Why Displacement-Time Graph Height Is Not Speed

A cart is sitting 1.2 m to the right of a starting mark. Its displacement-time graph is a horizontal line at +1.2 m. Another cart moves quickly past the starting mark, so its graph crosses zero. Which cart is moving faster?

The second cart. The first cart is stationary.

This exposes a common HSC Physics mistake: the height of a displacement-time graph tells you displacement, not speed. Speed is determined by the magnitude of the graph’s gradient.

The distinction becomes much easier when you compare displacement-time graphs with velocity-time graphs. Their vertical heights mean different things, and their areas mean different things too.

01Start with what the graph is measuring

Imagine a cart travelling along a straight track. We mark a starting point as the origin, \(x=0\), and choose the direction to the right as positive.

At any instant, the cart has a displacement \(x\), measured in metres.

A displacement-time graph plots:

• Horizontal axis: time, \(t\), in seconds (s).
• Vertical axis: displacement, \(x\), in metres (m).

Suppose the graph has a point at \(t=3.0\text{ s}\) and \(x=+5.0\text{ m}\).

This means the cart is 5.0 m to the right of the origin after 3.0 seconds. It does not mean the cart is travelling at 5.0 m/s.

The height gives you the cart’s location relative to the chosen origin. It tells you nothing directly about how quickly that location is changing.

A simple experiment: two carts, two very different motions

Consider an investigation using an ultrasonic motion sensor.

A straight track extends to the right from the sensor. A marked position along the track is chosen as the origin, and positive displacement is defined towards the right. The sensor sends out ultrasonic pulses, receives their reflections from the cart, and uses the return time to determine the cart’s distance.

A data logger records the cart’s position at regular time intervals. After accounting for the fixed distance between the sensor and the chosen origin, it displays signed displacement against time.

There is an important measurement problem here.

If we take just one ruler reading, we might find that a cart is 1.2 m from the origin. But is it stationary, moving towards the origin, or moving away from it?

One position measurement cannot tell us.

We need repeated measurements with known time intervals.

Consider two trials:

• Cart A remains stationary at \(x=+1.2\text{ m}\).
• Cart B moves to the right at a constant speed of \(0.80\text{ m s}^{-1}\), starting at \(x=-0.4\text{ m}\).

The following are illustrative data, not measurements from an actual experiment.

Notice two particularly interesting results.

At \(t=0.5\text{ s}\), Cart B passes through the origin. Its displacement is zero, but it is still moving at \(0.80\text{ m s}^{-1}\).

At \(t=2.0\text{ s}\), both carts have exactly the same displacement of \(+1.2\text{ m}\). However, Cart A is stationary while Cart B is moving at \(0.80\text{ m s}^{-1}\).

If graph height represented speed, these observations would make no sense.

The experiment instead supports a different interpretation: vertical position tells us displacement, while the rate at which the graph changes tells us velocity.

The original single-reading approach could not distinguish stationary and moving carts. Recording repeated, time-stamped positions solves that problem.

02What the height and gradient actually mean

Let’s give the two parts of the graph their proper physical meanings.

Height: displacement

The vertical coordinate is \(x\), the displacement from the chosen origin.

For example:

A positive displacement does not mean positive velocity. Likewise, negative displacement does not mean negative speed.

Displacement describes where an object is. Velocity describes how its displacement changes.

Gradient: velocity

The gradient measures the change in displacement divided by the change in time.

For a straight section of a displacement-time graph:

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

where:

• \(v\) is velocity, measured in \(\text{m s}^{-1}\).
• \(\Delta x\) is the change in displacement, measured in metres.
• \(\Delta t\) is the time interval, measured in seconds.

The units make sense:

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

Unlike displacement, velocity includes direction.

A positive gradient means positive velocity. A negative gradient means negative velocity. A horizontal section has zero gradient, meaning the object is stationary.

So where does speed fit in?

Speed is the magnitude of velocity. It tells you how quickly an object moves, without specifying its direction.

Therefore:

\[ \boxed{\text{Speed}=|v|} \]

For a straight section of a displacement-time graph:

\[ \boxed{\text{Speed}= \left|\frac{\Delta x}{\Delta t}\right|} \]

The absolute value means speed cannot be negative.

For example, if the gradient is \(-3.0\text{ m s}^{-1}\), the velocity is negative, but the speed is \(3.0\text{ m s}^{-1}\).

This is a useful distinction to remember:

One qualification: steepness must be calculated using the axis scales. A graph’s visual angle can change if you stretch or compress either axis. Always use the labelled values rather than judging speed by appearance alone.

What if the graph is curved?

The same principle applies, but the gradient changes throughout the motion.

For a curved displacement-time graph, the instantaneous velocity at a particular moment is the gradient of the tangent at that point.

