Can Two Objects Have the Same Speed but Different Velocities?

Two model rovers travel at exactly \(1.2\,\mathrm{m\,s^{-1}}\) along parallel tracks. One heads east, and the other heads west. A stationary observer measures the same speed for both.

Would a camera travelling with rover A see rover B moving at zero speed because their speeds match?

No. From rover A’s frame, rover B moves west at \(2.4\,\mathrm{m\,s^{-1}}\).

Yes, two objects can have the same speed but a large relative velocity. The important difference is direction. Speed tells us how fast something moves; velocity tells us how fast and in which direction.

In this example, the rovers have equal speeds relative to the laboratory floor, but their velocities are opposite. That produces a relative speed twice as large as either rover’s ground speed.

01Why identical speed measurements can be misleading

Imagine trying to work out how two moving objects behave using only their speedometer readings.

You measure \(5.0\,\mathrm{m\,s^{-1}}\) for both. What can you conclude about their movement relative to one another?

Surprisingly little.

They might travel alongside each other, maintaining the same separation. They might approach head-on, closing the gap quickly. They could even move at right angles.

All three situations are consistent with the same two speed measurements.

The problem is that speed is a scalar: it has magnitude but no direction. Velocity is a vector: it has both.

To compare two objects properly, we need their velocities measured in one common reference frame.

We’ll use two frames throughout the first example:

• Ground frame (G): An observer stationary relative to the laboratory floor measures both rovers.
• Rover A’s frame (A): An observer travelling with rover A measures the motion of rover B.

The important rule is that the speeds being compared must be measured relative to the same reference frame.

Two equal speedometer readings don’t necessarily establish that condition. For example, a boat’s speed relative to the water isn’t automatically its speed relative to the shore.

02An experiment: what can two light gates actually tell us?

Consider a proposed laboratory experiment designed to compare the motion of two motorised rovers.

This is an illustrative measurement setup, not a report of a particular historical experiment.

The original apparatus

Place two straight tracks alongside one another, running east-west.

On each track, install two light gates separated by \(0.60\,\mathrm{m}\). Connect the gates to a timer that displays the elapsed time between interruptions, but not which gate was triggered first.

As each rover passes through its pair of gates, the timer measures how long it takes to travel the known distance.

Suppose both timers display \(0.50\,\mathrm{s}\).

Both calculated speeds are

\[
\begin{aligned}
s&=\frac{\text{distance}}{\text{time}}\\
&=\frac{0.60\,\mathrm{m}}{0.50\,\mathrm{s}}\\
&=1.2\,\mathrm{m\,s^{-1}}
\end{aligned}
\]

Here, \(s\) is the average speed between the gates.

If the rovers have constant velocities, this also gives the magnitude of each rover’s velocity.

But the apparatus has a weakness.

A rover travelling west to east takes \(0.50\,\mathrm{s}\) to cross its gates. A rover travelling east to west at the same speed takes exactly the same time.

The displays cannot distinguish those situations because they discard direction.

The prediction that equal speeds mean zero relative motion therefore cannot be tested using these readings alone.

Revising the measurement

Now change the equipment so that each gate records its identity and trigger time.

If the west gate triggers first, the rover is travelling east. If the east gate triggers first, it is travelling west.

Add a position-tracking camera to measure each rover’s east-west coordinate over time. This also lets us check whether its velocity remains approximately constant.

Suppose the revised measurements indicate that rover A travels east and rover B travels west.

The two speed measurements are still equal, but their directions are now known.

The revised design answers the original measurement problem: equal speeds don’t imply equal velocities.

It also has an experimental limitation. A light-gate calculation gives average speed over the gate spacing. If a rover accelerates between the gates, that result isn’t necessarily its instantaneous speed. Continuous position measurements help reveal such changes.

03Calculating the velocity of B relative to A

Let’s turn the observation into mathematics.

Choose east as the positive direction.

The ground-frame velocities are

\[
\begin{aligned}
v_{A/G}&=+1.2\,\mathrm{m\,s^{-1}}\\
v_{B/G}&=-1.2\,\mathrm{m\,s^{-1}}
\end{aligned}
\]

The positive sign means east; the negative sign means west.

