What a Curved Velocity-Time Graph Reveals About Acceleration

A paper coffee filter is released beside a metre ruler. Its downward velocity increases quickly at first, then approaches a steady value. Plot velocity against time and the line bends, gradually becoming flatter.

What does that curve tell you?

Acceleration is changing with time. At any instant, the acceleration is the gradient of the tangent to the velocity-time curve. The direction and steepness of that tangent tell you the sign and magnitude of acceleration. The way the curve bends tells you how the acceleration itself is changing.

There is an important trap here: a curve bending upwards does not necessarily mean positive acceleration. You need to look at its gradient, not just its shape.

01Why does a velocity-time graph become curved?

Consider the falling coffee filter again. If gravity were the only force acting on it, what would you predict its velocity-time graph would look like?

It would be a straight line with a constant positive gradient, assuming downward is positive. Gravity would produce a constant acceleration of approximately \(9.8\ \text{m s}^{-2}\).

But a coffee filter experiences air resistance.

As it falls faster, air resistance increases. This reduces the resultant downward force, so its acceleration becomes smaller. Its velocity continues increasing, but at a decreasing rate.

The result is a rising curve that gradually flattens.

How the experiment produces the graph

The experiment begins with a simple measurement problem: how can we determine whether the filter’s acceleration stays constant as it falls?

Arrange the apparatus in this order:

1. Place a metre ruler vertically beside the filter’s falling path.
2. Position a smartphone camera side-on, with the ruler and filter in the same plane.
3. Release the filter from rest without pushing it.
4. Record its position at regular time intervals.
5. Use the video frames and ruler to estimate its displacement, then calculate its velocity over short intervals.

The camera measures position, not velocity directly. Velocity must be calculated from the changes in position over time.

The initial prediction is that the filter will accelerate at a constant rate if air resistance is negligible. The velocity-time graph should therefore be straight.

Instead, the measured velocity increases while the gradient becomes smaller.

The physical explanation is that the filter experiences two main vertical forces:

• Weight, \(mg\), acting downwards.
• Air resistance, \(F_D\), acting upwards.

Taking downward as positive, the resultant force is

\[ F_{\text{net}}=mg-F_D \]

and Newton’s second law gives

\[ a=\frac{mg-F_D}{m} \]

As the filter speeds up, \(F_D\) increases. The resultant force decreases, so the acceleration decreases.

Eventually, if the filter reaches terminal velocity, air resistance balances its weight. The resultant force becomes zero, and the velocity stops increasing.

The velocity-time graph approaches a horizontal line.

Illustrative experimental data

The following values are illustrative rather than measurements from an actual trial.

Downwards is positive.

Calculate the average acceleration over each interval using

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

The velocity increases throughout the experiment, but the average acceleration gets smaller.

There is one experimental limitation worth recognising. Each calculated velocity represents an average over a time interval, not the exact instantaneous velocity at one moment. If the intervals are too large, a changing gradient can be hidden.

A better measurement uses a fixed camera, a higher frame rate, a ruler in the same plane as the filter, and shorter time intervals. Repeating the drop also helps identify unusual results caused by the filter fluttering.

The experiment supports the explanation that air resistance changes the acceleration. It does not establish that every falling object follows exactly the same curve.

02What the gradient of a curved graph tells you

The vertical coordinate of a velocity-time graph is velocity, \(v\), measured in metres per second.

The horizontal coordinate is time, \(t\), measured in seconds.

Acceleration measures how quickly velocity changes with time. Its units are

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

For a curved velocity-time graph, the instantaneous acceleration is the gradient of the tangent at the particular instant.

\[ a=\frac{\text{gradient of tangent}}{} \]

In practical terms:

• An upward-sloping tangent means positive acceleration.
• A downward-sloping tangent means negative acceleration.
• A horizontal tangent means zero acceleration at that instant.
• A steeper tangent means a greater magnitude of acceleration.

The key is to examine the tangent, rather than trying to calculate the gradient directly from the curved section.

Compare these four situations

Notice the difference between the second and third situations.

In the second situation, acceleration is positive but decreasing.

In the third situation, acceleration is negative but increasing.

That second statement might sound contradictory. It isn’t.

An acceleration changing from \(-4.0\ \text{m s}^{-2}\) to \(-1.0\ \text{m s}^{-2}\) is increasing because its numerical value is becoming less negative. However, it remains negative throughout.

This is why the gradient matters more than the visual direction of the curve.

Concavity tells you how acceleration changes

You can go one step further by looking at the curvature of the graph.

A curve is described as concave up when its gradient becomes more positive as time increases.