The instantaneous speed is the magnitude of that tangent gradient:

\[ \text{Instantaneous speed} = \left|\frac{dx}{dt}\right| \]

This is important because the height of a curved graph still gives displacement, even while its gradient changes continuously.

For a broader explanation of reading gradients, direction, and changing motion, see Displacement-Time Graphs: How to Read Motion in HSC Physics.

03Why velocity-time graphs use height differently

Now let’s compare a displacement-time graph with a velocity-time graph.

This is where another common mistake appears.

On a velocity-time graph, the vertical axis represents velocity rather than displacement.

That means the height now tells you how quickly the object is moving, including its direction.

Notice the difference.

On a displacement-time graph, you use the gradient to find velocity.

On a velocity-time graph, you use the height to find velocity.

The axes determine what the graph tells you. Never assume that the same feature has the same meaning across different graph types.

What does the area under a velocity-time graph represent?

The signed area under a velocity-time graph represents displacement.

For constant velocity, the area is simply:

\[ \Delta x=v\Delta t \]

This is the area of a rectangle.

If the graph is above the time axis, the area is positive. If it is below the time axis, the area is negative.

Why does the sign matter?

Because displacement includes direction.

Suppose a cart travels 6 m to the right, then 6 m to the left. Its total displacement is zero, even though it has travelled 12 m.

The positive and negative areas cancel when calculating displacement.

To calculate total distance, add the magnitudes of the individual areas instead.

\[ \text{Distance} = \text{Sum of absolute areas} \]

This is why signed area and total distance are not interchangeable.

There is an important warning here.

The area under a displacement-time graph is not the speed, distance, or displacement you are looking for in ordinary HSC kinematics questions.

Its units would be:

\[ \text{m}\times\text{s} =\text{m s} \]

That is not the unit of speed (\(\text{m s}^{-1}\)) or displacement (m).

Keep the two graph types separate.

If you want to practise handling positive and negative areas when a velocity-time graph crosses zero, see How to Calculate Distance When a Velocity-Time Graph Crosses Zero.

04Worked examples

05Questions and solutions

Question 1

A cart remains stationary at a displacement of \(+4.0\text{ m}\) for three seconds.

What is its displacement and speed? Explain why the height of its displacement-time graph does not indicate its speed.

Solution 1

The displacement is \(+4.0\text{ m}\), and the speed is \(0\text{ m s}^{-1}\).

The cart remains at the same displacement throughout the three-second interval.

Therefore:

\[ \Delta x=0\text{ m} \]

Using the gradient:

\[ v=\frac{\Delta x}{\Delta t} =\frac{0}{3} =0\text{ m s}^{-1} \]

The speed is the magnitude of velocity, so it is also zero.

The graph’s height of \(+4.0\text{ m}\) tells us that the cart is 4.0 m from the origin in the positive direction. Its horizontal shape tells us that its displacement is not changing.

The tempting mistake is to interpret a higher graph as a faster object. A high horizontal line represents a stationary object that is simply far from the origin.

Question 2

An object’s displacement-time graph is a straight line joining the points:

• \(t=1.0\text{ s}\), \(x=+7.0\text{ m}\).
• \(t=4.0\text{ s}\), \(x=-2.0\text{ m}\).

Determine:

a) The velocity of the object.

b) Its speed.

c) Its displacement at \(t=2.0\text{ s}\).

d) Explain how the object can have positive displacement while moving with negative velocity.

Solution 2

The velocity is \(-3.0\text{ m s}^{-1}\), the speed is \(3.0\text{ m s}^{-1}\), and the displacement at \(t=2.0\text{ s}\) is \(+4.0\text{ m}\).

a) Velocity

Calculate the gradient:

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

\[ v=\frac{-2-7}{4-1} \]

\[ \boxed{v=-3.0\text{ m s}^{-1}} \]

The negative sign indicates motion in the negative direction.

b) Speed

Speed is the magnitude of velocity:

\[ \boxed{\text{Speed}=3.0\text{ m s}^{-1}} \]

Speed is not negative, even when the object moves in the negative direction.

c) Displacement at \(t=2.0\text{ s}\)

Between \(t=1.0\text{ s}\) and \(t=2.0\text{ s}\), one second passes.

The object moves with velocity \(-3.0\text{ m s}^{-1}\).

\[ x=7.0+(-3.0)(1.0) \]

\[ \boxed{x=+4.0\text{ m}} \]

d) Explain the signs

The object has positive displacement because it is still 4.0 m to the right of the origin.

However, it has negative velocity because it is moving towards the left.

Position and direction of motion are separate ideas. An object can be on the positive side of the origin while moving towards the negative side.

Question 3

A curved displacement-time graph passes through the point \(t=3.0\text{ s}\), \(x=+5.0\text{ m}\).

