Speed and Velocity: HSC Physics Explained Clearly

A car travels 100 m east, turns around, and travels 100 m west back to where it started. It took 20 s.

What was its speed? What was its velocity?

It definitely moved, so saying both are zero feels wrong. But because it finished exactly where it started, one of them actually is zero. This is the problem speed and velocity solve: they describe motion in two different ways.

To understand the difference, you need to keep two earlier ideas separate:

• distance is the total ground covered
• displacement is the change in position, including direction

If those two ideas are still fuzzy, review distance and displacement in HSC Physics first.

Speed is built from distance. Velocity is built from displacement.

That one connection explains almost everything.

01Speed tells you how fast. Velocity tells you how fast and which way.

Imagine you’re walking along a perfectly straight road.

You walk east at 2 m/s.

Someone asks, “How fast are you moving?”

The answer is 2 m/s. That’s your speed.

Now they ask, “How fast are you moving, and in which direction?”

The answer is 2 m/s east. That’s your velocity.

So:

\[
\text{average speed} = \frac{\text{total distance}}{\text{total time}}
\]

and

\[
\text{average velocity} = \frac{\text{displacement}}{\text{total time}}
\]

Speed is a scalar. It has magnitude only.

Velocity is a vector. It has magnitude and direction.

You can review that distinction more broadly in scalars and vectors in kinematics.

Predict this

Two students both move at 3 m/s. One walks east, and the other walks west.

Do they have the same speed?

Yes. Both have a speed of 3 m/s.

Do they have the same velocity?

No. Their directions are opposite, so their velocities are different.

If we choose east as the positive direction, we could write their velocities as:

• east-moving student: \(+3\text{ m/s}\)
• west-moving student: \(-3\text{ m/s}\)

The negative sign doesn’t mean the second student is moving “less”. It tells us the direction.

02Using positive and negative directions

In one-dimensional motion, objects can only move along one line. That makes vectors much easier to handle.

We choose one direction to be positive. The opposite direction is then negative.

For example:

\[
\text{east} = + \qquad \text{west} = –
\]

or

\[
\text{right} = + \qquad \text{left} = –
\]

You could choose the opposite convention and the physics would still work. The important thing is to stay consistent.

Think of it like deciding which end of a hallway counts as “forward”. The hallway doesn’t care which choice you make. Your calculations do.

This sign convention matters for velocity and displacement, because they have direction.

Distance and speed don’t need positive or negative signs in normal HSC kinematics problems. They describe how much ground is covered and how fast that ground is being covered.

Quick check

A cyclist moves west at 8 m/s. East has been defined as positive.

What is the cyclist’s velocity?

04What changes when you turn around?

Suppose you walk 10 m east, then 4 m west.

How far did you travel?

You covered all 10 m going east and another 4 m coming back:

\[
\text{distance} = 10 + 4 = 14\text{ m}
\]

But where did you finish compared with where you started?

You ended 6 m east of your starting position:

\[
\text{displacement} = +10 – 4 = +6\text{ m}
\]

That means speed and velocity will no longer have the same magnitude.

This is the point students often miss. Turning around adds to distance, but it can reduce displacement.

A slightly silly way to picture it is texting someone that you’re “100 metres into the relationship”, then walking 40 metres back towards where you started. Your feet have travelled 140 metres, but your actual change in position is only 60 metres. Physics cares about both numbers, depending on what you’re calculating.

The analogy stops being useful once motion becomes more complicated than position along a line. The actual definitions still come from distance and displacement.

06The return-to-start case

Now return to the puzzle from the beginning.

A car travels 100 m east, then 100 m west, returning to its starting point. The entire trip takes 20 s.

Before calculating anything, predict the answer.

The car has clearly been moving, so its average speed must be greater than zero.

But its final position is exactly the same as its initial position, so its displacement is zero.

Question: Calculate the average speed

Step 1: Find the total distance

\[
\text{distance} = 100\text{ m} + 100\text{ m} = 200\text{ m}
\]

Step 2: Divide by total time

\[
\text{average speed}
= \frac{200\text{ m}}{20\text{ s}}
= 10\text{ m/s}
\]

Question: Calculate the average velocity

Step 1: Find the displacement

The car returns to its initial position:

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

Step 2: Divide by total time

\[
v_{\text{avg}}
= \frac{0\text{ m}}{20\text{ s}}
= 0\text{ m/s}
\]

07The most tempting misconception: “Velocity is just speed with a direction added”

This is close enough to be useful at first, but it can become misleading.

At a particular instant, velocity does tell you the rate of motion and its direction. Its magnitude is the instantaneous speed.

But when calculating average values across a journey, you can’t simply calculate average speed and attach a direction to it.

Consider the runner from earlier:

• average speed = \(4.67\text{ m/s}\)
• average velocity = \(2.0\text{ m/s east}\)

Writing “4.67 m/s east” for the average velocity would be wrong.

The two calculations use different quantities:

So when you see an HSC question, don’t start by asking, “Which formula do I remember?”

Ask:

Am I interested in the total path travelled, or the change in position?

If it’s the total path, think distance and speed.

If it’s the change in position, think displacement and velocity.

08Can speed be negative?

No.

A speed such as:

\[
-5\text{ m/s}
\]

doesn’t make physical sense as a speed in this context.

Speed is the magnitude of motion, so we report it as a non-negative value.

Velocity can be negative because its sign represents direction.

For example, if right is positive:

\[
v = -5\text{ m/s}
\]

means the object is travelling left at a speed of 5 m/s.

Question

A car has a velocity of \(-12\text{ m/s}\). What is its speed?

09Average velocity doesn’t tell you everything that happened

Suppose two cars each have an average velocity of \(5\text{ m/s east}\) over 20 s.

They have the same displacement:

\[
\Delta x = v_{\text{avg}}\Delta t
\]

so:

\[
\Delta x = (5\text{ m/s})(20\text{ s}) = 100\text{ m east}
\]

But their journeys might have been completely different.

One car could have travelled steadily east the whole time.

The other could have accelerated, slowed down, briefly travelled west, and then finished 100 m east of where it started.

Average velocity only compares the starting and finishing positions over the time interval. It does not describe every moment of the motion.

That limitation leads naturally to the next idea in kinematics: instantaneous velocity.

Instead of asking how position changed over an entire interval, instantaneous velocity asks how quickly position is changing at one particular moment. Once that distinction is clear, velocity-time graphs and acceleration become much easier to interpret.