How to Find Average Velocity from a Nonlinear Position-Time Graph

At \(t=1.0\text{ s}\), a trolley is at position \(x=2.0\text{ m}\). Four seconds later, at \(t=5.0\text{ s}\), it is at \(x=14.0\text{ m}\). Its position-time graph curves upwards, becoming progressively steeper.

Suppose the tangent at the final point has a gradient of \(5.0\text{ m s}^{-1}\). Is the trolley’s average velocity over the four seconds therefore \(5.0\text{ m s}^{-1}\)?

No. Its average velocity is \(3.0\text{ m s}^{-1}\).

The key is to use the gradient of a straight line joining the starting and finishing points of the interval, rather than the gradient of the curve at one particular instant.

The calculation is:

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

\[ =\frac{14.0-2.0}{5.0-1.0} \]

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

This method works even when the position-time graph is curved, irregular, or changes gradient several times.

01Why the gradient of a nonlinear graph needs special attention

A position-time graph tells you how an object’s position changes as time passes.

Its gradient represents velocity. However, you need to distinguish between two different types of gradient.

Average velocity: the secant gradient

A secant is a straight line connecting two points on a curve.

Its gradient measures the overall change in position divided by the time taken between those points.

For a position-time graph:

\[ \boxed{\bar v=\frac{x_2-x_1}{t_2-t_1}} \]

where:

• \(\bar v\) is average velocity, measured in \(\text{m s}^{-1}\).
• \(x_1\) is the initial position, measured in metres.
• \(x_2\) is the final position, measured in metres.
• \(t_1\) is the initial time, measured in seconds.
• \(t_2\) is the final time, measured in seconds.

The average velocity is the gradient of the secant connecting the two points.

Instantaneous velocity: the tangent gradient

A tangent touches the curve at a particular point and follows its direction at that instant.

Its gradient gives the instantaneous velocity.

For example, consider a trolley whose position-time graph becomes increasingly steep.

• At the beginning, the tangent has a relatively small positive gradient.
• Later, the tangent becomes steeper.
• At the end, the tangent has a much larger positive gradient.

This tells you that the trolley’s instantaneous velocity is increasing.

However, the average velocity over the entire interval is still determined by the two endpoints.

If the position-time graph is a straight line, its gradient is constant. The average and instantaneous velocities are then equal.

On a nonlinear graph, they generally differ.

02A reliable method for finding average velocity

You can use the same method regardless of how complicated the curve looks.

Assume the object moves along a straight line, with the positive direction chosen to the right.

Step 1: Identify the time interval.

Find the initial and final times specified in the question.

Don’t automatically assume the interval starts at zero.

Step 2: Read the initial and final positions.

Use the vertical axis to determine the object’s position at each endpoint.

Be careful with negative positions.

Step 3: Calculate the displacement.

Subtract the initial position from the final position.

\[ \Delta x=x_f-x_i \]

Displacement is a signed quantity, so keep the signs of both positions.

Step 4: Calculate the elapsed time.

\[ \Delta t=t_f-t_i \]

Step 5: Divide displacement by elapsed time.

\[ \boxed{\bar v=\frac{x_f-x_i}{t_f-t_i}} \]

Include the correct units and interpret the sign.

• Positive average velocity means net displacement in the positive direction.
• Negative average velocity means net displacement in the negative direction.
• Zero average velocity means the final position equals the initial position.

One important point: you do not need to calculate the length of the curved line or measure the area beneath it. The two endpoint positions are enough.

03The area trap: position-time graphs versus velocity-time graphs

This is where two different graphical methods can easily become confused.

You might remember that the area beneath a graph can represent displacement. That’s correct for a velocity-time graph, but not a position-time graph.

The axes determine what the area means.

Consider the units.

For a position-time graph, the area has units:

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

That is not a unit of displacement.

For a velocity-time graph:

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

This gives displacement.

A contrasting example: finding average velocity using signed areas

Imagine a cart whose velocity decreases uniformly from \(+4.0\text{ m s}^{-1}\) at \(t=0\) to \(-2.0\text{ m s}^{-1}\) at \(t=6.0\text{ s}\).

Its velocity-time graph is a straight line that crosses the time axis at \(t=4.0\text{ s}\).

What is its average velocity?

