Why a Negative Velocity-Time Gradient Does Not Always Mean Slowing Down

A car is travelling to the right at 12 m/s. Five seconds later, it is still travelling to the right, but its velocity has fallen to 5 m/s. Its velocity-time graph has a negative gradient.

Now imagine a second car travelling to the left. Its velocity changes from −5 m/s to −12 m/s over the same five seconds. Its velocity-time graph also has a negative gradient.

Both cars have negative acceleration. But the first car is slowing down, while the second is speeding up.

**A negative gradient on a velocity-time graph means negative acceleration. It does not automatically mean decreasing speed.** To decide whether an object is speeding up or slowing down, you must consider both the sign of its velocity and the sign of its acceleration.

This is one of the most important distinctions when interpreting motion graphs in HSC Physics.

## 1. What does a negative gradient actually tell you?

The gradient of a velocity-time graph represents acceleration. It tells you how quickly velocity changes, including its direction.

The equation is:

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

where:

- \(a\) is acceleration, measured in \(\mathrm{m\,s^{-2}}\).
- \(v_i\) is initial velocity, measured in \(\mathrm{m\,s^{-1}}\).
- \(v_f\) is final velocity, measured in \(\mathrm{m\,s^{-1}}\).
- \(\Delta t\) is the time interval, measured in seconds.

A negative gradient means the acceleration is negative under your chosen sign convention. It tells you that velocity is becoming more negative.

It does not tell you whether the magnitude of velocity is increasing or decreasing.

Remember that speed is the magnitude of velocity:

\[
\text{speed}=|v|
\]

For example, changing velocity from \(+12\) to \(+5\ \mathrm{m\,s^{-1}}\) decreases speed from 12 to 5 m/s.

However, changing velocity from \(-5\) to \(-12\ \mathrm{m\,s^{-1}}\) increases speed from 5 to 12 m/s.

The negative sign describes direction. The magnitude describes how fast the object is moving.

[IMAGE: Two contrasting velocity-time graphs side by side, both with straight negative gradients. In the first, velocity decreases from +12 m/s to +5 m/s, remaining above the time axis, with a label showing speed decreases. In the second, velocity decreases from -5 m/s to -12 m/s, remaining below the time axis, with a label showing speed increases. Include clearly marked positive and negative velocity regions, time axes, and gradient arrows.]

### The sign test: compare velocity with acceleration

You can work out whether an object is speeding up or slowing down by comparing the signs of its velocity and acceleration.

Use this rule:

- Same signs: speed increases.
- Opposite signs: speed decreases.

For one-dimensional motion, this gives you four possible combinations.

| Velocity | Acceleration | Effect on speed |
|---|---|---|
| Positive | Positive | Increases |
| Positive | Negative | Decreases |
| Negative | Positive | Decreases |
| Negative | Negative | Increases |

Why does this work?

Imagine pushing a trolley towards the east. If the trolley is moving east and the acceleration is east, it gains speed. If the acceleration points west, it loses speed.

Now turn the trolley around. If it moves west and accelerates west, it gains speed again, even though its acceleration is negative under an east-positive convention.

Acceleration describes the direction of the change in velocity. Whether that change increases or decreases speed depends on the direction the object is already moving.

[IMAGE: Four small velocity-time graphs arranged in a clear two-by-two comparison. Show positive velocity with positive gradient, positive velocity with negative gradient, negative velocity with positive gradient, and negative velocity with negative gradient. Include direction arrows and short annotations indicating increasing or decreasing speed. Clearly distinguish graph height from gradient.]

### Worked example: A scooter slows down with negative acceleration

A scooter travels east along a straight road. East is defined as positive.

Its velocity decreases uniformly from \(+10\ \mathrm{m\,s^{-1}}\) to \(+2\ \mathrm{m\,s^{-1}}\) over \(4.0\) seconds.

Calculate its acceleration and displacement. Determine whether its speed increases or decreases.

[IMAGE: A velocity-time graph showing a straight line descending from (0 s, +10 m/s) to (4 s, +2 m/s). Keep the entire line above the horizontal axis. Label both endpoints and show the positive direction.]

**Step 1**: Calculate the acceleration using the gradient.

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

Substitute the values:

\[
\begin{aligned}
a&=\frac{2-10}{4.0}\\
&=\frac{-8}{4.0}\\
&=-2.0\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

The scooter has negative acceleration.

