How to Match a Written Journey to Three Motion Graphs

A trolley moves to the right, slows down, stops momentarily, and then rolls to the left. You are given three graphs and asked to match them to the journey. All three graphs have different shapes, but which features should you look for?

The reliable way to match a written journey to displacement-time, velocity-time, and acceleration-time graphs is to break the journey into time intervals and translate each interval into three quantities: position, velocity, and acceleration.

Remember these three relationships:

• The gradient of a displacement-time graph gives velocity.
• The gradient of a velocity-time graph gives acceleration.
• The signed area under a velocity-time graph gives displacement, while the signed area under an acceleration-time graph gives the change in velocity.

These relationships let you build all three graphs from the same description and check that they agree.

Before going further, make one prediction.

A trolley is still moving to the right, but it is slowing down. Does its displacement-time graph slope upwards or downwards?

It slopes upwards. The trolley is still moving in the positive direction, so its displacement is increasing. However, the gradient becomes progressively smaller because the trolley is slowing down.

Its velocity-time graph slopes downwards, and its acceleration-time graph lies below zero.

This is the first important distinction: the direction of motion is determined by velocity, not acceleration.

01Start by translating the written journey

Before drawing anything, choose a positive direction.

For the examples in this article, take right as positive. Therefore:

• Positive displacement means a position to the right of the origin.
• Positive velocity means moving right.
• Negative velocity means moving left.
• Positive acceleration means acceleration towards the right.
• Negative acceleration means acceleration towards the left.

We will use:

• \(x\) = position or displacement coordinate, in metres (m)
• \(v\) = velocity, in metres per second (\(\mathrm{m\,s^{-1}}\))
• \(a\) = acceleration, in metres per second squared (\(\mathrm{m\,s^{-2}}\))
• \(t\) = time, in seconds (s)

The position coordinate \(x\) is measured from a fixed origin. If the object starts at the origin, its position coordinate is also its displacement from the starting point.

Recognise the important words

Most written motion descriptions contain a small number of phrases that tell you what the graphs should look like.

Notice the difference between speeding up and slowing down.

An object speeds up when the magnitude of its velocity increases. Its velocity and acceleration have the same sign.

An object slows down when the magnitude of its velocity decreases. Its velocity and acceleration have opposite signs.

This gives you a useful rule:

\[
\begin{aligned}
v>0,\ a>0 &: \text{right, speeding up}\\
v>0,\ a<0 &: \text{right, slowing down}\\
v<0,\ a<0 &: \text{left, speeding up}\\
v<0,\ a>0 &: \text{left, slowing down}
\end{aligned}
\]

Be careful with the word negative. Negative acceleration does not automatically mean that an object is travelling left.

2. Understand what each graph actually shows

The three graphs describe different aspects of the same motion.

There is another useful difference: the areas under two of these graphs have physical meanings.

For a velocity-time graph:

\[
\boxed{\text{Signed area}=\Delta x}
\]

For an acceleration-time graph:

\[
\boxed{\text{Signed area}=\Delta v}
\]

Here, \(\Delta x\) is the change in position, and \(\Delta v\) is the change in velocity.

The units confirm these relationships.

\[
(\mathrm{m\,s^{-1}})(\mathrm{s})=\mathrm{m}
\]

\[
(\mathrm{m\,s^{-2}})(\mathrm{s})
=\mathrm{m\,s^{-1}}
\]

The area under a displacement-time graph is not used to find displacement. For that graph, focus on the gradient.

What should you look for in each graph?

Displacement-time graph

• Positive gradient: moving right.
• Negative gradient: moving left.
• Increasing positive gradient: moving right and speeding up.
• Decreasing positive gradient: moving right and slowing down.
• Horizontal section: stationary.
• Maximum or minimum point: an instant when velocity is zero, potentially marking a change of direction.

Velocity-time graph

• Above the horizontal axis: moving right.
• Below the horizontal axis: moving left.
• Horizontal section: constant velocity.
• Positive gradient: positive acceleration.
• Negative gradient: negative acceleration.
• Crossing the horizontal axis: velocity changes sign, so the object reverses direction if velocity is continuous.