The area above the time axis is a positive triangle:

\[ A_+=\frac12(4.0)(4.0)=+8.0\text{ m} \]

The area below the time axis is a negative triangle:

\[ A_-=-\frac12(2.0)(2.0)=-2.0\text{ m} \]

The total displacement is the sum of the signed areas:

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

Therefore:

\[ \bar v=\frac{6.0}{6.0} \]

\[ \boxed{\bar v=+1.0\text{ m s}^{-1}} \]

Notice that the cart actually reverses direction during the six seconds.

If its initial position is \(3.0\text{ m}\), its final position must be \(9.0\text{ m}\).

The secant gradient on its corresponding position-time graph is therefore also:

\[ \bar v=\frac{9.0-3.0}{6.0}=+1.0\text{ m s}^{-1} \]

Both graphical methods give the same average velocity.

The important distinction is:

• On a position-time graph, use the secant gradient to find average velocity.
• On a velocity-time graph, add the signed areas to find displacement, then divide by elapsed time to find average velocity.

If you also need to calculate total distance when velocity changes sign, see How to Calculate Distance When a Velocity-Time Graph Crosses Zero.

04Worked examples

05Common mistakes to avoid

Before attempting exam questions, make sure you can recognise these traps.

A particularly useful exam habit is to write the general formula before substituting any values:

\[ \bar v=\frac{x_f-x_i}{t_f-t_i} \]

This makes it much harder to accidentally use the wrong gradient or sign.

06Questions and solutions

The following questions increase in difficulty. The first tests the basic method, while the later questions combine nonlinear position-time graphs, direction changes, and signed areas.

Question 1

A particle moves along a straight track. Its nonlinear position-time graph passes through the following two points:

• At \(t=2.0\text{ s}\), its position is \(-3.0\text{ m}\).
• At \(t=8.0\text{ s}\), its position is \(+9.0\text{ m}\).

Calculate its average velocity over this interval.

Solution 1

The average velocity is \(+2.0\text{ m s}^{-1}\) in the positive direction.

Step 1: Calculate the displacement.

\[ \Delta x=x_f-x_i \]

\[ =9.0-(-3.0) \]

\[ =+12.0\text{ m} \]

Step 2: Calculate the elapsed time.

\[ \Delta t=8.0-2.0=6.0\text{ s} \]

Step 3: Calculate average velocity.

\[ \bar v=\frac{+12.0}{6.0} \]

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

The negative initial position does not mean the displacement is negative. The particle finishes at a greater position than it started, so its net displacement is positive.

Question 2

A runner’s smooth, nonlinear position-time graph has the following features:

• At \(t=1.0\text{ s}\), the runner is at \(x=-4.0\text{ m}\).
• The runner reaches a single maximum position of \(x=+8.0\text{ m}\) at \(t=3.0\text{ s}\).
• At \(t=7.0\text{ s}\), the runner is at \(x=0\text{ m}\).

Calculate:

a) The average velocity between \(t=1.0\text{ s}\) and \(t=7.0\text{ s}\).

b) The average velocity between \(t=3.0\text{ s}\) and \(t=7.0\text{ s}\).

c) Explain why these two average velocities have different signs.

Solution 2

a) The average velocity from \(1.0\text{ s}\) to \(7.0\text{ s}\) is \(+0.67\text{ m s}^{-1}\).

b) The average velocity from \(3.0\text{ s}\) to \(7.0\text{ s}\) is \(-2.0\text{ m s}^{-1}\).

c) The signs differ because the two intervals have different starting positions.

Part a: Average velocity from 1.0 s to 7.0 s

Step 1: Identify the endpoint positions.

\[ x_i=-4.0\text{ m} \]

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

Step 2: Calculate displacement and elapsed time.

\[ \Delta x=0-(-4.0)=+4.0\text{ m} \]

\[ \Delta t=7.0-1.0=6.0\text{ s} \]

Step 3: Calculate average velocity.

\[ \bar v=\frac{+4.0}{6.0} \]

\[ \boxed{\bar v=+0.67\text{ m s}^{-1}} \]

Part b: Average velocity from 3.0 s to 7.0 s

Step 1: Identify the new starting position.