A tangent drawn at this point passes through two additional points:

• \(t=2.0\text{ s}\), \(x=+8.0\text{ m}\).
• \(t=4.0\text{ s}\), \(x=+2.0\text{ m}\).

Calculate the instantaneous velocity and speed at \(t=3.0\text{ s}\).

Is the object moving towards or away from the origin at that instant? Explain why its positive displacement does not determine the direction of its motion.

Solution 3

The instantaneous velocity is \(-3.0\text{ m s}^{-1}\), and the instantaneous speed is \(3.0\text{ m s}^{-1}\). The object is moving towards the origin.

For a curved displacement-time graph, instantaneous velocity is determined by the gradient of the tangent at the required time.

Step 1: Calculate the tangent gradient.

\[ v=\frac{2-8}{4-2} \]

\[ v=\frac{-6}{2} \]

\[ \boxed{v=-3.0\text{ m s}^{-1}} \]

Step 2: Calculate speed.

\[ \text{Speed}=|-3.0| \]

\[ \boxed{\text{Speed}=3.0\text{ m s}^{-1}} \]

Step 3: Determine the direction of motion.

At \(t=3.0\text{ s}\), the object has a positive displacement of \(+5.0\text{ m}\).

However, its velocity is negative.

This means it is on the positive side of the origin but moving in the negative direction, towards the origin.

The misconception is to assume that positive displacement automatically means positive velocity. The displacement tells us which side of the origin the object occupies. The tangent gradient tells us its direction of motion.

Question 4

A cart has an initial displacement of \(+1.0\text{ m}\).

Its velocity-time graph consists of three sections:

• From \(t=0\) to \(t=2.0\text{ s}\), velocity is \(+4.0\text{ m s}^{-1}\).
• From \(t=2.0\) to \(t=5.0\text{ s}\), velocity is \(-2.0\text{ m s}^{-1}\).
• From \(t=5.0\) to \(t=6.0\text{ s}\), velocity is zero.

Calculate:

a) The net displacement.

b) The final displacement.

c) The total distance travelled.

d) The average velocity.

e) The average speed.

Explain why the negative area on the velocity-time graph reduces the net displacement but increases the total distance travelled.

Solution 4

The net displacement is \(+2.0\text{ m}\), the final displacement is \(+3.0\text{ m}\), the total distance is \(14.0\text{ m}\), the average velocity is \(+0.33\text{ m s}^{-1}\), and the average speed is \(2.33\text{ m s}^{-1}\).

a) Net displacement

Calculate the signed area of each section.

First section:

\[ A_1=(4.0)(2.0)=+8.0\text{ m} \]

Second section:

\[ A_2=(-2.0)(3.0)=-6.0\text{ m} \]

Third section:

\[ A_3=0\text{ m} \]

Add the signed areas:

\[ \Delta x=8.0-6.0+0 \]

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

b) Final displacement

The final displacement equals the initial displacement plus the change in displacement.

\[ x_f=x_i+\Delta x \]

\[ x_f=1.0+2.0 \]

\[ \boxed{x_f=+3.0\text{ m}} \]

c) Total distance travelled

For distance, we add the magnitudes of the areas.

\[ d=|8.0|+|-6.0|+|0| \]

\[ \boxed{d=14.0\text{ m}} \]

The negative area represents motion in the negative direction. It does not represent negative distance.

d) Average velocity

Average velocity is net displacement divided by total time.

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

\[ v_{\text{avg}}=\frac{2.0}{6.0} \]

\[ \boxed{v_{\text{avg}}=+0.33\text{ m s}^{-1}} \]

e) Average speed

Average speed is total distance divided by total time.

\[ \text{Average speed}=\frac{14.0}{6.0} \]

\[ \boxed{\text{Average speed}=2.33\text{ m s}^{-1}} \]

The negative area reduces net displacement because the cart travels back towards its starting point.

However, the cart still covers additional distance during that return journey. That is why we add the absolute areas when calculating distance rather than allowing positive and negative areas to cancel.

Question 5

A cart travels along a straight track.

Its initial displacement is \(+1.0\text{ m}\). It then:

• Travels in the positive direction at a constant velocity of \(+2.0\text{ m s}^{-1}\) for 2.0 seconds.
• Reverses direction and travels at a constant velocity of \(-1.0\text{ m s}^{-1}\) for another 2.0 seconds.

A student draws the displacement-time graph and calculates the signed area underneath it using trapeziums.

With the original origin, the graph passes through:

The student obtains an area of \(14\text{ m s}\).

Another student moves the origin to the cart’s original position, so every displacement is reduced by 1.0 m. The new graph passes through:

The second student calculates an area of \(10\text{ m s}\).

Both students claim that the area under their displacement-time graph represents the cart’s displacement.

Explain why their answers differ, even though the physical motion is identical.