Acceleration-time graph

• Above the horizontal axis: positive acceleration.
• Below the horizontal axis: negative acceleration.
• On the horizontal axis: zero acceleration.
• Horizontal section above or below zero: constant, non-zero acceleration.
• Area above or below the axis: positive or negative change in velocity.

A horizontal velocity-time graph does not necessarily mean the object is stationary. It means its velocity is constant. A horizontal acceleration-time graph also does not necessarily mean zero acceleration.

03Use one timeline to construct all three graphs

Let’s work through a complete written journey.

A trolley starts at position \(x=0\) and moves along a straight track.

Its journey is described as follows:

1. It moves right at a constant speed of \(3\ \mathrm{m\,s^{-1}}\) for 2 seconds.
2. It continues right but slows uniformly to rest over the next 2 seconds.
3. It accelerates left, reaching a velocity of \(2\ \mathrm{m\,s^{-1}}\) to the left after another 2 seconds.
4. It continues left at a constant speed of \(2\ \mathrm{m\,s^{-1}}\) for 2 seconds.
5. It continues left but slows uniformly to rest over the next 2 seconds.
6. It remains stationary for the final second.

All numerical data here are illustrative. Take right as positive and assume constant acceleration during each interval.

Step 1: Divide the journey into intervals

The important times are:

\[
t=0,\ 2,\ 4,\ 6,\ 8,\ 10,\ 11\text{ s}
\]

Each time marks a change in the trolley’s motion.

Now translate the written description into velocity and acceleration.

The acceleration values come from:

\[
a=\frac{v_f-v_i}{\Delta t}
\]

For example, between 2 and 4 seconds:

\[
a=\frac{0-3}{2}=-1.5\ \mathrm{m\,s^{-2}}
\]

Between 8 and 10 seconds:

\[
a=\frac{0-(-2)}{2}
=+1.0\ \mathrm{m\,s^{-2}}
\]

Notice that the trolley has negative acceleration between 2 and 6 seconds, but it does not travel left throughout that entire period. It first slows down while moving right, then reverses.

Step 2: Construct the velocity-time graph

The velocity-time graph is usually the easiest graph to construct directly from a written journey.

Follow the velocity in chronological order.

• From 0-2 seconds, velocity remains at \(+3\). Draw a horizontal line above zero.
• From 2-4 seconds, velocity decreases uniformly to zero. Draw a straight descending line.
• From 4-6 seconds, velocity becomes increasingly negative, reaching \(-2\). Continue the descending line.
• From 6-8 seconds, velocity remains at \(-2\). Draw a horizontal line below zero.
• From 8-10 seconds, velocity increases towards zero. Draw a straight ascending line.
• From 10-11 seconds, velocity remains zero.

The graph must cross the horizontal axis at \(t=4\text{ s}\). This represents the instant when the trolley stops and reverses direction.

Step 3: Construct the acceleration-time graph

Now use the gradient of the velocity-time graph.

The velocity is constant from 0-2 seconds, so acceleration is zero.

From 2-4 seconds, velocity decreases at a constant rate, so acceleration is \(-1.5\ \mathrm{m\,s^{-2}}\).

From 4-6 seconds, velocity continues decreasing, but at a different constant rate. Acceleration is now \(-1.0\ \mathrm{m\,s^{-2}}\).

Continue this process for the remaining intervals.

The acceleration-time graph consists of horizontal sections at:

\[
0,\ -1.5,\ -1,\ 0,\ +1,\ 0
\]

The values correspond to the six successive time intervals.

Step 4: Construct the displacement-time graph

This is where many students make mistakes.

Remember that the gradient of a displacement-time graph represents velocity.

Therefore:

• Positive velocity means the graph rises.
• Negative velocity means the graph falls.
• Greater speed means a steeper gradient.
• Zero velocity means a horizontal tangent.

Apply those rules to the trolley.

From 0-2 seconds, the trolley moves right at constant speed. Its displacement-time graph is a straight rising line.