A curve is concave down when its gradient becomes more negative, or less positive, as time increases.

This gives you information about how acceleration changes.

Crucially, concavity does not tell you the sign of acceleration.

For example, a velocity-time graph can be concave up while having a negative gradient throughout. The object has negative acceleration, but that acceleration is increasing towards zero.

Likewise, a concave-down graph can have a positive gradient throughout. The object has positive acceleration, but that acceleration is decreasing.

03Average acceleration versus instantaneous acceleration

There are two different gradients you might be asked to calculate.

Average acceleration describes the overall change in velocity across a time interval.

It is the gradient of the straight line joining the two endpoints of the curved section. This line is called a secant.

\[ a_{\text{avg}}=\frac{v_f-v_i}{t_f-t_i} \]

Here, \(v_i\) is initial velocity, \(v_f\) is final velocity, and \(t_f-t_i\) is the elapsed time.

Instantaneous acceleration describes the acceleration at one particular moment.

It is the gradient of the tangent to the curve at that moment.

The distinction matters because the secant and tangent gradients can be different.

Imagine a cyclist whose velocity increases from \(2.0\ \text{m s}^{-1}\) to \(10.0\ \text{m s}^{-1}\) over four seconds.

The average acceleration is

\[ a_{\text{avg}}=\frac{10.0-2.0}{4.0} =2.0\ \text{m s}^{-2} \]

However, if the graph is curved, the instantaneous acceleration might be \(3.0\ \text{m s}^{-2}\) early in the journey and \(1.0\ \text{m s}^{-2}\) later.

The average is still \(2.0\ \text{m s}^{-2}\).

Do not assume that a curved graph has one constant acceleration simply because you can calculate an average acceleration for the entire interval.

What if the tangent is horizontal?

A horizontal tangent means the instantaneous acceleration is zero.

It does not mean the velocity is zero.

For example, imagine a car moving forwards at \(9.0\ \text{m s}^{-1}\) at the instant its velocity-time graph reaches a maximum.

At that instant, the tangent is horizontal, so

\[ a=0\ \text{m s}^{-2} \]

but

\[ v=+9.0\ \text{m s}^{-1} \]

The car is still moving forwards.

There is also an important difference between a horizontal tangent at one instant and a horizontal graph over an entire interval. The first means acceleration is zero momentarily. The second means velocity is constant and acceleration remains zero throughout that interval.

04Does the area under a curved graph still represent displacement?

Yes.

The signed area between a velocity-time graph and the time axis gives displacement, whether the graph is straight or curved.

The units confirm this:

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

The area therefore represents a change in position.

For a curved graph, you might not be able to calculate the area using one familiar geometric shape. However, you can estimate it by dividing the region into smaller trapezia.

For consecutive velocity values \(v_1\) and \(v_2\), separated by a time interval \(\Delta t\), the trapezium approximation is

\[ \Delta x\approx \frac{v_1+v_2}{2}\Delta t \]

The signs of the velocities must be retained.

• Area above the time axis represents positive displacement.
• Area below the time axis represents negative displacement.
• Positive and negative areas can cancel.

This gives you a useful distinction:

Gradient tells you acceleration. Signed area tells you displacement.

A steep curve does not automatically have a large displacement, and a large area does not automatically mean large acceleration.

A useful calculation from the coffee-filter experiment

Return to the illustrative data from earlier.

Estimate the downward displacement during the first second using trapezia.

For each interval:

\[ \begin{aligned} \Delta x_1&=\frac{0+1.32}{2}(0.2)=0.132\\ \Delta x_2&=\frac{1.32+1.92}{2}(0.2)=0.324\\ \Delta x_3&=\frac{1.92+2.18}{2}(0.2)=0.410\\ \Delta x_4&=\frac{2.18+2.30}{2}(0.2)=0.448\\ \Delta x_5&=\frac{2.30+2.36}{2}(0.2)=0.466 \end{aligned} \]

Adding the signed areas:

\[ \Delta x=0.132+0.324+0.410+0.448+0.466 \]

\[ \Delta x\approx1.78\ \text{m} \]

The filter’s estimated displacement is \(1.78\ \text{m}\) downwards.

Notice that we used the changing velocities to calculate displacement, not acceleration. The curvature affects the shape of the regions, but it does not change what their area represents.

05Worked examples

06A reliable method for interpreting curved velocity-time graphs

When you encounter an unfamiliar curved graph in an HSC question, use this sequence.