**Step 2**: Compare its initial and final speeds.

Its initial speed is \(10\ \mathrm{m\,s^{-1}}\), and its final speed is \(2\ \mathrm{m\,s^{-1}}\).

Its speed has decreased by \(8\ \mathrm{m\,s^{-1}}\).

This makes sense because its velocity is positive while its acceleration is negative. They have opposite signs.

**Step 3**: Calculate its displacement using the signed area under the graph.

The graph forms a trapezium above the time axis.

\[
\begin{aligned}
\Delta x&=\frac{1}{2}(10+2)(4.0)\\
&=24\ \mathrm{m}
\end{aligned}
\]

The displacement is \(+24\ \mathrm{m}\), meaning the scooter finishes 24 metres east of its starting position.

Notice that the gradient and area tell you different things:

- Gradient: acceleration is \(-2.0\ \mathrm{m\,s^{-2}}\).
- Signed area: displacement is \(+24\ \mathrm{m}\).

Neither result contradicts the other.

### Worked example: A cart speeds up with negative acceleration

An electric cart travels west along a straight track. West is defined as negative.

Its velocity changes uniformly from \(-3.0\ \mathrm{m\,s^{-1}}\) to \(-9.0\ \mathrm{m\,s^{-1}}\) over \(3.0\) seconds.

Calculate its acceleration and displacement. Explain whether it speeds up or slows down.

[IMAGE: A velocity-time graph entirely below the horizontal axis, with a straight descending line from (0 s, -3 m/s) to (3 s, -9 m/s). Show the negative velocity region and label both endpoints.]

**Step 1**: Calculate the acceleration.

\[
\begin{aligned}
a&=\frac{v_f-v_i}{\Delta t}\\
&=\frac{-9.0-(-3.0)}{3.0}\\
&=\frac{-6.0}{3.0}\\
&=-2.0\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

The cart has negative acceleration.

**Step 2**: Determine whether its speed increases or decreases.

The initial speed is \(3.0\ \mathrm{m\,s^{-1}}\).

The final speed is \(9.0\ \mathrm{m\,s^{-1}}\).

Its speed has increased by \(6.0\ \mathrm{m\,s^{-1}}\).

Both velocity and acceleration are negative. They have the same sign, so the cart speeds up.

**Step 3**: Calculate the displacement using the signed area.

The graph forms a trapezium below the time axis.

\[
\begin{aligned}
\Delta x&=\frac{1}{2}(-3.0-9.0)(3.0)\\
&=-18\ \mathrm{m}
\end{aligned}
\]

The displacement is \(-18\ \mathrm{m}\), meaning the cart finishes 18 metres west of its starting position.

Because the cart never changes direction, its total distance is also 18 metres.

The important result is that both this cart and the scooter in the previous example had an acceleration of \(-2.0\ \mathrm{m\,s^{-2}}\).

One slowed down. The other sped up.

The difference was their initial direction of motion.

## 3. What happens when the graph crosses the time axis?

There is one more situation worth understanding.

Suppose a car starts with a velocity of \(+8\ \mathrm{m\,s^{-1}}\). Its velocity then decreases uniformly to \(-8\ \mathrm{m\,s^{-1}}\) over four seconds.

Its velocity-time graph has a constant negative gradient throughout.

Does this mean the car slows down for all four seconds?

No. It slows down until it reaches zero velocity, then speeds up in the opposite direction.

Consider the following illustrative data:

| Time (s) | Velocity (m/s) | Speed (m/s) |
|---:|---:|---:|
| 0 | +8 | 8 |
| 1 | +4 | 4 |
| 2 | 0 | 0 |
| 3 | −4 | 4 |
| 4 | −8 | 8 |

The acceleration is:

\[
a=\frac{-8-8}{4}=-4.0\ \mathrm{m\,s^{-2}}
\]

The car reaches zero velocity after two seconds.

Before that instant, its velocity is positive and its acceleration is negative. Its speed decreases.

After that instant, its velocity and acceleration are both negative. Its speed increases.

[IMAGE: Two aligned graphs showing the same motion. The upper velocity-time graph is a straight descending line from +8 m/s at 0 s, through zero at 2 s, to -8 m/s at 4 s. The lower speed-time graph is V-shaped, falling from 8 to zero and then rising to 8. Use matching time axes and clearly mark the instant of zero velocity.]