From 2-4 seconds, it continues moving right but slows down. The graph continues rising, but becomes progressively flatter.

At \(t=4\text{ s}\), the graph reaches a maximum because the trolley momentarily stops before reversing.

From 4-6 seconds, the trolley moves left and speeds up. The graph falls and becomes progressively steeper.

From 6-8 seconds, it moves left at constant speed. The graph is a straight falling line.

From 8-10 seconds, it continues left but slows down. The graph still falls, but becomes progressively flatter.

Finally, from 10-11 seconds, it is stationary, so the graph becomes horizontal.

Step 5: Check the signed areas

The displacement-time graph should agree with the signed areas under the velocity-time graph.

For this journey:

The total displacement is:

\[
\begin{aligned}
\Delta x&=6+3-2-4-2\\
&=+1\text{ m}
\end{aligned}
\]

The trolley finishes 1 metre to the right of its starting position.

Its total distance is different because distance adds the magnitudes of all the areas:

\[
d=6+3+2+4+2
\]

\[
d=17\text{ m}
\]

This gives you a useful consistency check. If your displacement-time graph finishes at a position other than \(+1\text{ m}\), something is wrong with your graph or calculations.

For more practice with this distinction, see How to Calculate Distance When a Velocity-Time Graph Crosses Zero.

04The important contrast: negative acceleration does not always mean slowing down

Consider two different journeys.

Journey A: A cyclist travels right but slows down.

Journey B: A cyclist travels left and speeds up.

Both cyclists have negative acceleration.

However, their displacement-time graphs look quite different.

The reason is simple: acceleration tells you how velocity changes, not the direction in which the object is currently travelling.

There are three particularly common graph-reading mistakes to avoid:

1. Confusing a negative gradient with negative velocity. A negative gradient on a velocity-time graph means negative acceleration. A negative gradient on a displacement-time graph means negative velocity.
2. Confusing a horizontal graph with zero motion. A horizontal displacement-time graph means the object is stationary. A horizontal velocity-time graph can represent an object moving at constant speed.
3. Confusing zero velocity with zero displacement. An object can have zero velocity while being several metres from its starting position.

That last point is particularly important when a velocity-time graph crosses zero. The object may be momentarily stationary at its turning point without having returned to its starting position.

05Worked examples

05Questions and solutions

The following questions move from straightforward graph matching to more challenging problems that require you to check whether three different representations describe the same physical journey.

Question 1

A delivery robot starts at position \(x=0\).

Its journey is:

• From 0-3 seconds, it travels right at a constant velocity of \(+2.0\ \mathrm{m\,s^{-1}}\).
• From 3-5 seconds, it continues right but slows uniformly to rest.
• From 5-6 seconds, it remains stationary.

Take right as positive.

Describe the displacement-time, velocity-time, and acceleration-time graphs.

Calculate the robot’s acceleration during the braking interval and its final position.

Solution 1

The robot finishes at \(x=+8.0\text{ m}\). Its acceleration during braking is \(-1.0\ \mathrm{m\,s^{-2}}\).

Step 1: Calculate the acceleration.

The initial velocity during braking is \(+2.0\ \mathrm{m\,s^{-1}}\), and the final velocity is zero.

\[
\begin{aligned}
a&=\frac{v_f-v_i}{\Delta t}\\
&=\frac{0-2.0}{2.0}\\
&=-1.0\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

The negative acceleration opposes the positive velocity, so the robot slows down.

Step 2: Calculate the displacement.

From 0-3 seconds:

\[
\Delta x_1=(2.0)(3.0)=+6.0\text{ m}
\]

From 3-5 seconds, the velocity decreases uniformly from \(+2.0\) to zero.

The area under the velocity-time graph is a triangle:

\[
\Delta x_2=\frac12(2.0)(2.0)=+2.0\text{ m}
\]

From 5-6 seconds, displacement is zero.

Therefore:

\[
\Delta x=6.0+2.0=+8.0\text{ m}
\]

Step 3: Describe the graphs.

