Acceleration-Time Graphs: Area and Velocity Change

A car accelerates at \(3.0\ \text{m s}^{-2}\) for 4 seconds, then its acceleration drops to zero. What has definitely changed after those 4 seconds: its position, its velocity, or both?

Its position has changed, but the acceleration-time graph does not tell you that change directly. What it gives you directly is the change in velocity. In this case,

\[
\Delta v = (3.0\ \text{m s}^{-2})(4.0\ \text{s}) = 12\ \text{m s}^{-1}
\]

Notice what we just calculated: acceleration multiplied by time. On an acceleration-time graph, that multiplication appears as an area.

That is the central idea of this guide:

The signed area between an acceleration-time graph and the time axis gives the change in velocity.

The word signed matters. Area above the time axis is positive. Area below it is negative.

01What an acceleration-time graph actually shows

An acceleration-time graph has:

• time \(t\), measured in seconds (s), on the horizontal axis
• acceleration \(a\), measured in metres per second squared (\(\text{m s}^{-2}\)), on the vertical axis

Suppose a graph is a horizontal line at \(a = 2.0\ \text{m s}^{-2}\).

Before going further, make a prediction. Does that mean the object is moving at a constant velocity?

No. It means its velocity is changing at a constant rate.

An acceleration of \(2.0\ \text{m s}^{-2}\) means that every second, the velocity changes by \(2.0\ \text{m s}^{-1}\).

So if the velocity begins at \(5.0\ \text{m s}^{-1}\):

• after 1 s, it is \(7.0\ \text{m s}^{-1}\)
• after 2 s, it is \(9.0\ \text{m s}^{-1}\)
• after 3 s, it is \(11.0\ \text{m s}^{-1}\)

That is why acceleration-time graphs connect so closely to velocity-time graphs. Acceleration tells you how the velocity graph is changing.

02Why area gives change in velocity

Start with the familiar equation for constant acceleration:

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

Here:

• \(a\) is acceleration in \(\text{m s}^{-2}\)
• \(\Delta v\) is the change in velocity in \(\text{m s}^{-1}\)
• \(\Delta t\) is the elapsed time in seconds

Rearranging,

\[
\Delta v = a\Delta t
\]

Now picture a constant acceleration on an acceleration-time graph. The region underneath it is a rectangle.

The rectangle has:

• height \(a\)
• width \(\Delta t\)

Its area is therefore

\[
\text{area} = a\Delta t
\]

But \(a\Delta t = \Delta v\).

So the area is not just a convenient graph trick. It comes directly from the definition of acceleration.

The units confirm it:

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

Those are units of velocity.

03The graph gives a change, not automatically the final velocity

This is an easy place to lose marks.

Suppose the area under an acceleration-time graph is \(+8\ \text{m s}^{-1}\). A student writes:

Final velocity \(= 8\ \text{m s}^{-1}\).

That is only true if the object started from rest.

The graph has told us

\[
\Delta v = +8\ \text{m s}^{-1}
\]

not necessarily \(v_f = 8\ \text{m s}^{-1}\).

In general,

\[
v_f = v_i + \Delta v
\]

If \(v_i = -3\ \text{m s}^{-1}\), for example,

\[
v_f = -3 + 8 = +5\ \text{m s}^{-1}
\]

The acceleration history changed the velocity by \(+8\ \text{m s}^{-1}\), enough in this case to reverse the object’s direction.

04Areas below the time axis are negative

Imagine the acceleration graph lies at

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

for 2.0 s.

Its signed area is

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

So the velocity decreases by \(8.0\ \text{m s}^{-1}\).

Be careful with the phrase “velocity decreases”. It does not always mean the object slows down.

Suppose an object begins with

\[
v_i = -5.0\ \text{m s}^{-1}
\]

and undergoes a velocity change of

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

Then

\[
v_f = -5.0 – 8.0 = -13\ \text{m s}^{-1}
\]

Its velocity has become more negative, but its speed has increased from \(5.0\ \text{m s}^{-1}\) to \(13\ \text{m s}^{-1}\).