To calculate the velocity of B as measured by an observer travelling with A, subtract A’s ground velocity from B’s ground velocity:

\[
v_{B/A}=v_{B/G}-v_{A/G}
\]

where:

• \(v_{B/A}\) is the velocity of B relative to A.
• \(v_{B/G}\) is the velocity of B relative to the ground.
• \(v_{A/G}\) is the velocity of A relative to the ground.

Substituting:

\[
\begin{aligned}
v_{B/A}&=-1.2-(+1.2)\\
&=-2.4\,\mathrm{m\,s^{-1}}
\end{aligned}
\]

The negative sign indicates west.

So B’s relative velocity is \(2.4\,\mathrm{m\,s^{-1}}\) west, and its relative speed is \(2.4\,\mathrm{m\,s^{-1}}\).

The speed is simply the magnitude of the relative velocity. It cannot be negative.

Why does subtracting make sense?

Imagine a measuring tape that moves east with rover A.

Every second, A travels \(1.2\,\mathrm{m}\) east across the laboratory floor. During that same second, B travels \(1.2\,\mathrm{m}\) west.

From A’s moving measuring tape, B therefore shifts \(2.4\,\mathrm{m}\) west every second.

This moving-ruler picture is useful for straight-line motion. It becomes incomplete when the reference frame rotates or accelerates, or when motion needs two spatial coordinates.

04Three equal-speed situations, three different relative velocities

Keep rover A travelling east at \(1.2\,\mathrm{m\,s^{-1}}\). Change only B’s direction.

Predict which arrangement produces the largest velocity of B relative to A.

The answer is opposite directions, because the velocity vectors point directly away from one another when compared.

For two-dimensional motion, use \(x\) for east and \(y\) for north. An ordered pair \((v_x,v_y)\) gives the two velocity components.

All three rows have exactly the same ground-speed readings for both objects.

Yet the relative speeds range from zero to twice either object’s speed.

More generally, if both objects have speed \(s\) in the same ground frame, then

\[
0\leq s_{\mathrm{rel}}\leq 2s
\]

Here, \(s_{\mathrm{rel}}\) means the magnitude of their relative velocity.

The minimum occurs when their velocity vectors are identical. The maximum occurs when they point in opposite directions.

This is a useful limit: relative velocity can be large compared with each object’s speed, but it isn’t unlimited when their speeds are fixed.

05What if the objects move at right angles?

Opposite-direction motion is easy to picture. Two-dimensional motion is where subtracting speeds becomes especially misleading.

Consider two drones. Drone A flies east at \(8.0\,\mathrm{m\,s^{-1}}\), while drone B flies north at \(8.0\,\mathrm{m\,s^{-1}}\), both relative to the ground.

Someone might subtract \(8.0-8.0\) and conclude that their relative speed is zero.

But an observer travelling east with A would see B shifting west and north at the same time.

The observer has changed, so the velocity components must change with it.

06A large relative speed doesn’t always mean a collision

There’s one more distinction worth making.

Imagine the two rovers passing each other on neighbouring parallel tracks. At the instant they are directly alongside one another, the line connecting them points sideways across the gap between the tracks.

Their relative velocity, however, points along the tracks.

Those directions are perpendicular.

At that instant, a rangefinder measuring their straight-line distance would record no instantaneous change in distance, even though their relative speed is still \(2.4\,\mathrm{m\,s^{-1}}\).

That’s because three quantities are different:

Closing speed is sometimes used informally to mean relative speed, especially for head-on motion. In two dimensions, the distinction becomes important.

At the point of closest approach, the rate of change of distance can be zero while the relative speed remains large.

So when a question asks whether two objects meet, relative speed alone isn’t enough. You also need to know their relative positions and directions of travel.

07Questions and solutions

The questions below begin with a familiar straight-line comparison, then test what happens when measurements lose direction, paths cross, reference frames change, and an observer can choose their own velocity.

Assume classical, non-relativistic motion throughout. Unless stated otherwise, velocities are constant, east is positive \(x\), north is positive \(y\), and speeds refer to a fixed ground observer.