The interval now begins at the maximum position:

\[ x_i=+8.0\text{ m} \]

The final position remains:

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

Step 2: Calculate displacement and elapsed time.

\[ \Delta x=0-8.0=-8.0\text{ m} \]

\[ \Delta t=7.0-3.0=4.0\text{ s} \]

Step 3: Calculate average velocity.

\[ \bar v=\frac{-8.0}{4.0} \]

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

Part c: Explain the different signs

The first interval begins at a negative position and finishes at zero. Its net displacement is positive.

The second interval begins at the runner’s maximum position and finishes at zero. Its net displacement is negative.

The same nonlinear curve can therefore produce different average velocities depending on the selected time interval.

This is why you must identify the interval before calculating anything.

Question 3

A cart’s position-time graph is nonlinear. Its corresponding velocity-time graph consists of three sections:

• From \(t=0\) to \(t=2.0\text{ s}\), velocity increases linearly from \(+2.0\text{ m s}^{-1}\) to \(+6.0\text{ m s}^{-1}\).
• From \(t=2.0\) to \(t=4.0\text{ s}\), velocity remains constant at \(+6.0\text{ m s}^{-1}\).
• From \(t=4.0\) to \(t=8.0\text{ s}\), velocity decreases linearly from \(+6.0\text{ m s}^{-1}\) to \(-2.0\text{ m s}^{-1}\).

Calculate:

a) The total displacement over the eight seconds.

b) The average velocity.

c) The average speed.

Solution 3

a) The total displacement is \(+28\text{ m}\).

b) The average velocity is \(+3.5\text{ m s}^{-1}\).

c) The average speed is \(3.75\text{ m s}^{-1}\).

The key is to use signed areas for displacement, then absolute areas for total distance.

Part a: Total displacement

The displacement equals the signed area beneath the velocity-time graph.

Step 1: Calculate the area from 0 to 2 seconds.

This section is a trapezium.

\[ A_1=\frac12(2.0+6.0)(2.0) \]

\[ A_1=+8.0\text{ m} \]

Step 2: Calculate the area from 2 to 4 seconds.

This section is a rectangle.

\[ A_2=6.0(2.0) \]

\[ A_2=+12.0\text{ m} \]

Step 3: Calculate the signed area from 4 to 8 seconds.

This section is a trapezium extending below the time axis.

\[ A_3=\frac12(6.0+(-2.0))(4.0) \]

\[ A_3=+8.0\text{ m} \]

Step 4: Add the signed areas.

\[ \Delta x=A_1+A_2+A_3 \]

\[ =8.0+12.0+8.0 \]

\[ \boxed{\Delta x=+28.0\text{ m}} \]

Part b: Average velocity

The elapsed time is \(8.0\text{ s}\).

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

\[ =\frac{28.0}{8.0} \]

\[ \boxed{\bar v=+3.5\text{ m s}^{-1}} \]

This is also the gradient of the secant joining the initial and final positions on the corresponding position-time graph.

Part c: Average speed

For average speed, we need total distance rather than net displacement.

The velocity reaches zero during the final section.

Step 1: Find when the velocity reaches zero.

Velocity decreases from \(+6.0\text{ m s}^{-1}\) to \(-2.0\text{ m s}^{-1}\) over four seconds.

The decrease is \(8.0\text{ m s}^{-1}\), so the velocity reaches zero after three seconds.

Therefore:

\[ t=4.0+3.0=7.0\text{ s} \]

Step 2: Calculate the positive distance from 4 to 7 seconds.

This is a triangle above the time axis.

\[ d_+=\frac12(3.0)(6.0) \]

\[ d_+=9.0\text{ m} \]

Step 3: Calculate the distance from 7 to 8 seconds.

This is a triangle below the time axis. For distance, use its magnitude.

\[ d_-=\frac12(1.0)(2.0) \]

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

Step 4: Calculate total distance.

\[ d_{\text{total}}=8.0+12.0+9.0+1.0 \]

\[ =30.0\text{ m} \]

Step 5: Calculate average speed.

\[ \text{Average speed}=\frac{30.0}{8.0} \]

\[ \boxed{3.75\text{ m s}^{-1}} \]

The average speed is greater than the magnitude of average velocity because the cart reverses direction.

Question 4

Two carts travel along the same straight track for \(6.0\text{ s}\).

