How to Compare Accelerations on Graphs with Different Scales

Two velocity-time graphs can show different accelerations even when their lines appear to have the same steepness. In fact, a line that looks flatter on the page can represent a greater acceleration than one that looks much steeper.

The reason is simple: the graphs may use different scales on their horizontal and vertical axes.

To compare accelerations correctly, calculate the numerical gradient of each velocity-time graph using its actual axis scales. Never compare the visual angles of the lines.

Consider two graphs:

• Graph A has a horizontal scale of 2 s per division and a vertical scale of 2 m s⁻¹ per division. Its line rises 4 divisions over 4 divisions.
• Graph B has a horizontal scale of 1 s per division and a vertical scale of 5 m s⁻¹ per division. Its line rises 2 divisions over 4 divisions.

Graph A looks steeper. However, its acceleration is only 1.0 m s⁻², while Graph B has an acceleration of 2.5 m s⁻².

Graph B therefore has the greater acceleration, despite looking flatter.

01Why visual steepness can be misleading

Acceleration is the rate at which velocity changes with time.

On a velocity-time graph, this means acceleration is represented by the gradient:

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

where:

• \(a\) is acceleration in metres per second squared (m s⁻²).
• \(\Delta v\) is the change in velocity in metres per second (m s⁻¹).
• \(\Delta t\) is the time interval in seconds (s).

The important word is change. You need to calculate the difference between two velocity values and divide it by the difference between their corresponding times.

For Graph A:

\[ \begin{aligned} \Delta v &= 8-0=8\text{ m s}^{-1}\\ \Delta t &= 8-0=8\text{ s}\\ a_A &= \frac{8}{8}\\ &=1.0\text{ m s}^{-2} \end{aligned} \]

For Graph B:

\[ \begin{aligned} \Delta v &= 10-0=10\text{ m s}^{-1}\\ \Delta t &= 4-0=4\text{ s}\\ a_B &= \frac{10}{4}\\ &=2.5\text{ m s}^{-2} \end{aligned} \]

Graph B accelerates 2.5 times as much as Graph A.

The physical gradient depends on the actual velocity and time values, not the angle a line makes with the page.

Converting grid divisions into actual values

If a graph uses different scales, convert the number of grid divisions into their corresponding physical values before calculating the gradient.

You can use:

\[ a=\frac{N_v S_v}{N_t S_t} \]

where:

• \(N_v\) is the signed number of vertical grid divisions.
• \(S_v\) is the vertical scale in m s⁻¹ per division.
• \(N_t\) is the number of horizontal grid divisions.
• \(S_t\) is the horizontal scale in seconds per division.

For example, suppose a line rises 3 vertical divisions over 4 horizontal divisions.

The vertical scale is 2 m s⁻¹ per division, and the horizontal scale is 0.5 s per division.

The actual changes are:

\[ \Delta v=3\times2=6\text{ m s}^{-1} \]

\[ \Delta t=4\times0.5=2\text{ s} \]

Therefore:

\[ a=\frac{6}{2}=3.0\text{ m s}^{-2} \]

Notice that counting divisions alone would give you a ratio of \(3/4\). That is not the physical acceleration.

A useful comparison table

When comparing two graphs, organise the information before calculating.

This makes it much harder to accidentally compare the wrong quantities.

02A reliable method for comparing accelerations

Use this method whenever an HSC Physics question gives you two or more velocity-time graphs with different axis scales.

Step 1: Identify the graph type.

Check that you are looking at velocity against time. Acceleration is the gradient of this graph.

Step 2: Read both axis scales.

Write down how much velocity each vertical division represents and how much time each horizontal division represents.

Step 3: Choose two points on each line.

Use points that are easy to read, preferably where the line crosses grid intersections. Choose points reasonably far apart to reduce reading errors.

Step 4: Calculate the actual changes.

Multiply the number of divisions by the corresponding axis scale.

Remember to retain the sign of the velocity change.

Step 5: Calculate each acceleration.

Use:

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

Step 6: Compare the results.

Compare the numerical accelerations, including their signs. If the question asks which object has the greater acceleration magnitude, compare their absolute values.

