Uniform and Uniformly Accelerated Motion in HSC Physics

A car travels 20 m in the first second, another 20 m in the next second, and another 20 m in the third. A second car travels 5 m, then 15 m, then 25 m over the same three one-second intervals.

Which car is accelerating?

The first car is covering equal displacements in equal times. Its velocity is staying constant. The second car covers more distance during each successive second, which tells us its velocity is changing.

That difference gives us two of the most useful motion models in HSC Physics:

• uniform motion, where velocity stays constant
• uniformly accelerated motion, where acceleration stays constant

The names sound annoyingly similar. The trick is not to memorise them as definitions. Instead, ask one question:

What quantity is staying constant?

01Uniform motion: velocity stays constant

Picture a car moving along a straight road at \(12\text{ m s}^{-1}\).

After 1 second, it travels 12 m. After another second, another 12 m. After another second, another 12 m.

Before reading further, predict what happens to its velocity.

It stays at \(12\text{ m s}^{-1}\).

That is uniform motion.

More precisely:

An object is in uniform motion when its velocity is constant.

This means both the magnitude and direction of the velocity remain unchanged.

If you’re not completely comfortable with that distinction, it is worth reviewing speed and velocity before going further.

What stays constant?

For uniform motion:

Because velocity does not change,

\[
a = 0
\]

where \(a\) is acceleration in \(\text{m s}^{-2}\).

For motion in one dimension,

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

where \(v\) is velocity, \(\Delta x\) is displacement, and \(\Delta t\) is the time interval.

Rearranging gives:

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

This equation is useful because constant velocity means the object gains the same displacement during every equal time interval.

A quick prediction

A cyclist moves east at a constant \(6.0\text{ m s}^{-1}\). What is the cyclist’s acceleration?

Check your understanding

A runner moves west at \(4.0\text{ m s}^{-1}\) for 15 s. What is the runner’s displacement?

02Uniformly accelerated motion: acceleration stays constant

Now imagine a car whose velocity changes like this:

What pattern do you notice?

The velocity increases by \(3\text{ m s}^{-1}\) every second.

So the velocity is definitely not constant. But the rate at which it changes is constant.

The acceleration is:

\[
a = 3\text{ m s}^{-2}
\]

This is uniformly accelerated motion.

An object undergoes uniformly accelerated motion when its acceleration remains constant.

Another way to say the same thing is:

The velocity changes by equal amounts in equal time intervals.

That second version is often easier to recognise in a table.

03Do equal time intervals mean equal distances?

This is where a tempting mistake appears.

Suppose a car accelerates constantly. It travels for one second, then another second, then another.

Would it cover equal distances in each second?

No.

If it is speeding up, each one-second interval begins with a greater velocity than the previous one. It therefore covers more displacement during later intervals.

Constant acceleration means equal changes in velocity, not equal changes in position.

That distinction is worth remembering.

04The four constant-acceleration equations

Once acceleration is constant, several useful relationships become available.

For one-dimensional motion:

\[
v = u + at
\]

\[
\Delta x = ut + \frac{1}{2}at^2
\]

\[
v^2 = u^2 + 2a\Delta x
\]

\[
\Delta x = \frac{u+v}{2}t
\]

where:

• \(u\) is initial velocity in \(\text{m s}^{-1}\)
• \(v\) is final velocity in \(\text{m s}^{-1}\)
• \(a\) is constant acceleration in \(\text{m s}^{-2}\)
• \(t\) is time in seconds
• \(\Delta x\) is displacement in metres

These equations are sometimes called the constant-acceleration equations or SUVAT equations.

Do not choose one because it “looks familiar”. Look at the information you have and the quantity you need.

For example, if a question gives \(u\), \(a\), and \(t\), and asks for \(v\), then

\[
v = u + at
\]

contains exactly those four quantities.

05Why displacement contains \(t^2\)

The equation

\[
\Delta x = ut + \frac{1}{2}at^2
\]

can look like something you simply have to memorise. There is a useful physical reason for its shape.

Start with a car moving at initial velocity \(u\). If it did not accelerate at all, after time \(t\) it would travel:

\[
ut
\]

But acceleration adds extra velocity as time passes. That extra motion contributes:

\[
\frac{1}{2}at^2
\]

So the total displacement is:

\[
\Delta x = \underbrace{ut}_{\text{motion from initial velocity}} + \underbrace{\frac{1}{2}at^2}_{\text{extra motion caused by acceleration}}
\]

The \(t^2\) matters. If an object starts from rest with constant acceleration, doubling the time does not simply double its displacement. It makes the displacement four times as large.

Prediction

An object starts from rest and accelerates uniformly. It travels 10 m in the first 2 seconds. If the same acceleration continues for 4 seconds in total, will it have travelled 20 m?

No.

Starting from rest means \(u=0\), so:

\[
\Delta x = \frac{1}{2}at^2
\]

Displacement is proportional to \(t^2\).

Changing time from 2 s to 4 s multiplies time by 2, so displacement is multiplied by:

\[
2^2 = 4
\]

The object would therefore travel:

\[
4(10\text{ m}) = 40\text{ m}
\]

08Negative acceleration does not always mean slowing down

This is one of the most common kinematics mistakes.

