Relative Velocity in One Dimension for HSC Physics

Two cars can travel along the same road at almost 25 m/s, yet each driver can see the other car barely moving. How can something moving that fast look nearly stationary from another car?

Suppose car A moves right at \(25\text{ m s}^{-1}\), while car B beside it moves right at \(23\text{ m s}^{-1}\). Before calculating, predict what a passenger in car B would measure for car A. Would A still be moving at \(25\text{ m s}^{-1}\)?

No. From B’s moving frame, A creeps forward at only \(2\text{ m s}^{-1}\). Relative velocity is the tool that lets us calculate that change of viewpoint properly.

01The frame changes what you measure

Velocity always has to be measured relative to something.

A roadside observer might measure a car travelling at \(25\text{ m s}^{-1}\) to the right. A passenger in another car might measure that same car travelling at only \(2\text{ m s}^{-1}\) to the right. Neither measurement is wrong. The observers are using different frames of reference.

A frame of reference is the viewpoint from which positions and velocities are measured.

For one-dimensional motion, we can picture every object on the same straight number line.

Imagine attaching the zero of a long measuring tape to car B. As B moves, the whole measuring tape moves with it. Car A’s velocity across that moving tape is its velocity relative to B.

That gives us the central idea:

To find how A moves from B’s frame, subtract B’s velocity from A’s velocity.

02The relative velocity equation

The velocity of object A relative to object B is

\[
v_{A/B}=v_A-v_B
\]

where:

• \(v_{A/B}\) is the velocity of A relative to B,
• \(v_A\) is the velocity of A in your original reference frame, and
• \(v_B\) is the velocity of B in that same original reference frame.

The order matters. \(v_{A/B}\) means A as seen from B.

You might also see the notation \(v_{AB}\), but the slash makes the meaning harder to mix up.

There is a simple reason subtraction appears. If the positions of the two objects are \(x_A\) and \(x_B\), then A’s position measured from B is

\[
x_{A/B}=x_A-x_B.
\]

Velocity tells us how quickly position changes. So the rate at which that separation changes is

\[
v_{A/B}=v_A-v_B.
\]

This isn’t just a trick for memorising the formula. Relative velocity measures how quickly one object’s position is changing compared with the other object’s position.

If your understanding of signed velocity is shaky, it is worth revisiting Speed and Velocity: HSC Physics Explained Clearly before going further.

03Choose a positive direction first

The most common relative velocity mistake is to work with speeds instead of velocities.

Velocity has direction. In one dimension, we represent that direction using positive and negative signs.

For example, choose right as positive:

• \(+12\text{ m s}^{-1}\) means \(12\text{ m s}^{-1}\) to the right.
• \(-12\text{ m s}^{-1}\) means \(12\text{ m s}^{-1}\) to the left.

Then use

\[
v_{A/B}=v_A-v_B
\]

without changing the rule.

This is safer than trying to memorise instructions such as “subtract when they go the same way” and “add when they go opposite ways”. Those shortcuts can work, but signed velocities handle every case with one equation.

04Opposite directions are where signs earn their keep

Now suppose two vehicles approach each other.

A tempting prediction is that we should subtract their speeds. If each vehicle moves at \(15\text{ m s}^{-1}\), that would give zero.

But zero would mean the distance between them isn’t changing. That clearly isn’t what happens when they drive towards each other.

The issue is that \(15\text{ m s}^{-1}\) and \(15\text{ m s}^{-1}\) are speeds. Once we include direction, one velocity is positive and the other is negative.

Suppose A moves right at \(15\text{ m s}^{-1}\), while B moves left at \(15\text{ m s}^{-1}\):

\[
v_A=+15\text{ m s}^{-1}, \qquad v_B=-15\text{ m s}^{-1}.
\]

Then

\[
v_{A/B}=15-(-15)=30\text{ m s}^{-1}.
\]

From B’s frame, A approaches at \(30\text{ m s}^{-1}\).

Subtracting a negative number produces the familiar “sum of the speeds”, but the underlying rule never changed.

05A negative relative velocity is not a negative speed

Suppose you calculate

\[
v_{A/B}=-4\text{ m s}^{-1}.
\]

That does not mean something has gone wrong.

It means A moves at \(4\text{ m s}^{-1}\) in the direction you labelled negative when viewed from B.

For example, suppose both cars move east:

\[
v_A=16\text{ m s}^{-1}, \qquad v_B=20\text{ m s}^{-1}.
\]

Then

\[
v_{A/B}=16-20=-4\text{ m s}^{-1}.
\]

From B’s frame, A appears to move backwards at \(4\text{ m s}^{-1}\).

A is not actually reversing relative to the road. B is simply pulling ahead faster.