1. Read the vertical coordinate. This gives the velocity at the selected instant.
2. Draw or identify the tangent. Use its gradient to find instantaneous acceleration.
3. Check the gradient’s sign. An upward tangent means positive acceleration; a downward tangent means negative acceleration.
4. Compare tangent gradients. This tells you whether acceleration is increasing or decreasing.
5. Check velocity and acceleration together. If they have the same sign, speed increases. If they have opposite signs, speed decreases.
6. Calculate signed area. This gives displacement over the selected interval.

Keep the distinction between the graph’s height, gradient, and area clear.

07Questions and solutions

Question 1

A scooter travels east along a straight road. At \(t=2.0\ \text{s}\), its velocity is \(+6.0\ \text{m s}^{-1}\).

The velocity-time graph is curved. The tangent at this instant passes through the points \((1.0,4.0)\) and \((3.0,8.0)\), where time is measured in seconds and velocity in metres per second.

Calculate the instantaneous acceleration and determine whether the scooter is speeding up or slowing down.

Solution 1

The instantaneous acceleration is \(+2.0\ \text{m s}^{-2}\), and the scooter is speeding up.

The acceleration is the gradient of the tangent:

\[ \begin{aligned} a&=\frac{8.0-4.0}{3.0-1.0}\\ &=\frac{4.0}{2.0}\\ &=+2.0\ \text{m s}^{-2} \end{aligned} \]

The velocity is positive, and acceleration is positive.

Because velocity and acceleration have the same sign, the magnitude of velocity is increasing. Therefore, the scooter is speeding up in the eastward direction.

Question 2

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

Its velocity-time graph remains below the time axis throughout a four-second interval. The curve slopes downwards but is concave up.

The instantaneous acceleration is:

• At \(t=1.0\ \text{s}\): \(-2.5\ \text{m s}^{-2}\).
• At \(t=3.0\ \text{s}\): \(-0.5\ \text{m s}^{-2}\).

The corresponding velocities are \(-5.0\ \text{m s}^{-1}\) and \(-7.0\ \text{m s}^{-1}\).

Explain:

(a) Whether acceleration is positive or negative at each instant.

(b) Whether acceleration is increasing or decreasing.

(c) Whether the trolley is speeding up or slowing down.

(d) Why the description concave up does not mean positive acceleration.

Solution 2

The trolley has negative acceleration at both instants, its acceleration is increasing, and it is speeding up in the negative direction.

(a) Acceleration

At \(t=1.0\ \text{s}\):

\[ a_1=-2.5\ \text{m s}^{-2} \]

At \(t=3.0\ \text{s}\):

\[ a_3=-0.5\ \text{m s}^{-2} \]

Both tangent gradients are negative.

(b) Change in acceleration

Acceleration changes from \(-2.5\) to \(-0.5\ \text{m s}^{-2}\).

This means acceleration is increasing because its numerical value is becoming less negative.

The graph is concave up because its gradient becomes more positive as time increases.

(c) Change in speed

At both instants, velocity and acceleration are negative.

Therefore, the trolley is speeding up in the negative direction.

Its velocity becomes more negative, so its speed increases.

(d) The concavity misconception

Concave up describes how the gradient changes, not the sign of the gradient itself.

The tangent can remain negative while becoming progressively less steep.

That is exactly what happens here. Acceleration is negative, even though it is increasing.

Question 3

A robot moves along a straight track. Its velocity-time graph is curved and passes through the following measured points.

For numerical estimates, assume straight-line interpolation between consecutive data points.

Calculate:

(a) The approximate time at which the robot changes direction.

(b) Its net displacement over four seconds.

(c) Its total distance travelled over four seconds.

Solution 3

The robot changes direction at approximately \(2.5\ \text{s}\), its net displacement is \(+2.5\ \text{m}\), and its total distance is approximately \(7.0\ \text{m}\).

(a) Direction change

Between \(t=2\) and \(t=3\ \text{s}\), velocity changes from \(+1.0\) to \(-1.0\ \text{m s}^{-1}\).

With straight-line interpolation, the graph crosses zero halfway through the interval.

\[ t=2.5\ \text{s} \]

(b) Net displacement

Calculate the signed trapezoidal areas.

From 0 to 1 second:

\[ A_1=\frac{4.0+2.0}{2}(1)=+3.0\ \text{m} \]

From 1 to 2 seconds:

\[ A_2=\frac{2.0+1.0}{2}(1)=+1.5\ \text{m} \]

From 2 to 3 seconds:

\[ A_3=\frac{1.0-1.0}{2}(1)=0 \]

From 3 to 4 seconds:

\[ A_4=\frac{-1.0-3.0}{2}(1)=-2.0\ \text{m} \]

Therefore,

\[ \begin{aligned} \Delta x&=3.0+1.5+0-2.0\\ &=+2.5\ \text{m} \end{aligned} \]

The robot finishes \(2.5\ \text{m}\) east of its starting position.