A student sees:

\[
a = -3\text{ m s}^{-2}
\]

and says, “The object is slowing down.”

Not necessarily.

The minus sign only tells us the direction of the acceleration.

To decide whether the object is speeding up or slowing down, compare the directions of velocity and acceleration.

Here is a slightly silly way to picture it. Velocity and acceleration are like two people deciding where to go on a date. If they agree on the direction, the relationship gets more intense and the speed increases. If they pull in opposite directions, the speed decreases.

Where does the analogy break? Velocity and acceleration are vectors, not people with intentions. The useful part is simply whether their directions match.

Check your understanding

An object has velocity \(-8\text{ m s}^{-1}\) and acceleration \(-2\text{ m s}^{-2}\). Is it speeding up or slowing down?

09How to recognise the two models from words

HSC questions do not always say “this object undergoes uniform motion”. More often, you have to recognise the model from the description.

Clues for uniform motion

Look for wording such as:

• “constant velocity”
• “moves at \(15\text{ m s}^{-1}\) east”
• “travels equal displacements in equal time intervals”
• “moves in a straight line at constant speed”

The key idea is:

\[
\boxed{v = \text{constant}}
\]

and therefore:

\[
\boxed{a=0}
\]

Clues for uniformly accelerated motion

Look for:

• “constant acceleration”
• “accelerates at \(4.0\text{ m s}^{-2}\)”
• “slows at a constant rate”
• “velocity increases by the same amount each second”
• “freely falling”, when air resistance is neglected and gravitational acceleration is being treated as constant near Earth’s surface

The key idea is:

\[
\boxed{a = \text{constant}}
\]

The velocity usually changes.

10How to recognise them from graphs

Graphs make the distinction especially clear.

11Position-time graphs

The gradient of a position-time graph is velocity.

For uniform motion, velocity is constant, so the gradient is constant.

That produces a straight line.

A steeper straight line means a greater magnitude of velocity.

A horizontal line means:

\[
v=0
\]

so the object is stationary.

For uniformly accelerated motion, velocity changes with time. The gradient of the position-time graph therefore changes.

The graph is curved rather than straight.

Check your understanding

A position-time graph becomes progressively steeper as time passes. Is the object in uniform motion?

12Velocity-time graphs

The gradient of a velocity-time graph is acceleration:

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

This gives a very useful recognition rule.

Uniform motion

Constant velocity gives a horizontal line on a velocity-time graph.

Its gradient is zero:

\[
a=0
\]

Uniformly accelerated motion

Constant acceleration gives a straight sloping line.

The line is straight because its gradient stays constant.

Positive gradient means positive acceleration.

Negative gradient means negative acceleration.

Do not confuse “straight line” with “constant velocity”. On a velocity-time graph, a sloping straight line means the acceleration is constant.

Check your understanding

A velocity-time graph is a straight line rising from \(4\text{ m s}^{-1}\) at \(t=0\) to \(16\text{ m s}^{-1}\) at \(t=6\text{ s}\). Find the acceleration.

13Acceleration-time graphs

An acceleration-time graph is even more direct.

For uniform motion:

\[
a=0
\]

so the graph lies along the time axis.

For uniformly accelerated motion, acceleration is some constant value, so the graph is a horizontal line above or below the time axis.

This gives us a useful comparison:

14One important complication: constant speed is not always uniform motion

Suppose a car travels around a circular track at a constant speed of \(10\text{ m s}^{-1}\).

Is its velocity constant?

No.

Velocity includes direction. The car’s direction is continually changing as it travels around the circle.

That means its velocity is changing, so it has acceleration even though its speed stays constant.

This is why “constant speed” and “constant velocity” are not always interchangeable.

For straight-line HSC kinematics problems, constant speed in a fixed direction does correspond to constant velocity. Once motion becomes two-dimensional, direction becomes much more important. If this feels unfamiliar, scalars and vectors in kinematics is the useful prerequisite.

15A decision rule for exam questions

When you meet a motion problem, do not immediately hunt for an equation.

First ask:

1. What quantity is constant?

If velocity is constant:

\[
a=0
\]

You are dealing with uniform motion.

If acceleration is constant:

\[
a=\text{constant}
\]

You can use the constant-acceleration equations.

2. What direction have I chosen as positive?

Assign signs before substituting values.

3. What quantities do I know?

List \(u\), \(v\), \(a\), \(t\), and \(\Delta x\).

4. Which equation contains what I know and what I need?

For example:

The constant-acceleration equations are only valid when acceleration is constant over the interval you are analysing.

That condition is not a small technical detail. It is the reason the equations work.

16Final check: can you tell the models apart?

Consider these four descriptions.

A. A train travels north at \(22\text{ m s}^{-1}\) for 30 s.

B. A car increases its velocity from \(5\text{ m s}^{-1}\) to \(17\text{ m s}^{-1}\) over 4 s, gaining \(3\text{ m s}^{-1}\) each second.

C. A ball’s velocity changes irregularly as it is blown around by gusting wind.

D. A car travels around a bend at constant speed.

Which model applies to each?