03What if the graph is curved?

So far, we have used straight lines. These represent constant acceleration because their gradients do not change.

A curved velocity-time graph represents changing acceleration. You need to be more precise about which acceleration the question asks for.

There are two possibilities.

Average acceleration: Calculate the gradient between two points on the curve. This is the gradient of a straight line joining those points, called a secant.

\[ a_{\text{avg}}=\frac{v_2-v_1}{t_2-t_1} \]

Instantaneous acceleration: Calculate the gradient of a tangent to the curve at the particular time.

A tangent is a straight line that matches the curve’s direction at that instant.

04Do not confuse gradients with signed areas

There is another graph-reading mistake worth avoiding.

Acceleration is the gradient of a velocity-time graph, but it is the vertical value (height) of an acceleration-time graph.

The two graphs have different jobs.

This distinction matters when the axes have different scales.

For an acceleration-time graph, the signed area is calculated using the actual acceleration and time values, not simply the number of grid squares.

For example, consider an object with:

• Acceleration of \(+3.0\text{ m s}^{-2}\) for 2.0 s.
• Acceleration of \(-1.0\text{ m s}^{-2}\) for the following 3.0 s.

The first section produces a positive velocity change:

\[ \Delta v_1=(3.0)(2.0)=+6.0\text{ m s}^{-1} \]

The second section produces a negative velocity change:

\[ \Delta v_2=(-1.0)(3.0)=-3.0\text{ m s}^{-1} \]

The total velocity change is:

\[ \begin{aligned} \Delta v&=+6.0-3.0\\ &=+3.0\text{ m s}^{-1} \end{aligned} \]

Notice that the height of the graph tells you the acceleration at a particular time, while the signed area tells you how much the velocity changes over an interval.

If the acceleration-time graph also has different axis scales, you must convert both the horizontal and vertical divisions before calculating its area.

For a rectangular region, for example:

\[ \Delta v=(\text{height})(\text{width}) \]

The units confirm the result:

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

For a fuller explanation of signed areas, including triangular regions, negative acceleration, and changes in direction, see Acceleration-Time Graphs: Area and Velocity Change.

05Common mistakes to avoid

A particularly useful check is to look at the units.

If you are calculating acceleration from a velocity-time graph, your answer must have units of m s⁻².

If you obtain m s⁻¹, you have probably calculated a velocity change instead.

06Questions and solutions

Question 1

A trolley’s velocity-time graph is a straight line passing through the following points:

• At \(t=0\) s, its velocity is \(+2.0\text{ m s}^{-1}\).
• At \(t=4.0\) s, its velocity is \(+10.0\text{ m s}^{-1}\).

The graph uses a horizontal scale of 1.0 s per division and a vertical scale of 2.0 m s⁻¹ per division.

Calculate the trolley’s acceleration.

Solution 1

The trolley’s acceleration is \(+2.0\text{ m s}^{-2}\).

Use the gradient of the velocity-time graph:

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

The change in velocity is:

\[ \Delta v=10.0-2.0=8.0\text{ m s}^{-1} \]

The change in time is:

\[ \Delta t=4.0-0=4.0\text{ s} \]

Therefore:

\[ \begin{aligned} a&=\frac{8.0}{4.0}\\ &=+2.0\text{ m s}^{-2} \end{aligned} \]

The positive sign indicates acceleration in the chosen positive direction.

Notice that we used the change in velocity, not the final velocity divided by the time. The trolley already had an initial velocity of 2.0 m s⁻¹.

Question 2

Two objects, P and Q, have their velocities plotted against time on separate graphs.

Both lines rise by 2 vertical divisions over 4 horizontal divisions.

However, their axis scales are different:

Object P

• Vertical scale: 3.0 m s⁻¹ per division.
• Horizontal scale: 0.50 s per division.
• Initial velocity: 2.0 m s⁻¹.

Object Q

• Vertical scale: 1.0 m s⁻¹ per division.
• Horizontal scale: 1.0 s per division.
• Initial velocity: 4.0 m s⁻¹.

Calculate the acceleration of each object.

Determine how many times greater P’s acceleration is than Q’s, and explain why the two lines’ identical visual gradients do not indicate equal accelerations.