(c) Total distance

Distance requires adding the magnitudes of all areas.

The interval from 2 to 3 seconds contains equal positive and negative triangles.

Each triangle has:

\[ A=\frac12(0.5)(1.0)=0.25\ \text{m} \]

The total distance is therefore

\[ \begin{aligned} d&=3.0+1.5+0.25+0.25+2.0\\ &=7.0\ \text{m} \end{aligned} \]

The robot travels \(7.0\ \text{m}\), even though its net displacement is only \(2.5\ \text{m}\).

The difference occurs because the robot reverses direction and travels back along part of its original path.

Question 4

An object moves along a straight track, with east defined as positive.

Its velocity-time graph is smooth and curved. It begins at \(+5.0\ \text{m s}^{-1}\), reaches a maximum velocity of \(+9.0\ \text{m s}^{-1}\) at \(t=2.0\ \text{s}\), and returns to \(+5.0\ \text{m s}^{-1}\) at \(t=6.0\ \text{s}\).

At the maximum velocity, the tangent is horizontal.

The following measurements are supplied:

For the displacement estimate, use trapezia between consecutive measurements.

(a) Calculate the average acceleration over the six seconds.

(b) Estimate the total displacement.

(c) Explain how the object can have zero average acceleration but non-zero instantaneous acceleration during most of the interval.

(d) Describe how its speed changes.

Solution 4

The average acceleration is zero, the estimated displacement is \(41.0\ \text{m}\) east, and the object has non-zero instantaneous acceleration except at the instant of maximum velocity.

(a) Average acceleration

Use the initial and final velocities.

\[ \begin{aligned} a_{\text{avg}}&=\frac{v_f-v_i}{\Delta t}\\ &=\frac{5.0-5.0}{6.0}\\ &=0\ \text{m s}^{-2} \end{aligned} \]

The velocity has no net change over the six seconds.

(b) Displacement

Estimate the area using consecutive trapezia.

\[ \begin{aligned} A_1&=\frac{5.0+7.0}{2}(1)=6.0\\ A_2&=\frac{7.0+9.0}{2}(1)=8.0\\ A_3&=\frac{9.0+8.0}{2}(1)=8.5\\ A_4&=\frac{8.0+6.5}{2}(1)=7.25\\ A_5&=\frac{6.5+5.5}{2}(1)=6.0\\ A_6&=\frac{5.5+5.0}{2}(1)=5.25 \end{aligned} \]

Adding the areas:

\[ \begin{aligned} \Delta x&=6.0+8.0+8.5\\ &\quad+7.25+6.0+5.25\\ &=41.0\ \text{m} \end{aligned} \]

The estimated displacement is \(41.0\ \text{m}\) east.

(c) Average versus instantaneous acceleration

The average acceleration is zero because the initial and final velocities are equal.

However, the velocity-time curve rises before \(t=2.0\ \text{s}\) and falls afterwards.

This means:

• Before \(t=2.0\ \text{s}\), the tangent has a positive gradient, so acceleration is positive.
• At \(t=2.0\ \text{s}\), the tangent is horizontal, so acceleration is zero.
• After \(t=2.0\ \text{s}\), the tangent has a negative gradient, so acceleration is negative.

The positive and negative changes in velocity cancel over the complete interval.

Zero average acceleration does not mean zero instantaneous acceleration throughout the journey.

(d) Change in speed

The velocity remains positive throughout.

The object speeds up until \(t=2.0\ \text{s}\), then slows down.

It never stops or reverses direction because its velocity remains above zero.

Question 5

Two carts, A and B, travel along the same straight track. East is defined as positive.

Both carts begin at the same position with an initial velocity of \(+6.0\ \text{m s}^{-1}\).

Cart A maintains a constant velocity of \(+6.0\ \text{m s}^{-1}\) for six seconds.

Cart B follows a smooth curved velocity-time graph. The following values are measured:

The graph for B is above A’s horizontal velocity line for \(0<t<3\ \text{s}\), and below it for \(3<t<6\ \text{s}\). At \(t=3.0\ \text{s}\), B’s curve crosses A’s line with a negative gradient.

Assume trapezoidal approximation between consecutive measurements.

(a) Calculate the displacement of each cart over the six seconds. Do they finish at the same position?

(b) Calculate how far ahead of A cart B is at \(t=3.0\ \text{s}\).

(c) Do the carts meet at \(t=3.0\ \text{s}\)? If not, determine when they meet again.