Their position-time graphs are both nonlinear.

Cart A moves continuously in the positive direction, passing through these points:

Cart B passes through these points:

Cart B moves forwards, then backwards, then forwards again. Assume there are no additional direction changes between the supplied points.

Calculate the average velocity and average speed of each cart.

Explain why their average velocities are equal even though their average speeds are different.

Solution 4

Both carts have an average velocity of \(+1.0\text{ m s}^{-1}\).

However, Cart A has an average speed of \(1.0\text{ m s}^{-1}\), while Cart B has an average speed of \(4.0\text{ m s}^{-1}\).

Step 1: Calculate the average velocity of Cart A.

Its initial and final positions are:

\[ x_i=2\text{ m} \]

\[ x_f=8\text{ m} \]

Therefore:

\[ \Delta x=8-2=+6\text{ m} \]

Its elapsed time is \(6.0\text{ s}\).

\[ \bar v_A=\frac{+6}{6.0} \]

\[ \boxed{\bar v_A=+1.0\text{ m s}^{-1}} \]

Step 2: Calculate the average velocity of Cart B.

Notice that Cart B has exactly the same initial and final positions.

Therefore:

\[ \Delta x=8-2=+6\text{ m} \]

\[ \bar v_B=\frac{+6}{6.0} \]

\[ \boxed{\bar v_B=+1.0\text{ m s}^{-1}} \]

The average velocities are identical because both carts have the same net displacement over the same time interval.

Step 3: Calculate the average speed of Cart A.

Cart A moves continuously in the positive direction.

Its total distance is:

\[ d_A=8-2=6\text{ m} \]

Therefore:

\[ \text{Average speed}_A=\frac{6}{6.0} \]

\[ \boxed{1.0\text{ m s}^{-1}} \]

Step 4: Calculate the average speed of Cart B.

Cart B changes direction twice.

Its first movement is from \(2\text{ m}\) to \(10\text{ m}\):

\[ d_1=10-2=8\text{ m} \]

Its second movement is from \(10\text{ m}\) to \(1\text{ m}\):

\[ d_2=10-1=9\text{ m} \]

Its third movement is from \(1\text{ m}\) to \(8\text{ m}\):

\[ d_3=8-1=7\text{ m} \]

Total distance:

\[ d_B=8+9+7=24\text{ m} \]

Therefore:

\[ \text{Average speed}_B=\frac{24}{6.0} \]

\[ \boxed{4.0\text{ m s}^{-1}} \]

Step 5: Explain the difference.

Both carts have the same starting position, final position, and elapsed time.

Consequently, they must have the same average velocity.

However, Cart B travels a much longer total distance because it reverses direction twice.

This demonstrates an important property of average velocity:

The average velocity over an interval depends only on the initial and final positions, not on the exact path taken between them.

The shape of the curve matters when investigating changes in velocity and total distance, but it does not change the endpoint calculation for average velocity.

Question 5

A cart travels along a straight track. Its position-time graph is smooth, and its velocity is continuous.

However, the experiment only records four positions:

There is no additional information about the exact shape of the position-time curve between these measurements.

Two students make different claims:

• Student A argues that the cart’s average velocity over the entire eight seconds is zero, so it must have remained stationary.
• Student B argues that the cart must have stopped exactly once because the recorded positions first increase and then decrease.

Determine:

a) The average velocity over the entire eight seconds.

b) The average velocity over each of the three individual time intervals.

c) Whether the cart must have been momentarily at rest at some point.

d) Whether the measurements prove that the cart stopped exactly once.

Support your reasoning using the signed areas on a possible velocity-time graph. Your answer should distinguish what the measurements prove from what they cannot establish.

Solution 5

Student A is incorrect, and Student B is not justified.

The cart’s average velocity over the full interval is zero, but it cannot have remained stationary. The measurements guarantee at least one instant at which its velocity is zero, but they do not establish exactly how many times this happens.

Part a: Average velocity over the entire interval

Step 1: Identify the initial and final positions.

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

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

Step 2: Calculate displacement.

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

Step 3: Calculate average velocity.

\[ \bar v=\frac{0}{8.0} \]

\[ \boxed{\bar v=0\text{ m s}^{-1}} \]

The cart finishes at the same position at which it started.