There is also an interesting result from the signed areas.

The positive triangular area is:

\[
A_+=\frac12(2)(8)=8\ \mathrm{m}
\]

The negative triangular area has a magnitude of:

\[
A_-=\frac12(2)(8)=8\ \mathrm{m}
\]

Therefore, the displacement is:

\[
\Delta x=8-8=0\ \mathrm{m}
\]

The car finishes where it started.

However, its total distance is:

\[
d=8+8=16\ \mathrm{m}
\]

A zero displacement does not mean the car remained stationary. It travelled 16 metres before returning to its starting position.

This example connects two important ideas:

- The gradient tells you the acceleration.
- The signed area tells you the displacement.

For a more detailed explanation of zero crossings and reversals, read [Can a Velocity-Time Graph Cross Zero While an Object Keeps Moving?](https://tutorgum.com/learn/resources/hsc-physics-can-a-velocity-time-graph-cross-zero-while-an-object-keeps-moving/).

## 4. Do not confuse gradient, velocity and signed area

These three features of a velocity-time graph answer different questions.

| Feature | What it tells you | What it does not tell you |
|---|---|---|
| Gradient | Acceleration | Whether speed is increasing or decreasing on its own |
| Position above or below the axis | Sign and direction of velocity | The direction of acceleration |
| Signed area | Displacement | Whether speed is increasing or decreasing |

A particularly common mistake is interpreting a negative area as negative distance.

An area below the time axis contributes negative displacement because the object moves in the negative direction. However, distance is always positive.

Another mistake is assuming that a negative velocity means negative acceleration. These are independent properties. An object can have negative velocity and positive acceleration, or positive velocity and negative acceleration.

When a question asks whether an object is speeding up or slowing down, use this decision process:

1. Read the velocity from the vertical position of the graph.
2. Find the sign of the gradient to determine acceleration.
3. Compare the signs of velocity and acceleration.
4. If the signs match, speed increases. If they differ, speed decreases.
5. If the graph crosses zero, reassess the signs after the crossing.

For a curved velocity-time graph, remember that acceleration can change throughout the motion. Use the gradient of the tangent to determine instantaneous acceleration at a particular time. The gradient between two separated points gives average acceleration over that interval.

A negative gradient between the endpoints of a curved graph does not necessarily mean the instantaneous acceleration is negative throughout the interval.

## Questions and solutions

The following questions build from direct gradient calculations to more demanding interpretations of velocity-time graphs. Assume straight-line graph segments wherever stated.

### Question 1

A runner moves east along a straight track. East is defined as positive.

The runner's velocity decreases uniformly from \(+8.0\ \mathrm{m\,s^{-1}}\) to \(+2.0\ \mathrm{m\,s^{-1}}\) over \(3.0\) seconds.

Calculate:

a) The runner's acceleration.

b) Whether the runner is speeding up or slowing down.

c) The runner's displacement over the three seconds.

[IMAGE: A simple velocity-time graph with a straight descending line from +8 m/s at t=0 to +2 m/s at t=3 s. Show the positive velocity region, time axis and both labelled endpoints.]

### Solution 1

The runner has an acceleration of \(-2.0\ \mathrm{m\,s^{-2}}\), slows down, and travels \(15\ \mathrm{m}\) east.

**a) Acceleration**

Use the gradient of the velocity-time graph.

\[
\begin{aligned}
a&=\frac{v_f-v_i}{\Delta t}\\
&=\frac{2.0-8.0}{3.0}\\
&=-2.0\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

The negative acceleration indicates that velocity is decreasing in the positive direction.

**b) Change in speed**

The speed decreases from \(8.0\) to \(2.0\ \mathrm{m\,s^{-1}}\).

Velocity is positive, but acceleration is negative. Their signs are opposite, so the runner slows down.

**c) Displacement**

Calculate the area of the trapezium under the graph.

\[
\begin{aligned}
\Delta x&=\frac12(8.0+2.0)(3.0)\\
&=15\ \mathrm{m}
\end{aligned}
\]

The positive displacement means the runner finishes 15 metres east of the starting position.

### Question 2

A tram is travelling west along a straight track at \(18\ \mathrm{m\,s^{-1}}\). It experiences a constant eastward acceleration of \(3.0\ \mathrm{m\,s^{-2}}\) for eight seconds.