This is much like sitting in a train when the train beside you starts moving forward. For a moment, you may feel as though your own train has started rolling backwards. Relative motion tells you what is happening between the two trains. It doesn’t, by itself, tell you which train is moving relative to the ground.

That last point is where the analogy breaks: visual impressions can be misleading, while a relative velocity calculation uses measured velocities and a clearly defined frame.

06Swapping the observer reverses the sign

Suppose

\[
v_{A/B}=+7\text{ m s}^{-1}.
\]

What should B’s velocity relative to A be?

It must be

\[
v_{B/A}=-7\text{ m s}^{-1}.
\]

Mathematically,

\[
v_{B/A}=v_B-v_A=-(v_A-v_B).
\]

Therefore,

\[
\boxed{v_{B/A}=-v_{A/B}}.
\]

The magnitude stays the same, but the direction reverses.

If A sees B moving left at \(5\text{ m s}^{-1}\), then B sees A moving right at \(5\text{ m s}^{-1}\).

This is a useful error check. If you swap the two objects but your answer doesn’t change sign, check your working.

07Relative velocity can also connect several frames

Sometimes the relative velocity is given, and you need to recover an object’s velocity in another frame.

Starting from

\[
v_{A/B}=v_A-v_B,
\]

we can rearrange:

\[
v_A=v_{A/B}+v_B.
\]

More generally, if we name the frames explicitly,

\[
v_{A/C}=v_{A/B}+v_{B/C}.
\]

This says:

A’s velocity relative to C equals A’s velocity relative to B, plus B’s velocity relative to C.

The signs still carry the directions.

08The most useful decision rule

For one-dimensional relative velocity, use the same process every time:

1. Choose one direction as positive.
2. Give every velocity a sign.
3. Identify exactly which object is being viewed from which frame.
4. Write \(v_{A/B}=v_A-v_B\) in that order.
5. Substitute the signed velocities.
6. Use the sign of the result to state the direction.

If you find yourself deciding whether you should “add or subtract the speeds”, stop. That is usually the point where mistakes begin.

Work with velocities instead.

If you need a refresher on why direction changes a physical quantity from a scalar to a vector, see Scalars and Vectors in Kinematics for HSC Physics.

09Relative velocity can be zero without the objects being stationary

If

\[
v_{A/B}=0,
\]

then

\[
v_A=v_B.
\]

The two objects have the same velocity at that instant.

That does not necessarily mean either object has zero velocity.

Two cars could both be travelling east at \(30\text{ m s}^{-1}\). Their velocity relative to each other is zero even though both are moving rapidly relative to the road.

There is another subtle point. If two accelerating objects happen to have the same velocity at one instant, their relative velocity is zero at that instant. If their accelerations are different, that will not remain true.

This distinction becomes important when relative motion changes with time.

10Questions and solutions

Question 1

A delivery van travels east at \(17\text{ m s}^{-1}\). A car behind it travels east at \(21\text{ m s}^{-1}\).

Find the velocity of the car relative to the van.

Solution 1

The car moves at \(\boxed{4\text{ m s}^{-1}\text{ east}}\) relative to the van.

Choose east as positive:

\[
v_C=+21\text{ m s}^{-1}, \qquad v_V=+17\text{ m s}^{-1}.
\]

The car’s velocity relative to the van is

\[
v_{C/V}=v_C-v_V.
\]

Substituting,

\[
v_{C/V}=21-17=+4\text{ m s}^{-1}.
\]

The positive sign means east. Although the car travels at \(21\text{ m s}^{-1}\) relative to the road, it gains on the van at only \(4\text{ m s}^{-1}\).

Question 2

A small boat moves north along a straight section of river at \(8.0\text{ m s}^{-1}\), while another boat moves south along the same line at \(5.0\text{ m s}^{-1}\).

Take north as positive. Find the velocity of the first boat relative to the second.

Solution 2

The first boat moves at \(\boxed{13\text{ m s}^{-1}\text{ north}}\) relative to the second boat.

Using north as positive,

\[
v_1=+8.0\text{ m s}^{-1}, \qquad v_2=-5.0\text{ m s}^{-1}.
\]

Therefore,

\[
\begin{aligned}
v_{1/2}
&=v_1-v_2\\
&=8.0-(-5.0)\\
&=+13\text{ m s}^{-1}.
\end{aligned}
\]

The negative velocity of the second boat is important. Subtracting it makes the relative speed \(13\text{ m s}^{-1}\), not \(3\text{ m s}^{-1}\).

Question 3

A bus travels west at \(14\text{ m s}^{-1}\). A motorbike also travels west at \(20\text{ m s}^{-1}\).

Take east as positive.

Find:

(a) the velocity of the bus relative to the motorbike, and
(b) the velocity of the motorbike relative to the bus.

Explain why your two answers have different signs.