• Displacement-time: a straight rising line from 0-3 seconds, followed by a rising curve that gradually flattens, then a horizontal section.
• Velocity-time: a horizontal line at \(+2.0\), followed by a straight descending line to zero, then a line along zero.
• Acceleration-time: zero from 0-3 seconds, \(-1.0\ \mathrm{m\,s^{-2}}\) from 3-5 seconds, and zero from 5-6 seconds.

The robot never reverses direction. Its velocity approaches zero but never becomes negative.

Question 2

An elevator is moving downwards at \(1.0\ \mathrm{m\,s^{-1}}\) at \(t=0\).

Take upwards as positive. Its initial position is \(x=+4.0\text{ m}\) relative to a fixed reference floor.

Its acceleration is:

• \(-0.50\ \mathrm{m\,s^{-2}}\) from 0-2 seconds.
• \(+1.0\ \mathrm{m\,s^{-2}}\) from 2-4 seconds.

Determine:

1. Its velocity at 2 seconds.
2. Its velocity at 4 seconds.
3. Its total displacement.
4. Its final position.
5. The shape of each motion graph.

Explain whether the elevator reverses direction.

Solution 2

The elevator finishes at \(x=-1.0\text{ m}\), with zero velocity. It does not reverse direction during the four-second interval.

Step 1: Calculate its velocity at 2 seconds.

Initial velocity is negative because the elevator moves downwards:

\[
v_i=-1.0\ \mathrm{m\,s^{-1}}
\]

Using:

\[
v_f=v_i+at
\]

\[
\begin{aligned}
v_2&=-1.0+(-0.50)(2.0)\\
&=-2.0\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

It is moving downwards at \(2.0\ \mathrm{m\,s^{-1}}\).

Step 2: Calculate its velocity at 4 seconds.

During the second interval, acceleration is positive.

\[
\begin{aligned}
v_4&=-2.0+(1.0)(2.0)\\
&=0\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

The elevator reaches zero velocity at \(t=4.0\text{ s}\).

Although its acceleration is now upwards, its velocity remains negative throughout the second interval. It is still travelling downwards, but slowing down.

Step 3: Calculate displacement during each interval.

For 0-2 seconds, average velocity is:

\[
v_{\mathrm{av}}=\frac{-1.0-2.0}{2}
=-1.5\ \mathrm{m\,s^{-1}}
\]

Therefore:

\[
\Delta x_1=(-1.5)(2.0)=-3.0\text{ m}
\]

For 2-4 seconds:

\[
v_{\mathrm{av}}=\frac{-2.0+0}{2}
=-1.0\ \mathrm{m\,s^{-1}}
\]

Therefore:

\[
\Delta x_2=(-1.0)(2.0)=-2.0\text{ m}
\]

Total displacement:

\[
\Delta x=-3.0-2.0=-5.0\text{ m}
\]

Step 4: Calculate final position.

\[
\begin{aligned}
x_f&=x_i+\Delta x\\
&=4.0-5.0\\
&=-1.0\text{ m}
\end{aligned}
\]

The elevator finishes 1 metre below the reference floor.

Step 5: Describe the graphs.

• Displacement-time: falls with an increasingly negative gradient during the first interval, then continues falling while gradually flattening.
• Velocity-time: decreases linearly from \(-1.0\) to \(-2.0\), then increases linearly to zero.
• Acceleration-time: remains at \(-0.50\ \mathrm{m\,s^{-2}}\), then changes to \(+1.0\ \mathrm{m\,s^{-2}}\).

The tempting mistake is to see positive acceleration during the second interval and assume the elevator must be moving upwards. Acceleration determines how velocity changes, not the current direction of motion.

Question 3

Two laboratory carts, A and B, travel along the same straight track.

Both start at \(x=0\), but their initial velocities are different:

\[
v_{A,0}=+5.0\ \mathrm{m\,s^{-1}}
\]

\[
v_{B,0}=+1.0\ \mathrm{m\,s^{-1}}
\]

Both experience exactly the same acceleration:

• \(-1.0\ \mathrm{m\,s^{-2}}\) from 0-4 seconds.
• Zero acceleration from 4-6 seconds.