Take east as positive.

Calculate:

a) The tram's velocity after eight seconds.

b) The time at which it momentarily stops.

c) Its displacement over the eight seconds.

d) Its total distance travelled.

e) Explain when the tram is slowing down and when it is speeding up.

[IMAGE: A tram travelling along a straight east-west track, with west to the left and east to the right. Include an eastward acceleration arrow and a small sequence of positions showing the tram initially travelling west, momentarily stopping, and subsequently travelling east. Do not label calculated times or velocities.]

### Solution 2

The tram finishes travelling east at \(6.0\ \mathrm{m\,s^{-1}}\), momentarily stops after 6.0 seconds, and has a displacement of \(-48\ \mathrm{m}\). Its total distance is 60 m.

**a) Final velocity**

The initial velocity is:

\[
u=-18\ \mathrm{m\,s^{-1}}
\]

The acceleration is:

\[
a=+3.0\ \mathrm{m\,s^{-2}}
\]

Use:

\[
v=u+at
\]

Substitute:

\[
\begin{aligned}
v&=-18+(3.0)(8.0)\\
&=-18+24\\
&=+6.0\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

The positive final velocity means the tram is travelling east.

**b) Time when the tram stops**

At the instant of stopping, \(v=0\).

Rearrange the velocity equation:

\[
t=\frac{v-u}{a}
\]

\[
\begin{aligned}
t&=\frac{0-(-18)}{3.0}\\
&=6.0\ \mathrm{s}
\end{aligned}
\]

The tram momentarily stops after six seconds and then reverses direction.

**c) Displacement**

Use the signed area under the velocity-time graph.

From 0 to 6 seconds, the tram moves west. The triangular area is:

\[
A_1=-\frac12(6)(18)=-54\ \mathrm{m}
\]

From 6 to 8 seconds, it moves east:

\[
A_2=\frac12(2)(6)=+6\ \mathrm{m}
\]

Therefore:

\[
\begin{aligned}
\Delta x&=-54+6\\
&=-48\ \mathrm{m}
\end{aligned}
\]

The negative displacement means the tram finishes 48 metres west of its starting position.

**d) Total distance**

Distance requires adding the magnitudes of both areas.

\[
\begin{aligned}
d&=54+6\\
&=60\ \mathrm{m}
\end{aligned}
\]

The tram travels 60 metres altogether.

**e) Change in speed**

During the first six seconds, velocity and acceleration have opposite signs. The tram slows down.

After it reverses direction, both velocity and acceleration are positive. Its speed increases.

The acceleration never changes direction. What changes is the tram's direction of motion.

### Question 3

A ball is thrown vertically upwards from ground level with an initial velocity of \(+19.6\ \mathrm{m\,s^{-1}}\).

Take upwards as positive, assume gravitational acceleration is constant at \(-9.8\ \mathrm{m\,s^{-2}}\), and ignore air resistance.

Calculate:

a) The time taken to reach maximum height.

b) The velocity after three seconds.

c) The displacement after three seconds.

d) The total distance travelled after three seconds.

Explain how the ball's speed changes throughout this motion.

[IMAGE: A ball moving vertically upwards and then downwards, shown at several positions along its trajectory. Include an upward initial velocity arrow, the highest point where instantaneous velocity is zero, and a downward direction arrow during the descent. Show the vertical positive direction and gravity acting downwards.]

### Solution 3

The ball reaches maximum height after 2.0 seconds. It is moving downwards at \(9.8\ \mathrm{m\,s^{-1}}\) after three seconds, has a displacement of \(+14.7\ \mathrm{m}\), and has travelled a total distance of \(24.5\ \mathrm{m}\).

**a) Time to maximum height**

At maximum height, the instantaneous vertical velocity is zero.

Use:

\[
v=u+at
\]

Substitute:

\[
0=19.6-9.8t
\]

Rearrange:

\[
\begin{aligned}
9.8t&=19.6\\
t&=2.0\ \mathrm{s}
\end{aligned}
\]

The ball reaches its highest point after two seconds.

**b) Velocity after three seconds**

Use the same equation:

\[
\begin{aligned}
v&=u+at\\
&=19.6+(-9.8)(3.0)\\
&=19.6-29.4\\
&=-9.8\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

The negative velocity indicates that the ball is travelling downwards at \(9.8\ \mathrm{m\,s^{-1}}\).

