represent continuously changing motion, not just a collection of diary entries.

04A horizontal line means the object is at rest

Imagine the graph stays at \(x = 5\text{ m}\) from \(t = 2\text{ s}\) to \(t = 7\text{ s}\).

During those five seconds, time keeps passing, but displacement does not change.

So:

\[
\Delta x = 0
\]

and therefore:

\[
v = \frac{0}{\Delta t} = 0
\]

The object is stationary.

Notice what the horizontal line does not mean. It does not mean the object has returned to the origin.

If the line is horizontal at \(+5\text{ m}\), the object is resting 5 m from the origin.

Question: what if the graph lies along the time axis?

Suppose the displacement remains \(0\text{ m}\) for four seconds.

What is the object doing?

Answer

It is stationary at the origin.

A horizontal line anywhere means zero velocity. Its vertical position simply tells you where the object is resting.

05Worked example: reading constant velocity from a graph

An object’s displacement changes from \(-3\text{ m}\) at \(t = 2\text{ s}\) to \(+9\text{ m}\) at \(t = 8\text{ s}\). The section of the displacement-time graph between these points is a straight line.

Calculate the object’s velocity and describe its motion.

Step 1Find the change in displacement

\[
\Delta x = x_f – x_i
= 9 – (-3)
= 12\text{ m}
\]

The object has changed its displacement by \(+12\text{ m}\).

Step 2Find the change in time

\[
\Delta t = t_f – t_i
= 8 – 2
= 6\text{ s}
\]

Step 3Calculate the slope

\[
v = \frac{\Delta x}{\Delta t}
= \frac{12}{6}
= 2.0\text{ m s}^{-1}
\]

Step 4Interpret the result

The velocity is:

\[
\boxed{v = +2.0\text{ m s}^{-1}}
\]

The positive sign means the object is moving in the positive direction.

Because the graph is a straight line, the slope is constant, so the object’s velocity is constant throughout this interval.

The fact that the object starts at \(-3\text{ m}\) does not mean it has negative velocity. Position and velocity are different quantities.

06Crossing the displacement axis is not the same as changing direction

Suppose an object moves from \(-5\text{ m}\) to \(+5\text{ m}\) in a straight line on the graph.

At some point, the line crosses \(x = 0\).

What happened physically?

The object passed through the origin.

Did it change direction there?

No.

If the slope stays positive before and after crossing \(x = 0\), its velocity remains positive. It simply passes through the origin and keeps going.

This gives an important rule:

Changing the sign of displacement does not necessarily mean changing direction. Changing the sign of velocity does.

You can see this more clearly once displacement, speed, and velocity are kept separate. Speed and Velocity: HSC Physics Explained Clearly develops that distinction further.

07How to recognise a change in direction

An object changes direction when its velocity changes sign.

Since velocity is the slope of a displacement-time graph, look for the slope changing from:

positive to negative, or
negative to positive

Imagine a graph that rises, reaches a maximum displacement, then falls.

Before the highest point:

slope is positive
velocity is positive

At the highest point:

the tangent is horizontal
instantaneous velocity is zero

After the highest point:

slope is negative
velocity is negative

So the object has stopped momentarily and reversed direction.

You might picture someone walking towards a friend, realising they have approached the wrong person, stopping briefly, then awkwardly walking the other way. The graph records their position, not the social damage.

Question: must every point where velocity is zero mean the object stays stopped?

Answer

No.

A horizontal section means the object remains stopped for an interval of time.

A single point with a horizontal tangent can mean the object is stationary only momentarily before reversing direction.

That distinction becomes important on curved graphs.

08Curved displacement-time graphs mean velocity is changing

So far, we have mostly used straight lines.

A straight line has constant slope, which means constant velocity.

A curved line is different. Its slope changes from one moment to the next, so its velocity is changing.

For example, imagine a displacement-time curve that becomes progressively steeper upwards.

Early in the motion, its slope might be:

\[
1\text{ m s}^{-1}
\]

Later, it might be:

\[
3\text{ m s}^{-1}
\]

and later still:

\[
5\text{ m s}^{-1}
\]

The object is speeding up in the positive direction.

Prediction: what if the graph rises but becomes flatter?

Suppose displacement is still increasing, but the curve gradually becomes less steep.

Is the object moving forwards or backwards?

It is still moving in the positive direction, because the slope is still positive.

But its velocity is becoming smaller. It is slowing down.

This is where students often confuse where the graph is going with how steep it is.

The vertical height tells you displacement.

The slope tells you velocity.

09Average velocity and instantaneous velocity

On a straight section of a displacement-time graph, the slope is the same everywhere. There is no problem.

On a curved graph, however, the slope changes continuously.

If you draw a straight line between two points on the curve, its slope gives the average velocity over that time interval:

\[
v_{\text{avg}} = \frac{\Delta x}{\Delta t}
\]

If you want the velocity at one particular instant, you need the slope of the tangent to the curve at that point.

That gives instantaneous velocity.

A tangent is a straight line that matches the direction of the curve at the chosen instant.

The idea is explored in more detail in Average and Instantaneous Velocity in HSC Physics.