Solution 2

Object P accelerates at \(+3.0\text{ m s}^{-2}\), while Q accelerates at \(+0.50\text{ m s}^{-2}\). P’s acceleration is six times greater.

For P, the actual velocity change is:

\[ \Delta v_P=2\times3.0=6.0\text{ m s}^{-1} \]

The actual time change is:

\[ \Delta t_P=4\times0.50=2.0\text{ s} \]

Therefore:

\[ a_P=\frac{6.0}{2.0} =+3.0\text{ m s}^{-2} \]

For Q:

\[ \Delta v_Q=2\times1.0=2.0\text{ m s}^{-1} \]

\[ \Delta t_Q=4\times1.0=4.0\text{ s} \]

Therefore:

\[ a_Q=\frac{2.0}{4.0} =+0.50\text{ m s}^{-2} \]

Now compare the accelerations:

\[ \frac{a_P}{a_Q} =\frac{3.0}{0.50} =6 \]

P’s acceleration is six times Q’s acceleration.

The tempting mistake is to say that the objects have equal accelerations because their lines are parallel and have identical visual gradients.

However, P’s vertical divisions represent three times as much velocity, while its horizontal divisions represent half as much time.

Both scale differences increase its physical gradient. The appearance of the lines alone tells us nothing reliable about the ratio of their actual accelerations.

Question 3

An electric go-kart has a curved velocity-time graph.

Between \(t=2.0\) s and \(t=6.0\) s, its velocity increases from 4.0 m s⁻¹ to 16.0 m s⁻¹.

A tangent drawn at \(t=4.0\) s passes through:

\[ (3.5\text{ s},7.0\text{ m s}^{-1}) \]

and

\[ (4.5\text{ s},13.0\text{ m s}^{-1}) \]

Calculate:

a. The average acceleration between 2.0 s and 6.0 s.

b. The instantaneous acceleration at 4.0 s.

c. Explain why these two accelerations are different.

Solution 3

The average acceleration is \(+3.0\text{ m s}^{-2}\), while the instantaneous acceleration at 4.0 s is \(+6.0\text{ m s}^{-2}\).

Part a: Average acceleration

Use the two endpoints of the interval:

\[ \begin{aligned} a_{\text{avg}} &=\frac{16.0-4.0}{6.0-2.0}\\ &=\frac{12.0}{4.0}\\ &=+3.0\text{ m s}^{-2} \end{aligned} \]

Part b: Instantaneous acceleration

Use the gradient of the tangent.

\[ \begin{aligned} a_{\text{instantaneous}} &=\frac{13.0-7.0}{4.5-3.5}\\ &=\frac{6.0}{1.0}\\ &=+6.0\text{ m s}^{-2} \end{aligned} \]

Part c: Explanation

The accelerations differ because the velocity-time graph is curved.

Its gradient changes throughout the interval. The secant represents the average rate of change over four seconds, whereas the tangent represents the rate of change at exactly four seconds.

The instantaneous acceleration is twice the average acceleration in this example.

Using the tangent for the entire interval would give an incorrect average acceleration.

Question 4

A delivery robot moves along a straight track. Its acceleration-time graph has the following behaviour:

• From \(t=0\) s to \(t=2.0\) s, acceleration increases uniformly from \(+2.0\text{ m s}^{-2}\) to \(+6.0\text{ m s}^{-2}\).
• From \(t=2.0\) s to \(t=4.0\) s, acceleration remains constant at \(-3.0\text{ m s}^{-2}\).

The robot’s initial velocity is \(-3.0\text{ m s}^{-1}\).

Taking the positive direction as right, calculate:

a. The acceleration at \(t=1.0\) s.

b. The total change in velocity over the four seconds.

c. The final velocity.

d. The average acceleration over the four seconds.

e. Whether the robot changes direction during the interval.

Solution 4

The acceleration at 1.0 s is \(+4.0\text{ m s}^{-2}\), the total velocity change is \(+2.0\text{ m s}^{-1}\), and the final velocity is \(-1.0\text{ m s}^{-1}\). The robot changes direction twice.