Negative acceleration does not mean slowing down.

Whether an object speeds up or slows down depends on the signs of both velocity and acceleration. If you need to refresh that distinction, see Speed and Velocity: HSC Physics Explained Clearly.

05A useful decision rule for speeding up and slowing down

Compare the signs of velocity and acceleration.

Why?

Velocity tells you which way the object is moving. Acceleration tells you which way its velocity is being pushed.

Think of velocity and acceleration as two people arguing about where to go. If they agree on the direction, the magnitude of the velocity grows. If they disagree, the magnitude initially shrinks.

The analogy breaks once the velocity reaches zero. The object can then reverse direction, after which the same acceleration may make it speed up again.

06Changing acceleration means changing area shapes

Acceleration does not need to be constant.

Suppose acceleration rises steadily from \(0\) to \(6.0\ \text{m s}^{-2}\) over 4.0 s.

The region beneath the graph is a triangle, so

\[
\Delta v = \frac{1}{2}bh
\]

where \(b\) is the time interval and \(h\) is the maximum acceleration.

\[
\Delta v = \frac{1}{2}(4.0\ \text{s})(6.0\ \text{m s}^{-2})
= 12\ \text{m s}^{-1}
\]

Even though the acceleration was continuously changing, the area still gives the total velocity change.

More generally,

\[
\Delta v = \int_{t_1}^{t_2} a\,dt
\]

The integral symbol means “add up all the tiny contributions \(a\,dt\) across the time interval”. For HSC graph problems with straight-line sections, you can usually find that area using rectangles, triangles, and trapeziums rather than performing calculus.

07Positive and negative areas can cancel

Now consider an object with this acceleration history:

• from \(0\) to \(4.0\ \text{s}\), acceleration increases uniformly from \(0\) to \(4.0\ \text{m s}^{-2}\)
• from \(4.0\) to \(6.0\ \text{s}\), acceleration is \(-3.0\ \text{m s}^{-2}\)

Predict what happens to the velocity over the entire 6.0 s. Does it increase, decrease, or return to its starting value?

Work with the signed areas.

From \(0\) to \(4.0\ \text{s}\), the triangular positive area is

\[
\Delta v_1
= \frac{1}{2}(4.0)(4.0)
= 8.0\ \text{m s}^{-1}
\]

From \(4.0\) to \(6.0\ \text{s}\), the rectangular negative area is

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

Therefore,

\[
\Delta v_{\text{total}}
= 8.0 – 6.0
= 2.0\ \text{m s}^{-1}
\]

The velocity ends \(2.0\ \text{m s}^{-1}\) higher than it started.

The negative section did not “erase the motion”. It removed some of the earlier increase in velocity.

08Zero acceleration does not mean zero velocity

A horizontal section of an acceleration-time graph at \(a = 0\) can look suspiciously like “nothing is happening”.

Something is happening. The object can still be moving.

If

\[
a = 0
\]

then the velocity is not changing.

So an object travelling at \(18\ \text{m s}^{-1}\) with zero acceleration continues at \(18\ \text{m s}^{-1}\), assuming the one-dimensional model applies.

An object already at rest also remains at rest.

The acceleration graph alone cannot tell you which of those situations you have unless you also know the object’s velocity at some point.

09An acceleration-time graph does not directly give displacement

There are two different “area rules” students often mix up:

Why can’t the area under an acceleration-time graph give displacement?

Check the units:

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

That is velocity, not displacement.

To get displacement from an acceleration history, you first need enough information to determine the velocity as a function of time. Then you find the area under the corresponding velocity-time graph.

This is also why the initial velocity matters. Two objects can have identical acceleration-time graphs but completely different displacements because they started with different velocities.

10How to move between acceleration-time and velocity-time graphs

An acceleration-time graph tells you the slope behaviour of a velocity-time graph.

Remember that

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

so acceleration is the gradient of a velocity-time graph.