Then determine:

a) The net displacement.

b) The total distance travelled.

c) The average velocity.

d) The average speed.

Finally, explain how the signed areas under the velocity-time graph give the correct displacement and distance without depending on the choice of origin.

Solution 5

The two areas differ because the area under a displacement-time graph depends on the chosen origin. Neither area represents the cart’s displacement.

The actual motion is unchanged by shifting the origin. The correct net displacement is \(+2.0\text{ m}\), the total distance is \(6.0\text{ m}\), the average velocity is \(+0.50\text{ m s}^{-1}\), and the average speed is \(1.50\text{ m s}^{-1}\).

This question exposes an important limitation of interpreting graph areas: changing the coordinate origin changes the height of a displacement-time graph without changing the object’s motion.

Step 1: Calculate the actual motion.

During the first two seconds, the cart travels in the positive direction.

\[ \Delta x_1=(+2.0)(2.0) \]

\[ \Delta x_1=+4.0\text{ m} \]

During the next two seconds, it travels in the negative direction.

\[ \Delta x_2=(-1.0)(2.0) \]

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

Therefore, the net displacement is:

\[ \Delta x=4.0-2.0 \]

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

Step 2: Calculate total distance.

Distance adds the magnitudes of both sections:

\[ d=|4.0|+|-2.0| \]

\[ \boxed{d=6.0\text{ m}} \]

Step 3: Calculate average velocity.

The total time is:

\[ \Delta t=2.0+2.0=4.0\text{ s} \]

Therefore:

\[ v_{\text{avg}}=\frac{2.0}{4.0} \]

\[ \boxed{v_{\text{avg}}=+0.50\text{ m s}^{-1}} \]

Step 4: Calculate average speed.

\[ \text{Average speed}=\frac{6.0}{4.0} \]

\[ \boxed{\text{Average speed}=1.50\text{ m s}^{-1}} \]

Step 5: Understand why the displacement-time areas differ.

For the original origin, calculate the first trapezium:

\[ A_1=\frac{1+5}{2}\times2 \]

\[ A_1=6\text{ m s} \]

The second trapezium is:

\[ A_2=\frac{5+3}{2}\times2 \]

\[ A_2=8\text{ m s} \]

Total:

\[ A=6+8=14\text{ m s} \]

For the shifted origin, the first trapezium is:

\[ A'_1=\frac{0+4}{2}\times2 =4\text{ m s} \]

The second trapezium is:

\[ A'_2=\frac{4+2}{2}\times2 =6\text{ m s} \]

Total:

\[ A'=4+6=10\text{ m s} \]

The areas differ by \(4\text{ m s}\).

This happens because moving the origin down by 1.0 m shifts the entire displacement-time graph vertically by 1.0 m. Over four seconds, this changes its area by:

\[ (1.0)(4.0)=4.0\text{ m s} \]

But the cart has not changed its motion at all.

Its velocity and speed remain exactly the same.

This is the crucial reasoning: if a supposed measure of motion changes simply because we choose a different origin, it cannot represent the cart’s displacement, distance, or speed.

Step 6: Verify using the velocity-time graph.

The velocity-time graph has:

• \(+2.0\text{ m s}^{-1}\) from 0 to 2 seconds.
• \(-1.0\text{ m s}^{-1}\) from 2 to 4 seconds.

Its signed areas are:

\[ A_1=(+2.0)(2.0)=+4.0\text{ m} \]

\[ A_2=(-1.0)(2.0)=-2.0\text{ m} \]

Therefore:

\[ \Delta x=4.0-2.0 \]

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

For distance, add the magnitudes:

\[ d=|4.0|+|-2.0| \]

\[ \boxed{d=6.0\text{ m}} \]

The velocity-time graph gives the same physical result regardless of where we place the origin because velocity is determined by the change in displacement, not its absolute value.

That is why the area under a velocity-time graph is useful for finding displacement, while the height of a displacement-time graph is not a measure of speed.

06The decision to make in an HSC exam

When an exam question gives you a motion graph, resist the temptation to calculate something before identifying the axes.

Use this sequence:

1. Read the vertical axis. Is it displacement or velocity?
2. If it is a displacement-time graph, read the height as displacement.
3. Calculate the gradient to find velocity. Take its magnitude if the question asks for speed.
4. If it is a velocity-time graph, read the height as velocity.
5. Use signed areas under a velocity-time graph to find net displacement.
6. Add the magnitudes of the areas to find total distance.

The key relationships are:

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

\[ \boxed{\text{Speed}=|v|} \]

\[ \boxed{\Delta x=\text{Signed area under a }v\text{-}t\text{ graph}} \]

Once these distinctions are clear, the next useful skill is interpreting what happens when velocity changes sign. That is where signed areas, total distance, and changes in direction become particularly important.