Part a: Acceleration at 1.0 s

The acceleration increases uniformly from 2.0 to 6.0 m s⁻² over two seconds.

At the halfway point:

\[ a=2.0+\frac{6.0-2.0}{2} \]

\[ a=+4.0\text{ m s}^{-2} \]

Remember that acceleration is the vertical value of an acceleration-time graph, not its gradient.

Part b: Total change in velocity

The first section is a trapezium above the time axis.

Its area gives:

\[ \begin{aligned} \Delta v_1 &=\frac{1}{2}(2.0+6.0)(2.0)\\ &=+8.0\text{ m s}^{-1} \end{aligned} \]

The second section is a rectangle below the time axis:

\[ \begin{aligned} \Delta v_2 &=(-3.0)(2.0)\\ &=-6.0\text{ m s}^{-1} \end{aligned} \]

Adding the signed areas:

\[ \begin{aligned} \Delta v_{\text{total}} &=8.0-6.0\\ &=+2.0\text{ m s}^{-1} \end{aligned} \]

Part c: Final velocity

The initial velocity is \(-3.0\text{ m s}^{-1}\).

Using:

\[ v_f=v_i+\Delta v \]

\[ \begin{aligned} v_f&=-3.0+2.0\\ &=-1.0\text{ m s}^{-1} \end{aligned} \]

The robot finishes moving in the negative direction.

Part d: Average acceleration

Average acceleration is the total velocity change divided by the total time:

\[ \begin{aligned} a_{\text{avg}} &=\frac{+2.0}{4.0}\\ &=+0.50\text{ m s}^{-2} \end{aligned} \]

Part e: Change in direction

Yes. The robot changes direction twice.

At \(t=2.0\) s, its velocity is:

\[ v=-3.0+8.0=+5.0\text{ m s}^{-1} \]

It has therefore moved from negative velocity to positive velocity during the first section.

During the second section, its velocity decreases from \(+5.0\) to \(-1.0\text{ m s}^{-1}\), so it crosses zero again.

This is an important distinction: the robot starts and finishes moving in the negative direction, but that does not mean it travelled in that direction throughout the interval.

Question 5

Two experimental carts, A and B, move along a straight horizontal track.

Both start from rest at \(t=0\) and are observed until \(t=4.0\) s.

Their acceleration-time graphs are plotted using the same horizontal scale of 1.0 s per division, but different vertical scales.

Cart A:

• Vertical scale: 2.0 m s⁻² per division.
• Acceleration is constant at two divisions above the time axis from 0 to 2.0 s.
• Acceleration is constant at one division below the time axis from 2.0 to 4.0 s.

Cart B:

• Vertical scale: 0.50 m s⁻² per division.
• Acceleration is constant at two divisions above the time axis throughout the four seconds.

A student observes that the positive sections of both graphs have the same visual height and claims:

“The carts have the same acceleration. They also have the same final velocity because the signed areas under their acceleration-time graphs are equal. Therefore, they must travel the same distance.”

Evaluate the student’s reasoning.

Calculate:

a. The actual acceleration of each cart during each section.

b. The total change in velocity and final velocity of each cart.

c. The average acceleration of each cart over the four seconds.

d. The displacement of each cart over the four seconds.

e. The difference between their displacements.

Assume the accelerations are constant within each stated interval, and ignore friction and other external effects.

Solution 5

The student is only partly correct.

The carts have different acceleration histories, but they finish with the same velocity and have the same average acceleration. However, they do not travel the same distance.

Cart A travels 20 m, while Cart B travels 8.0 m. Cart A therefore travels 12 m further.

The key is that equal signed areas under acceleration-time graphs establish equal changes in velocity, not equal instantaneous accelerations or equal displacements.

Part a: Actual accelerations

For Cart A, the vertical scale is 2.0 m s⁻² per division.

During the first two seconds:

\[ a_{A1}=2\times2.0 =+4.0\text{ m s}^{-2} \]

During the next two seconds:

\[ a_{A2}=-1\times2.0 =-2.0\text{ m s}^{-2} \]

For Cart B, the vertical scale is 0.50 m s⁻² per division.