**c) Displacement after three seconds**

Use the displacement equation:

\[
\Delta y=ut+\frac12at^2
\]

Substitute:

\[
\begin{aligned}
\Delta y&=(19.6)(3.0)\\
&\quad+\frac12(-9.8)(3.0)^2\\
&=58.8-44.1\\
&=+14.7\ \mathrm{m}
\end{aligned}
\]

The ball is 14.7 metres above its starting position.

**d) Total distance travelled**

First calculate the distance travelled on the way up.

The ball travels for two seconds before reaching maximum height:

\[
\begin{aligned}
d_{\text{up}}&=(19.6)(2.0)\\
&\quad+\frac12(-9.8)(2.0)^2\\
&=39.2-19.6\\
&=19.6\ \mathrm{m}
\end{aligned}
\]

It then falls for one second.

During this interval, its velocity changes from zero to \(-9.8\ \mathrm{m\,s^{-1}}\).

Using the magnitude of the triangular area:

\[
d_{\text{down}}=\frac12(1.0)(9.8)
\]

\[
d_{\text{down}}=4.9\ \mathrm{m}
\]

Therefore:

\[
\begin{aligned}
d_{\text{total}}&=19.6+4.9\\
&=24.5\ \mathrm{m}
\end{aligned}
\]

**Why does the ball speed up on the way down?**

Gravity provides a constant downward acceleration.

While the ball travels upwards, its velocity is positive and its acceleration is negative. Its speed decreases.

At maximum height, its instantaneous velocity is zero.

As it falls, both its velocity and acceleration are negative. Its speed increases.

This is a particularly useful real-world example of why negative acceleration cannot automatically be interpreted as slowing down.

### Question 4

A motorised trolley moves west along a straight track. West is defined as negative.

Its velocity-time graph consists of two straight-line segments connecting these points:

| Time (s) | Velocity (m/s) |
|---:|---:|
| 0 | −6 |
| 2 | −2 |
| 5 | −5 |

Calculate:

a) The acceleration during each interval.

b) Whether the trolley speeds up or slows down during each interval.

c) Its total displacement over the five seconds.

A student claims that the trolley must be slowing down throughout because its velocity is negative and the graph stays below the time axis. Explain why this reasoning is incorrect.

[IMAGE: A velocity-time graph below the horizontal axis with three labelled points: (0 s, -6 m/s), (2 s, -2 m/s), and (5 s, -5 m/s). Connect consecutive points with straight lines. Clearly show that the first segment rises and the second segment falls, without adding acceleration or speed annotations.]

### Solution 4

The trolley first slows down with an acceleration of \(+2.0\ \mathrm{m\,s^{-2}}\), then speeds up with an acceleration of \(-1.0\ \mathrm{m\,s^{-2}}\). Its total displacement is \(-18.5\ \mathrm{m}\).

**a) Acceleration during each interval**

For the first interval:

\[
\begin{aligned}
a_1&=\frac{-2-(-6)}{2-0}\\
&=\frac{4}{2}\\
&=+2.0\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

For the second interval:

\[
\begin{aligned}
a_2&=\frac{-5-(-2)}{5-2}\\
&=\frac{-3}{3}\\
&=-1.0\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

The acceleration changes direction at \(t=2\ \mathrm{s}\).

**b) Change in speed**

During the first interval, velocity changes from \(-6\) to \(-2\ \mathrm{m\,s^{-1}}\).

Its speed decreases from 6 to 2 m/s.

The velocity is negative, while acceleration is positive. Their signs are opposite, so the trolley slows down.

During the second interval, velocity changes from \(-2\) to \(-5\ \mathrm{m\,s^{-1}}\).

Its speed increases from 2 to 5 m/s.

Both velocity and acceleration are negative, so the trolley speeds up.

**c) Displacement**

Calculate the signed area of each trapezium.

First interval:

\[
\begin{aligned}
A_1&=\frac12(-6-2)(2)\\
&=-8\ \mathrm{m}
\end{aligned}
\]

Second interval:

\[
\begin{aligned}
A_2&=\frac12(-2-5)(3)\\
&=-10.5\ \mathrm{m}
\end{aligned}
\]

Total displacement:

\[
\begin{aligned}
\Delta x&=-8-10.5\\
&=-18.5\ \mathrm{m}
\end{aligned}
\]

The trolley finishes 18.5 metres west of its starting position.