(d) Compare the instantaneous accelerations of the carts at \(t=3.0\ \text{s}\).

(e) Explain why B reaching a higher velocity than A does not guarantee that B finishes further east.

Solution 5

Both carts finish at the same position, \(36.0\ \text{m}\) east of their starting point. However, B is \(5.0\ \text{m}\) ahead of A at \(t=3.0\ \text{s}\). They do not meet at that instant; they meet again at \(t=6.0\ \text{s}\).

(a) Calculate the displacement of each cart.

For cart A, velocity is constant:

\[ \begin{aligned} \Delta x_A&=v\Delta t\\ &=6.0(6.0)\\ &=36.0\ \text{m} \end{aligned} \]

For cart B, calculate the trapezoidal areas.

\[ \begin{aligned} A_1&=\frac{6.0+8.0}{2}(1)=7.0\\ A_2&=\frac{8.0+9.0}{2}(1)=8.5\\ A_3&=\frac{9.0+6.0}{2}(1)=7.5\\ A_4&=\frac{6.0+3.0}{2}(1)=4.5\\ A_5&=\frac{3.0+4.0}{2}(1)=3.5\\ A_6&=\frac{4.0+6.0}{2}(1)=5.0 \end{aligned} \]

Therefore,

\[ \begin{aligned} \Delta x_B&=7.0+8.5+7.5\\ &\quad+4.5+3.5+5.0\\ &=36.0\ \text{m} \end{aligned} \]

Both carts have the same displacement.

Since they started at the same position, they finish at the same position.

(b) Find B’s separation from A at three seconds.

Compare the areas under their graphs from 0 to 3 seconds.

Cart A travels:

\[ \Delta x_A=6.0(3.0)=18.0\ \text{m} \]

Cart B travels:

\[ \begin{aligned} \Delta x_B&=7.0+8.5+7.5\\ &=23.0\ \text{m} \end{aligned} \]

The separation is

\[ \begin{aligned} \text{separation}&=23.0-18.0\\ &=5.0\ \text{m} \end{aligned} \]

Cart B is \(5.0\ \text{m}\) ahead of A.

(c) Determine when they meet.

At three seconds, B is ahead by \(5.0\ \text{m}\).

Now compare their displacements from three to six seconds.

For B:

\[ \Delta x_B=4.5+3.5+5.0=13.0\ \text{m} \]

For A:

\[ \Delta x_A=6.0(3.0)=18.0\ \text{m} \]

A travels \(5.0\ \text{m}\) further than B during this final interval.

Therefore, A closes the \(5.0\ \text{m}\) gap exactly at \(t=6.0\ \text{s}\).

They meet again at six seconds.

They do not meet between three and six seconds because B’s velocity remains below A’s throughout that interval. B is steadily losing its lead until the separation reaches zero at the end.

(d) Compare instantaneous accelerations at three seconds.

For cart A, the velocity-time graph is horizontal.

Therefore,

\[ a_A=0\ \text{m s}^{-2} \]

For cart B, its curve crosses the horizontal velocity line with a negative gradient.

Therefore,

\[ a_B<0 \]

Cart B has negative instantaneous acceleration, while cart A has zero acceleration.

Both carts have the same velocity at three seconds, but their accelerations are different.

(e) Explain the apparent contradiction.

The tempting prediction is that B must finish further east because it reaches a higher velocity than A.

That prediction ignores the rest of the journey.

B travels further than A during the first three seconds because its velocity is greater. However, it travels less than A during the final three seconds because its velocity falls below A’s.

The positive difference in displacement during the first interval is exactly cancelled by the negative difference during the second interval.

There is another subtle point.

At \(t=3.0\ \text{s}\), the carts have equal velocities, but they are not at the same position. Their velocity difference is zero at that instant, while their accumulated displacement difference is \(5.0\ \text{m}\).

In fact, B is furthest ahead at that instant. Its velocity changes from greater than A’s to less than A’s, so its relative motion changes from moving ahead to falling behind.

This is why you must distinguish the instantaneous information provided by a graph’s height and gradient from the accumulated information provided by its area.

08Where to go next

A curved velocity-time graph is really a record of changing acceleration. Once you can read its tangent gradients, the next useful step is to connect those gradients to an acceleration-time graph.

The relationship works in reverse, too: the height of an acceleration-time graph gives acceleration, while its signed area gives the change in velocity.

For a broader explanation of the underlying graph skills, see Velocity-Time Graphs: Slope, Area and Motion Explained.

To explore the reverse relationship, continue with Acceleration-Time Graphs: Area and Velocity Change.