That gives these connections:

There is one subtle point here. The height of an acceleration-time graph tells you the slope of the velocity-time graph. The area of the acceleration-time graph tells you how much the velocity itself has changed.

Those are different jobs.

11A reliable method for HSC graph questions

When you are given an acceleration-time graph, work in this order:

1. Choose a positive direction. Usually the graph has already done this through its signs.
2. Record the initial velocity. If it is not given, be careful about what can actually be determined.
3. Split the graph into simple regions. Use rectangles, triangles, or trapeziums where appropriate.
4. Give each area its sign. Above the time axis is positive. Below is negative.
5. Add the signed areas. This gives \(\Delta v\).
6. Use \(v_f = v_i + \Delta v\).
7. If asked about speed or direction, inspect the velocity rather than the acceleration alone.
8. Check the units. Area under an acceleration-time graph must have units of \(\text{m s}^{-1}\).

That final unit check catches a surprising number of mistakes.

12Questions and solutions

Question 1

A cyclist has an acceleration of \(1.5\ \text{m s}^{-2}\) for 4.0 s. The cyclist’s velocity at the start of the interval is \(6.0\ \text{m s}^{-1}\).

Find the cyclist’s velocity after 4.0 s.

Solution 1

The cyclist’s final velocity is \(\boxed{12\ \text{m s}^{-1}}\).

The area beneath the acceleration-time graph gives the change in velocity:

\[
\Delta v
= a\Delta t
= (1.5\ \text{m s}^{-2})(4.0\ \text{s})
= 6.0\ \text{m s}^{-1}
\]

Now add that change to the initial velocity:

\[
v_f
= v_i + \Delta v
= 6.0 + 6.0
= 12\ \text{m s}^{-1}
\]

The common trap is to report \(6.0\ \text{m s}^{-1}\), the area, as the final velocity. It is only the change in velocity.

Question 2

An object moves initially at \(-7.0\ \text{m s}^{-1}\). It then experiences a constant acceleration of \(+2.0\ \text{m s}^{-2}\) for 5.0 s.

Calculate its final velocity and state whether its direction of motion has changed.

Solution 2

The final velocity is \(\boxed{+3.0\ \text{m s}^{-1}}\), so the object has changed direction.

The velocity change is

\[
\Delta v
= a\Delta t
= (2.0\ \text{m s}^{-2})(5.0\ \text{s})
= +10\ \text{m s}^{-1}
\]

Therefore,

\[
v_f
= v_i + \Delta v
= -7.0 + 10
= +3.0\ \text{m s}^{-1}
\]

Because the velocity changed from negative to positive, it must have passed through zero during the interval. At that instant the object was momentarily at rest, then reversed direction.

The tempting mistake is to say positive acceleration means the object was moving in the positive direction for the whole 5.0 s. Acceleration describes how velocity changes, not necessarily the current direction of motion.

Question 3

An object’s acceleration changes uniformly from \(+2.0\ \text{m s}^{-2}\) at \(t=0\) to \(+6.0\ \text{m s}^{-2}\) at \(t=4.0\ \text{s}\).

Its initial velocity is \(5.0\ \text{m s}^{-1}\).

Find its velocity at \(t=4.0\ \text{s}\).

Solution 3

The object’s velocity at \(4.0\ \text{s}\) is \(\boxed{21\ \text{m s}^{-1}}\).

Because the acceleration changes linearly, the area under the graph is a trapezium. Its area is

\[
\Delta v
= \frac{1}{2}(a_1+a_2)\Delta t
\]

where \(a_1 = 2.0\ \text{m s}^{-2}\), \(a_2 = 6.0\ \text{m s}^{-2}\), and \(\Delta t = 4.0\ \text{s}\).

Substituting,

\[
\Delta v
= \frac{1}{2}(2.0+6.0)(4.0)
= 16\ \text{m s}^{-1}
\]

Then

\[
v_f
= 5.0 + 16
= 21\ \text{m s}^{-1}
\]

The acceleration is not constantly \(6.0\ \text{m s}^{-2}\), so multiplying \(6.0\) by the full 4.0 s would overestimate the change in velocity.