Determine:

1. The final velocity of each cart at 4 seconds.
2. The total displacement of each cart from 0-6 seconds.
3. Whether either cart reverses direction.
4. How their three motion graphs compare.

Solution 3

The carts have identical acceleration-time graphs, but their velocity-time and displacement-time graphs are different. Cart A finishes at \(+14\text{ m}\), while cart B finishes at \(-10\text{ m}\).

Step 1: Calculate the velocities at 4 seconds.

For cart A:

\[
\begin{aligned}
v_{A,4}&=5.0+(-1.0)(4.0)\\
&=+1.0\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

For cart B:

\[
\begin{aligned}
v_{B,4}&=1.0+(-1.0)(4.0)\\
&=-3.0\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

Cart A continues moving right. Cart B has reversed direction.

Step 2: Calculate the displacement of cart A.

From 0-4 seconds:

\[
\begin{aligned}
\Delta x_{A,1}
&=\frac{5.0+1.0}{2}(4.0)\\
&=+12\text{ m}
\end{aligned}
\]

From 4-6 seconds, velocity remains at \(+1.0\):

\[
\Delta x_{A,2}=(1.0)(2.0)=+2.0\text{ m}
\]

Therefore:

\[
\Delta x_A=12+2=+14\text{ m}
\]

Step 3: Calculate the displacement of cart B.

From 0-4 seconds:

\[
\begin{aligned}
\Delta x_{B,1}
&=\frac{1.0+(-3.0)}{2}(4.0)\\
&=-4.0\text{ m}
\end{aligned}
\]

From 4-6 seconds:

\[
\Delta x_{B,2}=(-3.0)(2.0)
=-6.0\text{ m}
\]

Therefore:

\[
\Delta x_B=-4.0-6.0=-10\text{ m}
\]

Step 4: Compare the graphs.

The acceleration-time graphs are identical because both carts experience the same acceleration over the same intervals.

The velocity-time graphs have the same gradient from 0-4 seconds, but start at different velocities. They are parallel straight lines, followed by horizontal sections at different values.

The displacement-time graphs are also different.

Cart A continues moving right throughout the journey. Its displacement-time graph rises while becoming progressively flatter, then continues rising in a straight line.

Cart B initially moves right, reaches zero velocity at \(t=1.0\text{ s}\), and then moves left. Its displacement-time graph rises to a maximum before falling, then continues downwards in a straight line.

The key lesson is that an acceleration-time graph alone cannot tell you the complete journey. You also need the initial velocity and initial position.

Question 4

A cart starts at:

\[
x_0=-1.0\text{ m}
\]

\[
v_0=+4.0\ \mathrm{m\,s^{-1}}
\]

Its acceleration is \(-2.0\ \mathrm{m\,s^{-2}}\) from 0-3 seconds, then zero from 3-5 seconds.

A student draws three graphs to represent the motion.

Their velocity-time graph:

• Starts at \(+4.0\ \mathrm{m\,s^{-1}}\).
• Falls linearly to \(-2.0\ \mathrm{m\,s^{-1}}\) at 3 seconds.
• Remains at \(-2.0\ \mathrm{m\,s^{-1}}\) until 5 seconds.

Their acceleration-time graph:

• Remains at \(-2.0\ \mathrm{m\,s^{-2}}\) from 0-3 seconds.
• Remains at zero from 3-5 seconds.

Their displacement-time graph:

• Reaches a maximum at 2 seconds.
• Passes through \(x=+2.0\text{ m}\) at 3 seconds.
• Then remains horizontal at \(x=+2.0\text{ m}\) until 5 seconds.

Is the student’s set of graphs consistent?

Identify which features are correct, explain the error, and calculate the correct final position.

Solution 4

The velocity-time graph and acceleration-time graph are correct. The displacement-time graph is correct up to 3 seconds, but its final horizontal section is wrong. The correct final position is \(-2.0\text{ m}\).