The student's mistake is confusing negative velocity with decreasing speed.

Both graph segments are below the time axis, so the trolley travels west throughout. However, its speed changes because the gradient changes.

The first segment has positive acceleration and decreasing speed. The second has negative acceleration and increasing speed.

The graph's position relative to the time axis tells you its direction. Its gradient tells you its acceleration.

### Question 5

A motion sensor records the velocity of a motorised cart moving along a straight east-west track. East is defined as positive.

The following illustrative measurements are obtained:

| Time (s) | Velocity (m/s) |
|---:|---:|
| 0 | +4 |
| 2 | +1 |
| 4 | +2 |
| 6 | −1 |
| 8 | −4 |

Assume that straight lines connect every pair of consecutive measurements.

A student draws a straight line joining only the first and last points. They conclude:

"The cart has a constant negative acceleration of \(-1.0\ \mathrm{m\,s^{-2}}\), so it must slow down continuously."

Evaluate this claim.

Calculate:

a) The average acceleration over the full eight seconds.

b) The acceleration during each of the four individual intervals.

c) Every interval during which the cart speeds up or slows down. Find the exact time at which its velocity reaches zero.

d) The total displacement and total distance travelled.

e) The average speed over the eight seconds.

Explain why the overall gradient does not provide enough information to describe the cart's speed throughout the motion.

[IMAGE: A complete velocity-time graph with time on the horizontal axis and velocity on the vertical axis. Plot the five supplied points (0, +4), (2, +1), (4, +2), (6, -1), and (8, -4), connecting consecutive points with straight segments. Show the horizontal zero-velocity axis. Do not label segment gradients, the zero-crossing time, areas, or speed behaviour.]

### Solution 5

The student's claim is incorrect. The average acceleration is \(-1.0\ \mathrm{m\,s^{-2}}\), but the cart does not have constant acceleration. It speeds up during part of the interval, slows down during other parts, and eventually reverses direction.

**a) Average acceleration**

Average acceleration uses the change in velocity over the entire time interval.

\[
a_{\text{avg}}=\frac{v_f-v_i}{\Delta t}
\]

Substitute the initial and final velocities:

\[
\begin{aligned}
a_{\text{avg}}&=\frac{-4-4}{8}\\
&=\frac{-8}{8}\\
&=-1.0\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

This result is correct.

However, it only describes the average change in velocity over eight seconds. It does not establish that acceleration is constant or negative at every instant.

**b) Acceleration during each interval**

Calculate the gradient of each individual segment.

From 0 to 2 seconds:

\[
\begin{aligned}
a_1&=\frac{1-4}{2}\\
&=-1.5\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

From 2 to 4 seconds:

\[
\begin{aligned}
a_2&=\frac{2-1}{2}\\
&=+0.5\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

From 4 to 6 seconds:

\[
\begin{aligned}
a_3&=\frac{-1-2}{2}\\
&=-1.5\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

From 6 to 8 seconds:

\[
\begin{aligned}
a_4&=\frac{-4-(-1)}{2}\\
&=-1.5\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

The results are:

| Time interval (s) | Acceleration (m/s²) |
|---|---:|
| 0-2 | −1.5 |
| 2-4 | +0.5 |
| 4-6 | −1.5 |
| 6-8 | −1.5 |

Notice the second interval.

The cart has positive acceleration even though its average acceleration over the entire eight seconds is negative.

The measured intermediate velocity values are what expose the student's incorrect assumption.

**c) Determine when the cart speeds up and slows down**

Now compare the velocity and acceleration signs in each interval.

From 0 to 2 seconds:

Velocity decreases from \(+4\) to \(+1\ \mathrm{m\,s^{-1}}\).

Acceleration is negative, while velocity is positive.

The cart slows down.

From 2 to 4 seconds:

Velocity increases from \(+1\) to \(+2\ \mathrm{m\,s^{-1}}\).

Both velocity and acceleration are positive.

The cart speeds up.

From 4 to 6 seconds:

Velocity changes from \(+2\) to \(-1\ \mathrm{m\,s^{-1}}\).

The cart initially slows down because its positive velocity and negative acceleration have opposite signs.