This does not mean it remained stationary. It means its net displacement is zero.

Part b: Average velocity over each interval

For the first interval, from 0 to 2 seconds:

\[ \bar v_1=\frac{6-0}{2-0} \]

\[ \boxed{\bar v_1=+3.0\text{ m s}^{-1}} \]

For the second interval, from 2 to 5 seconds:

\[ \bar v_2=\frac{1-6}{5-2} \]

\[ \boxed{\bar v_2=-1.67\text{ m s}^{-1}} \]

For the third interval, from 5 to 8 seconds:

\[ \bar v_3=\frac{0-1}{8-5} \]

\[ \boxed{\bar v_3=-0.33\text{ m s}^{-1}} \]

The cart has a positive average velocity over the first interval and negative average velocities over the following intervals.

Part c: Must the cart have been momentarily at rest?

Yes. At least one instant of zero velocity is guaranteed.

The first interval has a positive average velocity. This means the cart must have had positive velocity at some point.

The second interval has a negative average velocity. This means the cart must also have had negative velocity at some point.

Because the velocity is continuous, it must pass through zero when changing from positive to negative.

At that instant, the cart is momentarily at rest.

However, this only establishes that at least one such instant exists. It does not tell us the exact time or the total number of times the cart stops.

Part d: Can we prove that the cart stops exactly once?

No. The recorded positions do not contain enough information to determine the exact velocity history.

We can demonstrate this by constructing two different velocity-time graphs that both reproduce all four measured positions.

First, consider the signed areas required over each interval.

The change in position must equal the signed area under the velocity-time graph.

These areas give the recorded positions:

\[ 0+6=6\text{ m} \]

\[ 6-5=1\text{ m} \]

\[ 1-1=0\text{ m} \]

Now consider two possible velocity histories.

Possible history A: Exactly one zero-velocity instant

Suppose the velocity-time graph passes through these values, with straight lines joining consecutive points.

The first section has a trapezium area:

\[ A_1=\frac12(9-3)(2)=+6\text{ m} \]

The second section has:

\[ A_2=\frac12\left(-3-\frac13\right)(3) =-5\text{ m} \]

The third section has:

\[ A_3=-\frac13(3)=-1\text{ m} \]

This history has exactly one zero-velocity instant, between 0 and 2 seconds.

After that, velocity remains negative.

Possible history B: Two zero-velocity instants

Now consider a different velocity-time graph passing through:

Again, join consecutive points with straight lines.

The first section has area:

\[ A_1=\frac12(4+2)(2) =+6\text{ m} \]

The second section has area:

\[ A_2=\frac12\left(2-\frac{16}{3}\right)(3) =-5\text{ m} \]

The third section has area:

\[ A_3=\frac12\left(-\frac{16}{3}+\frac{14}{3}\right)(3) =-1\text{ m} \]

The signed areas are identical to those in history A.

However, this velocity-time graph crosses zero twice:

• Once between 2 and 5 seconds, as velocity changes from positive to negative.
• Again between 5 and 8 seconds, as velocity changes from negative to positive.

Both histories reproduce exactly the same four measured positions.

These two possible histories prove that the recorded position measurements cannot establish exactly how many times the cart was momentarily at rest.

The important distinction is between what the data determine and what they leave unknown:

• The overall average velocity is definitely zero.
• The average velocities over the three individual intervals are known.
• At least one instant of zero velocity is guaranteed.
• The exact number and timing of zero-velocity instants cannot be determined.

This is a useful limitation to remember in experimental questions. A few measurements of position do not necessarily reveal everything that happened between them.

07What to learn next

Once you can find average velocity using a secant on a nonlinear position-time graph, the next step is understanding how instantaneous velocity changes throughout the motion.

The tangent gradient gives instantaneous velocity at a particular time. The gradient of the velocity-time graph then tells you the acceleration.

You can build that connection in Acceleration-Time Graphs: Area and Velocity Change.

For HSC graph questions, keep the three operations separate:

• Position-time graph: use gradient to find velocity.
• Velocity-time graph: use gradient to find acceleration and signed area to find displacement.
• Acceleration-time graph: use signed area to find change in velocity.

Getting the graph type right is the first step towards getting the calculation right.