Step 1: Check the velocity-time graph.

The acceleration is \(-2.0\ \mathrm{m\,s^{-2}}\), so:

\[
v=4.0-2.0t
\]

At \(t=3\text{ s}\):

\[
v=4.0-2.0(3.0)
=-2.0\ \mathrm{m\,s^{-1}}
\]

After 3 seconds, acceleration is zero, so velocity remains at \(-2.0\ \mathrm{m\,s^{-1}}\).

The student’s velocity-time graph is correct.

Step 2: Check the turning point.

The cart reaches zero velocity when:

\[
0=4.0-2.0t
\]

\[
t=2.0\text{ s}
\]

Therefore, the displacement-time graph must have a maximum at 2 seconds.

The displacement by this time is the area under the velocity-time graph:

\[
\Delta x=\frac12(2.0)(4.0)
=+4.0\text{ m}
\]

The position is:

\[
x=-1.0+4.0=+3.0\text{ m}
\]

The cart reaches its maximum position at \(+3.0\text{ m}\).

Step 3: Check the position at 3 seconds.

Between 2 and 3 seconds, velocity changes from zero to \(-2.0\).

The signed triangular area is:

\[
\Delta x=-\frac12(1.0)(2.0)
=-1.0\text{ m}
\]

Therefore:

\[
x(3)=3.0-1.0=+2.0\text{ m}
\]

The student’s graph is correct at this point.

Step 4: Correct the final section.

From 3-5 seconds, velocity is constant at \(-2.0\ \mathrm{m\,s^{-1}}\).

Therefore, the displacement-time graph must be a straight descending line with gradient \(-2.0\ \mathrm{m\,s^{-1}}\).

The displacement during this interval is:

\[
\Delta x=(-2.0)(2.0)
=-4.0\text{ m}
\]

The final position is:

\[
x_f=2.0-4.0=-2.0\text{ m}
\]

The student’s mistake was treating zero acceleration as zero velocity. Zero acceleration means velocity remains constant. Since the cart is moving left at 3 seconds, it must continue moving left afterwards.

Question 5

A cart starts at the origin with zero velocity.

Take right as positive.

An idealised accelerometer records the following motion:

Assume the acceleration is constant within each interval and changes abruptly at the boundaries.

At \(t=4\text{ s}\), two sensors report:

• Position: \(x=0\).
• Velocity: \(v=0\).

A student argues that these measurements cannot all be correct because the signed area under the acceleration-time graph is zero, so the cart must have zero displacement.

Another student argues that the cart must return to the origin at \(t=2\text{ s}\), because its velocity is zero at that instant.

Are either of these students correct?

Determine:

1. Every time the cart has zero velocity.
2. Every time the cart is at the origin.
3. Its maximum position.
4. Its final velocity and displacement.
5. Its total distance travelled.

Describe the three motion graphs and explain why the two students’ arguments fail.

Solution 5

Both students are incorrect. All the sensor measurements are consistent.

The cart starts and finishes at the origin, but it travels a total distance of 4.0 metres. It reaches its maximum position of \(+2.0\text{ m}\) at \(t=2.0\text{ s}\).

This problem requires you to use both types of signed area.

Step 1: Determine the velocity at each important time.

The initial velocity is zero.

From 0-1 seconds:

\[
\begin{aligned}
\Delta v_1&=a\Delta t\\
&=(+2.0)(1.0)\\
&=+2.0\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

Therefore:

\[
v(1)=+2.0\ \mathrm{m\,s^{-1}}
\]

From 1-3 seconds:

\[
\begin{aligned}
\Delta v_2&=(-2.0)(2.0)\\
&=-4.0\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

Therefore:

\[
v(3)=2.0-4.0=-2.0\ \mathrm{m\,s^{-1}}
\]

From 3-4 seconds:

\[
\begin{aligned}
\Delta v_3&=(+2.0)(1.0)\\
&=+2.0\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

Therefore:

\[
v(4)=-2.0+2.0=0
\]

The cart finishes at rest.