It reaches zero velocity before the end of this interval.

To find the exact time, use:

\[
v=u+at
\]

For this segment, take the starting time as \(t=4\ \mathrm{s}\).

\[
0=2-1.5(t-4)
\]

Rearrange:

\[
\begin{aligned}
1.5(t-4)&=2\\
t-4&=\frac{2}{1.5}\\
t&=5.33\ \mathrm{s}
\end{aligned}
\]

The cart reaches zero velocity at approximately \(5.33\ \mathrm{s}\).

After this instant, velocity becomes negative while acceleration remains negative. The cart speeds up in the westward direction.

From 6 to 8 seconds:

Both velocity and acceleration are negative.

The cart continues speeding up westwards.

The complete motion is therefore:

| Time interval (s) | Speed behaviour |
|---|---|
| 0-2 | Decreasing |
| 2-4 | Increasing |
| 4-5.33 | Decreasing |
| 5.33-8 | Increasing |

The cart does not simply slow down throughout the motion, even though its overall average acceleration is negative.

**d) Calculate displacement and total distance**

Use the signed area under the velocity-time graph.

For the first interval, the area is:

\[
\begin{aligned}
A_1&=\frac12(4+1)(2)\\
&=5\ \mathrm{m}
\end{aligned}
\]

For the second interval:

\[
\begin{aligned}
A_2&=\frac12(1+2)(2)\\
&=3\ \mathrm{m}
\end{aligned}
\]

For the third interval, the graph crosses the horizontal axis.

This means we must separate the positive and negative regions.

The positive triangular area is:

\[
A_+=\frac12(2)\left(\frac43\right)
\]

\[
A_+=\frac43\ \mathrm{m}
\]

The negative triangular area has magnitude:

\[
A_-=\frac12(1)\left(\frac23\right)
\]

\[
A_-=\frac13\ \mathrm{m}
\]

The signed area for the entire third interval is:

\[
A_3=\frac43-\frac13=1\ \mathrm{m}
\]

For the final interval:

\[
\begin{aligned}
A_4&=\frac12(-1-4)(2)\\
&=-5\ \mathrm{m}
\end{aligned}
\]

Add all signed areas to calculate displacement:

\[
\begin{aligned}
\Delta x&=5+3+1-5\\
&=+4\ \mathrm{m}
\end{aligned}
\]

The cart finishes four metres east of its starting position.

Now calculate total distance by adding the magnitudes of all the individual areas.

\[
\begin{aligned}
d&=5+3+\frac43+\frac13+5\\
&=8+\frac53+5\\
&=\frac{44}{3}\ \mathrm{m}\\
&\approx14.67\ \mathrm{m}
\end{aligned}
\]

The cart travels approximately 14.67 metres altogether.

**e) Calculate average speed**

Average speed is total distance divided by total time.

\[
\text{Average speed}=\frac{\text{total distance}}{\text{total time}}
\]

\[
\begin{aligned}
\text{Average speed}&=\frac{44/3}{8}\\
&=\frac{11}{6}\\
&\approx1.83\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

The average speed is approximately \(1.83\ \mathrm{m\,s^{-1}}\).

**Why is this question different from simply calculating the overall gradient?**

The overall gradient describes the average acceleration between the first and last measurements.

It does not describe every change in motion between those measurements.

The intermediate data reveals that the cart actually has positive acceleration between two and four seconds. It also reveals that the cart reverses direction at approximately 5.33 seconds.

The signed areas provide another independent piece of information. They show that the cart finishes east of its starting position, despite travelling west during the latter part of the motion.

This is why you must interpret the actual graph rather than relying on its overall appearance or the line joining its endpoints.

## Where to go next

When interpreting a velocity-time graph, keep three questions separate:

- Which direction is the object moving? Read the sign of velocity.
- Is its velocity changing in the positive or negative direction? Read the gradient.
- How far has its position changed? Calculate the signed area.

Once you can distinguish these quantities, you can tackle more complicated motion graphs without confusing acceleration with deceleration.

For further practice, explore [how to calculate distance when a velocity-time graph crosses zero](https://tutorgum.com/learn/resources/hsc-physics-how-to-calculate-distance-when-a-velocity-time-graph-crosses-the-axis/). This builds directly on the distinction between signed displacement and total distance.