Solution 3

The bus moves at \(\boxed{6\text{ m s}^{-1}\text{ east}}\) relative to the motorbike, while the motorbike moves at \(\boxed{6\text{ m s}^{-1}\text{ west}}\) relative to the bus.

Because east is positive,

\[
v_B=-14\text{ m s}^{-1}, \qquad v_M=-20\text{ m s}^{-1}.
\]

For the bus relative to the motorbike,

\[
\begin{aligned}
v_{B/M}
&=v_B-v_M\\
&=-14-(-20)\\
&=+6\text{ m s}^{-1}.
\end{aligned}
\]

So the bus appears to move east at \(6\text{ m s}^{-1}\) from the motorbike’s frame.

For the motorbike relative to the bus,

\[
\begin{aligned}
v_{M/B}
&=v_M-v_B\\
&=-20-(-14)\\
&=-6\text{ m s}^{-1}.
\end{aligned}
\]

So the motorbike appears to move west at \(6\text{ m s}^{-1}\) from the bus’s frame.

The signs reverse because swapping the observer reverses the relative direction:

\[
v_{M/B}=-v_{B/M}.
\]

A common trap is to think the bus must appear to move west because it is physically travelling west relative to the road. From the faster motorbike’s frame, however, the slower bus falls behind, which is east relative to that moving frame.

Question 4

A train moves east at \(26\text{ m s}^{-1}\) relative to the ground. A passenger measures a trolley rolling west through the carriage at \(3.2\text{ m s}^{-1}\) relative to the train.

A student says, “The trolley’s velocity is west because it is rolling towards the rear of the train.”

Determine the trolley’s velocity relative to the ground, and explain what is wrong with the student’s reasoning.

Solution 4

The trolley moves at \(\boxed{22.8\text{ m s}^{-1}\text{ east relative to the ground}}\). The student’s mistake is confusing velocity relative to the train with velocity relative to the ground.

Choose east as positive:

\[
v_{T/G}=+26\text{ m s}^{-1}
\]

for the train, while the trolley’s velocity relative to the train is

\[
v_{R/T}=-3.2\text{ m s}^{-1}.
\]

The trolley’s ground velocity is

\[
v_{R/G}=v_{R/T}+v_{T/G}.
\]

Therefore,

\[
\begin{aligned}
v_{R/G}
&=-3.2+26\\
&=+22.8\text{ m s}^{-1}.
\end{aligned}
\]

The trolley rolls west relative to the carriage but still moves east relative to the ground because the train’s eastward velocity is much larger.

The phrase “the trolley moves west” is incomplete unless the reference frame is stated.

Question 5

Two vehicles A and B travel along the same straight road. Their velocities, measured relative to the road, are

\[
v_A=(4+2t)\text{ m s}^{-1}
\]

and

\[
v_B=(10-t)\text{ m s}^{-1},
\]

where \(t\) is measured in seconds.

Find the time when A has zero velocity relative to B.

A student then claims, “At that time, A will remain in the same position relative to B because their relative velocity is zero.”

Assess the claim.

Solution 5

A has zero velocity relative to B at \(\boxed{t=2.0\text{ s}}\), but the student’s conclusion is incorrect. Their relative velocity is zero only at that instant.

The relative velocity is

\[
v_{A/B}=v_A-v_B.
\]

Substitute the expressions:

\[
\begin{aligned}
v_{A/B}
&=(4+2t)-(10-t)\\
&=4+2t-10+t\\
&=3t-6.
\end{aligned}
\]

For zero relative velocity,

\[
\begin{aligned}
3t-6&=0\\
3t&=6\\
t&=2.0\text{ s}.
\end{aligned}
\]

At \(t=2.0\text{ s}\),

\[
v_A=4+2(2)=8\text{ m s}^{-1}
\]

and

\[
v_B=10-2=8\text{ m s}^{-1}.
\]

So the vehicles momentarily have identical velocities.

However, their velocities change differently. From the velocity equations, A’s acceleration is \(+2\text{ m s}^{-2}\), while B’s acceleration is \(-1\text{ m s}^{-2}\). Their relative acceleration is therefore

\[
a_{A/B}=2-(-1)=3\text{ m s}^{-2}.
\]

Their relative velocity immediately begins changing again after \(t=2.0\text{ s}\).

Zero relative velocity means their separation is not changing at that instant. It does not guarantee that the separation will remain constant.

11What relative velocity leads to next

In one dimension, relative velocity is controlled by a simple idea:

\[
\boxed{v_{A/B}=v_A-v_B}.
\]

The real skill is not the subtraction. It is defining the frame, choosing a sign convention, and interpreting the sign of the answer.

Once that is secure, the same idea extends naturally to changing velocities through

\[
a_{A/B}=a_A-a_B,
\]

and then to two-dimensional motion, where relative velocity becomes full vector subtraction rather than subtraction along a single line.