Distance and Displacement in HSC Physics Explained Clearly

You walk 30 m east to pick up your bag, realise you left your phone behind, then walk 30 m west back to where you started.

How far did you travel? 60 m.

How much did your position change? 0 m.

Those two answers describe the same journey, but they measure different things. Distance keeps track of the whole path. Displacement only compares where you finished with where you started.

That difference looks simple until directions and negative signs appear in a calculation. This guide builds a method for keeping them straight.

01Start by picturing the motion

Imagine a straight road with your house at one point and a bus stop 100 m east of it.

You walk from your house to the bus stop.

Your distance travelled is 100 m. Your displacement is also 100 m east.

So far, distance and displacement happen to have the same magnitude.

Now you walk 40 m back towards home.

Your total distance is now:

\[
100 + 40 = 140\text{ m}
\]

But you finish only 60 m east of your starting point.

Your displacement is therefore 60 m east.

The key idea is:

• Distance asks: how much ground did you cover?
• Displacement asks: where did you finish compared with where you started?

A useful picture is to imagine leaving a breadcrumb trail as you move. Distance measures the whole breadcrumb trail. Displacement draws one straight arrow from your starting point to your finishing point.

That picture is useful, but not exact in every situation. Displacement is not simply “the shortest route you could have travelled”. It is a vector change in position, so direction matters.

02Distance is total path length

Distance measures the total length of the path travelled.

If you move 20 m east and then 10 m west, you have still physically travelled:

\[
20 + 10 = 30\text{ m}
\]

You don’t subtract the second part when calculating distance because changing direction does not undo the ground you already covered.

Distance is a scalar quantity. It has magnitude, but no direction.

So it would make sense to write:

Distance travelled = 30 m

It would not make sense to write:

Distance travelled = 30 m east

Adding “east” would turn the description into something directional.

For a fuller explanation of scalar and vector quantities, see Scalars and Vectors in Kinematics for HSC Physics.

Check your understanding

A student walks 8 m north and then 3 m south. What distance have they travelled?

03Displacement is change in position

Displacement compares the final position with the initial position.

In one dimension:

\[
\Delta x = x_f – x_i
\]

where:

• \(\Delta x\) is displacement, measured in metres (m)
• \(x_f\) is final position
• \(x_i\) is initial position

The symbol \(\Delta\), pronounced “delta”, means “change in”.

So \(\Delta x\) literally means “change in position”.

Suppose you start at \(x = 2\text{ m}\) and finish at \(x = 9\text{ m}\).

Then:

\[
\Delta x = x_f – x_i = 9 – 2 = 7\text{ m}
\]

Your displacement is \(+7\text{ m}\).

If positive has been defined as east, that means 7 m east.

Now reverse the journey. Start at \(x = 9\text{ m}\) and finish at \(x = 2\text{ m}\).

\[
\Delta x = 2 – 9 = -7\text{ m}
\]

The negative sign does not mean you somehow moved “negative metres”. It tells you that the displacement points in the direction defined as negative.

If east is positive, west is negative.

04The sign comes from your coordinate system

Before solving a one-dimensional motion problem, choose a positive direction.

For example:

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

You could choose the opposite instead:

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

Both choices are valid.

Physics does not secretly prefer east.

The important thing is to choose one convention and use it consistently.

Suppose a person moves 12 m west. If east is positive, the change in position is:

\[
\Delta x = -12\text{ m}
\]

If west had instead been chosen as positive, the same physical displacement would be:

\[
\Delta x = +12\text{ m}
\]

The number changed because the coordinate system changed. The actual motion did not.

Prediction question

If east is defined as positive, a car has a displacement of \(-25\text{ m}\). Has it travelled a distance of \(-25\text{ m}\)?

06Why you should not automatically subtract

Students sometimes learn a shortcut such as:

“For displacement, subtract.”

That is only partly useful.

The safest method is to give each directed movement a sign and then add the signed changes.

For example, consider:

• 20 m west
• then 50 m east
• then 10 m west

Take east as positive.

The signed movements are:

\[
-20,\quad +50,\quad -10
\]

So:

\[
\Delta x = -20 + 50 – 10 = +20\text{ m}
\]

This method becomes much more reliable when there are several changes of direction.

For distance, ignore the direction signs and add the lengths:

\[
d = 20 + 50 + 10 = 80\text{ m}
\]

08Position is not the same as distance travelled

There is another distinction that becomes important in HSC kinematics.

Suppose an object is at:

\[
x = -4\text{ m}
\]

That does not mean it has travelled a distance of \(-4\text{ m}\).

It means the object is located 4 m on the negative side of the chosen origin.

Position describes where something is.

Distance describes how much path it has travelled.

Displacement describes how much its position has changed.

Consider an object that starts at \(x_i = -4\text{ m}\) and finishes at \(x_f = +6\text{ m}\).

Its displacement is:

\[
\Delta x = x_f – x_i
\]

\[
\Delta x = 6 – (-4) = 10\text{ m}
\]

Be careful with the double negative. Moving from \(-4\text{ m}\) to \(+6\text{ m}\) is a change of 10 m in the positive direction.

Check your understanding

An object starts at \(x_i = +7\text{ m}\) and finishes at \(x_f = -5\text{ m}\). Find its displacement.

10The most tempting misconception: “distance and displacement are basically the same”

They are equal in some simple journeys.

If you travel 50 m east in a straight line without turning around:

\[
d = 50\text{ m}
\]

and:

\[
|\Delta x| = 50\text{ m}
\]

So it is easy to start thinking that displacement is just another word for distance.

The difference appears as soon as your path includes a change of direction.

Suppose you travel 50 m east and then 50 m west.

Your distance is:

\[
d = 50 + 50 = 100\text{ m}
\]

But your final position is the same as your initial position, so:

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

You have moved a lot, but your net change in position is zero.

That is completely possible.

11Can displacement ever be larger than distance?

No.

The magnitude of displacement cannot be greater than the distance travelled.

In symbols:

\[
|\Delta x| \leq d
\]

where \(d\) is distance travelled.

Why?

Distance counts the whole route you actually followed. Displacement only compares the start and finish.

If you walk directly from A to B without changing direction, the magnitude of displacement can equal the distance.

If you take any detour or reverse direction, the distance becomes larger than the magnitude of displacement.

Check your understanding

A student travels a total distance of 80 m. Could their displacement have a magnitude of 100 m?

12A reliable method for HSC questions

When a question contains several movements, use this sequence.

1. Choose a positive direction.
   For example, east \(=+\) and west \(=-\).

2. For distance, add all path lengths as positive quantities.

3. For displacement, attach signs to movements or use

\[
\Delta x = x_f – x_i
\]

1. Interpret the sign.
   A negative displacement means the final change is in the direction you defined as negative.

2. Check whether the answer is physically possible.
   The magnitude of displacement should never exceed the distance travelled.

A compact comparison is:

The sign convention you practise here becomes important almost immediately when you move into velocity and acceleration. Average speed uses distance, while average velocity uses displacement. If you confuse the two at this stage, the same mistake tends to carry into later kinematics calculations.