Question 4

A robot moves along a straight rail. From \(t=0\) to \(t=3.0\ \text{s}\), its acceleration is \(+4.0\ \text{m s}^{-2}\). From \(t=3.0\) to \(t=7.0\ \text{s}\), its acceleration is \(-3.0\ \text{m s}^{-2}\).

A student says:

“The positive and negative parts have equal effects because \(4\times3 = 3\times4\), so the robot must finish where it started.”

The robot’s initial velocity is \(+5.0\ \text{m s}^{-1}\).

Determine the robot’s final velocity, and assess the student’s conclusion about its final position.

Solution 4

The robot’s final velocity is \(\boxed{+5.0\ \text{m s}^{-1}}\), but the claim that it must finish where it started is incorrect.

For the first interval,

\[
\Delta v_1
= (4.0)(3.0)
= +12\ \text{m s}^{-1}
\]

For the second interval,

\[
\Delta v_2
= (-3.0)(4.0)
= -12\ \text{m s}^{-1}
\]

Therefore,

\[
\Delta v_{\text{total}}
= +12 -12
= 0
\]

and

\[
v_f
= v_i + \Delta v_{\text{total}}
= 5.0 + 0
= 5.0\ \text{m s}^{-1}
\]

So the student is correct that the velocity returns to its initial value.

However, zero net change in velocity does not mean zero displacement. The robot has been moving during the interval, and for much of the motion its velocity is greater than \(5.0\ \text{m s}^{-1}\).

The area under an acceleration-time graph gives velocity change, not displacement. To determine the robot’s change in position, we would need to construct or calculate from its velocity-time behaviour.

Question 5

Two carts, A and B, travel along parallel straight tracks.

Both carts have exactly the same acceleration-time graph from \(t=0\) to \(t=8.0\ \text{s}\). The total signed area under that graph is \(-6.0\ \text{m s}^{-1}\).

Cart A begins with velocity \(+2.0\ \text{m s}^{-1}\), while cart B begins with velocity \(+10.0\ \text{m s}^{-1}\).

A student claims that because both carts have the same acceleration graph, they must have the same final velocity and travel the same displacement.

Evaluate both parts of the claim.

Solution 5

Neither part of the claim is correct. Cart A finishes at \(\boxed{-4.0\ \text{m s}^{-1}}\), cart B finishes at \(\boxed{+4.0\ \text{m s}^{-1}}\), and their displacements over the 8.0 s are not equal.

The common acceleration history gives both carts the same change in velocity:

\[
\Delta v = -6.0\ \text{m s}^{-1}
\]

For cart A,

\[
v_{A,f}
= v_{A,i}+\Delta v
= 2.0-6.0
= -4.0\ \text{m s}^{-1}
\]

For cart B,

\[
v_{B,f}
= v_{B,i}+\Delta v
= 10.0-6.0
= 4.0\ \text{m s}^{-1}
\]

So identical acceleration histories do not produce identical final velocities unless the initial velocities are also identical.

Their displacements are also different. At every instant, cart B’s velocity is \(8.0\ \text{m s}^{-1}\) greater than cart A’s because they begin \(8.0\ \text{m s}^{-1}\) apart in velocity and then undergo exactly the same acceleration.

Over 8.0 s, that constant difference in velocity produces a displacement difference of

\[
\Delta x_B-\Delta x_A
= (8.0\ \text{m s}^{-1})(8.0\ \text{s})
= 64\ \text{m}
\]

So cart B travels \(64\ \text{m}\) further in the positive direction than cart A over the interval.

The deeper point is that an acceleration-time graph tells you how velocity changes. To know the actual velocity, you also need an initial velocity. Once you know the velocity history, its area can then tell you displacement. That chain from acceleration to velocity to position is the next useful connection to master when interpreting motion graphs.