Its acceleration is:

\[ a_B=2\times0.50 =+1.0\text{ m s}^{-2} \]

Therefore:

The graphs’ identical positive heights are misleading because their vertical scales differ.

Part b: Total change in velocity and final velocity

For Cart A, calculate the signed area of each section.

First two seconds:

\[ \Delta v_{A1}=(4.0)(2.0) =+8.0\text{ m s}^{-1} \]

Next two seconds:

\[ \Delta v_{A2}=(-2.0)(2.0) =-4.0\text{ m s}^{-1} \]

Total change:

\[ \Delta v_A=8.0-4.0 =+4.0\text{ m s}^{-1} \]

For Cart B:

\[ \Delta v_B=(1.0)(4.0) =+4.0\text{ m s}^{-1} \]

Both carts start from rest, so:

\[ v_{fA}=0+4.0=+4.0\text{ m s}^{-1} \]

\[ v_{fB}=0+4.0=+4.0\text{ m s}^{-1} \]

The student is correct about their final velocities.

Part c: Average acceleration

For both carts:

\[ a_{\text{avg}}=\frac{\Delta v}{\Delta t} \]

\[ \begin{aligned} a_{\text{avg}} &=\frac{4.0}{4.0}\\ &=+1.0\text{ m s}^{-2} \end{aligned} \]

Both carts have the same average acceleration, even though their instantaneous accelerations are different.

Part d: Displacement of each cart

This is where the student’s reasoning fails.

To calculate displacement, we need to consider the velocity throughout the motion, rather than just the initial and final velocities.

For Cart A, consider the two intervals separately.

During the first two seconds, the cart starts from rest and accelerates at \(+4.0\text{ m s}^{-2}\).

Using:

\[ s=ut+\frac12at^2 \]

where \(s\) is displacement, \(u\) is initial velocity, \(a\) is acceleration, and \(t\) is elapsed time:

\[ \begin{aligned} s_{A1} &=0+\frac12(4.0)(2.0)^2\\ &=8.0\text{ m} \end{aligned} \]

At the end of this interval, its velocity is:

\[ v=0+(4.0)(2.0)=8.0\text{ m s}^{-1} \]

During the next two seconds, its initial velocity is 8.0 m s⁻¹ and its acceleration is -2.0 m s⁻².

\[ \begin{aligned} s_{A2} &=(8.0)(2.0) +\frac12(-2.0)(2.0)^2\\ &=16.0-4.0\\ &=12.0\text{ m} \end{aligned} \]

Total displacement:

\[ s_A=8.0+12.0 =20.0\text{ m} \]

Now consider Cart B.

It starts from rest and maintains an acceleration of \(+1.0\text{ m s}^{-2}\) for four seconds.

\[ \begin{aligned} s_B &=0+\frac12(1.0)(4.0)^2\\ &=8.0\text{ m} \end{aligned} \]

Part e: Difference in displacement

\[ \begin{aligned} \Delta s&=s_A-s_B\\ &=20.0-8.0\\ &=12.0\text{ m} \end{aligned} \]

Cart A travels 12.0 m further than Cart B.

The reason is that Cart A reaches a velocity of 8.0 m s⁻¹ after two seconds. Although it subsequently slows down, it spends much of the interval moving faster than Cart B.

Both carts have the same initial velocity, final velocity, and average acceleration. But their velocity-time graphs have different shapes, so the areas under those graphs are different.

This is the crucial distinction:

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

Equal areas on acceleration-time graphs do not guarantee equal displacements.

07What to remember in an HSC exam

When comparing accelerations on graphs with different axis scales, the safest approach is to calculate rather than estimate.

Keep these three rules separate:

1. Velocity-time graph: Calculate the gradient using the actual changes in velocity and time.
2. Curved velocity-time graph: Use a secant for average acceleration and a tangent for instantaneous acceleration.
3. Acceleration-time graph: Read acceleration from the vertical value and use signed area to calculate velocity change.

Always convert grid divisions into physical units before comparing gradients or areas.

Once you can move confidently between these three ideas, you can tackle more complicated motion questions involving changing acceleration, direction reversals, and multiple stages of motion without relying on how a graph happens to look on the page.