10Worked example: reading a multi-stage journey

A student’s displacement is described by the following displacement-time graph information:

at \(t = 0\text{ s}\), \(x = 0\text{ m}\)
at \(t = 4\text{ s}\), \(x = 12\text{ m}\)
from \(t = 4\text{ s}\) to \(t = 7\text{ s}\), displacement remains at \(12\text{ m}\)
at \(t = 13\text{ s}\), \(x = -6\text{ m}\)

Each section is a straight line.

Find the velocity during each stage, identify when the student is stationary, and determine whether the student passes the origin during the final stage.

Step 1Calculate the velocity from 0 s to 4 s

\[
v_1
= \frac{12 – 0}{4 – 0}
= \frac{12}{4}
= +3.0\text{ m s}^{-1}
\]

The student moves in the positive direction at a constant velocity of \(3.0\text{ m s}^{-1}\).

Step 2Analyse the section from 4 s to 7 s

The displacement stays at \(12\text{ m}\), so:

\[
\Delta x = 0
\]

Therefore:

\[
v_2 = 0\text{ m s}^{-1}
\]

The student is stationary for 3 s at a displacement of \(+12\text{ m}\).

Step 3Calculate the velocity from 7 s to 13 s

The displacement changes from \(+12\text{ m}\) to \(-6\text{ m}\):

\[
\Delta x = -6 – 12 = -18\text{ m}
\]

The time interval is:

\[
\Delta t = 13 – 7 = 6\text{ s}
\]

Therefore:

\[
v_3
= \frac{-18}{6}
= -3.0\text{ m s}^{-1}
\]

The student moves in the negative direction at a constant velocity of \(3.0\text{ m s}^{-1}\).

Step 4Determine whether the student passes the origin

Yes.

The displacement changes continuously from \(+12\text{ m}\) to \(-6\text{ m}\), so it must pass through \(x = 0\).

We can even find when.

From \(+12\text{ m}\), travelling at \(-3.0\text{ m s}^{-1}\), it takes:

\[
t = \frac{12\text{ m}}{3.0\text{ m s}^{-1}}
= 4.0\text{ s}
\]

The return journey begins at \(t = 7\text{ s}\), so the student reaches the origin at:

\[
7 + 4 = 11\text{ s}
\]

The graph would therefore cross \(x = 0\) at:

\[
\boxed{t = 11\text{ s}}
\]

Notice that the student does not change direction at \(t = 11\text{ s}\). The slope is still negative. They simply pass through the origin.

11Steeper means faster, but watch the sign

Suppose two straight sections have slopes:

\[
+4\text{ m s}^{-1}
\]

and

\[
-7\text{ m s}^{-1}
\]

Which represents the faster motion?

The second one.

Speed is the magnitude of velocity:

\[
|-7| = 7\text{ m s}^{-1}
\]

so an object travelling at \(-7\text{ m s}^{-1}\) moves faster than one travelling at \(+4\text{ m s}^{-1}\).

On the graph, the \(-7\text{ m s}^{-1}\) section would be steeper, even though it slopes downwards.

Question: compare these three slopes

Three objects have velocities:

\[
v_A = +2\text{ m s}^{-1}, \qquad
v_B = -5\text{ m s}^{-1}, \qquad
v_C = 0\text{ m s}^{-1}
\]

Which graph is steepest, and which object is stationary?

Answer

Object B’s graph is steepest because its speed is:

\[
|v_B| = 5\text{ m s}^{-1}
\]

Object C is stationary because:

\[
v_C = 0\text{ m s}^{-1}
\]

Its displacement-time graph is horizontal.

12A useful way to read any displacement-time graph

When you meet an unfamiliar graph, do not try to understand the whole thing at once.

Read it from left to right and ask the same questions for each section:

Where is the object?
Read the displacement from the vertical axis.
Is displacement increasing, decreasing, or constant?
This tells you the sign of the velocity.
How steep is the graph?
This tells you the magnitude of the velocity.
Is the slope constant or changing?
A constant slope means constant velocity. A changing slope means changing velocity.
Does the slope change sign?
If it does, the object changes direction.

This method is much safer than trying to memorise graph shapes.

13The misconception that causes the most trouble

Consider a displacement-time graph that rises steeply, then becomes horizontal.

A student might say:

“The object moved quickly, then its displacement became constant, so its velocity stayed constant.”

The first half is right. The second half confuses displacement with velocity.

A constant displacement means the object’s position is not changing.

Therefore:

\[
\Delta x = 0
\]

so:

\[
v = 0
\]

The object has stopped.

To show constant non-zero velocity, the graph must have a constant non-zero slope, which appears as a straight line tilted upwards or downwards.

Keep those two ideas separate:

constant displacement means zero velocity
constant slope means constant velocity

14From displacement-time graphs to acceleration

A displacement-time graph does not show acceleration directly, but it gives you the clue you need.

Acceleration means velocity is changing.

Since slope represents velocity, a displacement-time graph with a changing slope tells you that the object is accelerating.

A straight line means the slope is constant, so velocity is constant and acceleration is zero.

A curve means the slope changes, so the velocity changes and acceleration is non-zero.

That is the bridge to the next major graph in kinematics: the velocity-time graph. Once you can look at a displacement-time graph and mentally read its slope as velocity, velocity-time graphs become much easier to interpret.