Its velocity is zero at:

\[
\boxed{t=0,\ 2,\ 4\text{ s}}
\]

At 2 seconds, velocity changes from positive to negative. The cart momentarily stops and reverses direction.

Step 2: Calculate the displacement.

Use the signed areas under the velocity-time graph.

From 0-1 seconds, velocity increases from zero to \(+2.0\).

The area is a positive triangle:

\[
\Delta x_1=\frac12(1.0)(2.0)
=+1.0\text{ m}
\]

From 1-2 seconds, velocity decreases from \(+2.0\) to zero.

\[
\Delta x_2=\frac12(1.0)(2.0)
=+1.0\text{ m}
\]

From 2-3 seconds, velocity decreases from zero to \(-2.0\).

\[
\Delta x_3=-\frac12(1.0)(2.0)
=-1.0\text{ m}
\]

From 3-4 seconds, velocity increases from \(-2.0\) to zero.

\[
\Delta x_4=-\frac12(1.0)(2.0)
=-1.0\text{ m}
\]

The total displacement is:

\[
\begin{aligned}
\Delta x&=1.0+1.0-1.0-1.0\\
&=0\text{ m}
\end{aligned}
\]

The cart returns to its starting position.

Step 3: Determine its maximum position.

The cart moves right during the first two intervals, reaching its furthest position at \(t=2\text{ s}\).

Its displacement at this point is:

\[
\begin{aligned}
x(2)&=1.0+1.0\\
&=+2.0\text{ m}
\end{aligned}
\]

The cart is therefore 2.0 metres to the right of the origin when its velocity reaches zero.

It then reverses and travels back towards the origin.

This is why the second student’s reasoning fails: zero velocity identifies an instant of rest, not a position.

The cart returns to the origin at \(t=4\text{ s}\), not at \(t=2\text{ s}\).

Step 4: Calculate the total distance.

Distance adds the magnitudes of all four displacement contributions.

\[
\begin{aligned}
d&=1.0+1.0+1.0+1.0\\
&=4.0\text{ m}
\end{aligned}
\]

The cart travels 4.0 metres, even though its net displacement is zero.

Step 5: Explain the acceleration-time graph.

The signed area under the acceleration-time graph gives the change in velocity:

\[
\begin{aligned}
\Delta v&=(+2.0)(1.0)\\
&\quad+(-2.0)(2.0)\\
&\quad+(+2.0)(1.0)\\
&=0\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

This tells us that the final velocity equals the initial velocity.

It does not tell us the displacement.

To find displacement, we need the signed area under the velocity-time graph, which we calculated separately.

The important distinction is that the acceleration-time graph tells you how velocity changes, while the velocity-time graph tells you how position changes.

Even when an object finishes with exactly the same position and velocity it started with, it may have travelled a considerable distance in between.

06What to do when matching graphs in an exam

When you are given a written journey and three possible motion graphs, use this sequence.

1. Choose a positive direction. Write down which direction is positive before interpreting any negative values.
2. Split the journey into time intervals. Mark every change in direction, speed, or acceleration.
3. Determine the velocity. Decide whether the object is moving forwards, backwards, or is stationary, and whether its speed is increasing or decreasing.
4. Construct the velocity-time graph. Use the velocity signs and changes to establish its shape.
5. Construct the acceleration-time graph. Use the gradient of the velocity-time graph.
6. Construct the displacement-time graph. Use velocity to determine the direction and steepness of its gradient.
7. Check the areas. The signed area under the velocity-time graph must equal the change in position. The signed area under the acceleration-time graph must equal the change in velocity.
8. Check the endpoints. Make sure the initial and final positions and velocities agree with the written description.

For a broader explanation of how force, acceleration, velocity, and displacement connect, read Force, Motion and Graph Representations for HSC Physics.

Once you can reliably match all three motion graphs, the next useful step is to connect acceleration to the net force using Newton’s second law:

\[
F_{\mathrm{net}}=ma
\]

That lets you move from describing an object’s motion to explaining what causes its velocity to change.