Question 1

Two skaters, A and B, travel east at \(4.0\,\mathrm{m\,s^{-1}}\) along a straight rink. B is initially \(9.0\,\mathrm{m}\) east of A.

(a) Calculate B’s velocity relative to A.

(b) Find their separation after \(3.0\,\mathrm{s}\).

(c) At that instant, B reverses direction and travels west at \(4.0\,\mathrm{m\,s^{-1}}\), while A continues east. Calculate B’s new velocity relative to A.

Solution 1

B is initially stationary relative to A, their separation remains \(9.0\,\mathrm{m}\) after three seconds, and B’s relative velocity becomes \(8.0\,\mathrm{m\,s^{-1}}\) west after reversing.

For (a), both velocities are equal:

\[
\begin{aligned}
v_{B/A}&=v_{B/G}-v_{A/G}\\
&=4.0-4.0\\
&=0\,\mathrm{m\,s^{-1}}
\end{aligned}
\]

For (b), zero relative velocity means zero change in separation:

\[
\begin{aligned}
\Delta x_{B/A}&=v_{B/A}\Delta t\\
&=0(3.0)\\
&=0\,\mathrm{m}
\end{aligned}
\]

Their separation remains \(9.0\,\mathrm{m}\).

For (c), reversing direction changes B’s signed ground velocity from \(+4.0\) to \(-4.0\,\mathrm{m\,s^{-1}}\).

\[
\begin{aligned}
v_{B/A}&=-4.0-(+4.0)\\
&=-8.0\,\mathrm{m\,s^{-1}}
\end{aligned}
\]

B now moves west at \(8.0\,\mathrm{m\,s^{-1}}\) relative to A.

The tempting mistake is to subtract the two speedometer values again and obtain zero. The magnitudes haven’t changed, but B’s velocity vector has reversed.

Question 2

Two robotic carts travel along separate, parallel east-west tracks.

At \(t=0\), cart A is at \(x=0\,\mathrm{m}\) and cart B is at \(x=24\,\mathrm{m}\).

Each maintains a constant ground speed of \(4.0\,\mathrm{m\,s^{-1}}\). Each travels either east or west, but their directions are initially unknown.

A tracking system reports only the magnitude of their east-west coordinate difference, \(|x_B-x_A|\). It gives the following measurements:

(a) Determine the direction and signed ground velocity of each cart.

(b) When will their east-west coordinates be equal?

(c) Predict the tracking-system reading at \(t=4.0\,\mathrm{s}\). Explain why a single reading of that value cannot establish whether the carts are approaching or moving apart.

Solution 2

A travels east at \(+4.0\,\mathrm{m\,s^{-1}}\), B travels west at \(-4.0\,\mathrm{m\,s^{-1}}\), and they are level after \(3.0\,\mathrm{s}\). At \(4.0\,\mathrm{s}\), the tracker reads \(8.0\,\mathrm{m}\) again, but the carts are now moving apart.

For (a), define the signed relative position:

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

Initially, \(x_{B/A}=+24\,\mathrm{m}\).

Since each cart can move only east or west at \(4.0\,\mathrm{m\,s^{-1}}\), the possible constant relative velocities are \(0\), \(+8.0\), or \(-8.0\,\mathrm{m\,s^{-1}}\).

Only \(-8.0\,\mathrm{m\,s^{-1}}\) can produce the \(16\,\mathrm{m}\) separation measured after one second.

Therefore,

\[
\begin{aligned}
v_{B/A}&=v_{B/G}-v_{A/G}\\
&=-4.0-(+4.0)\\
&=-8.0\,\mathrm{m\,s^{-1}}
\end{aligned}
\]

The second measurement confirms the result.

For (b), the signed relative position is

\[
x_{B/A}=24-8.0t
\]

with position in metres and \(t\) in seconds.

The carts are level when \(x_{B/A}=0\):

\[
\begin{aligned}
0&=24-8.0t\\
t&=\frac{24}{8.0}\\
&=3.0\,\mathrm{s}
\end{aligned}
\]

For (c), at \(t=4.0\,\mathrm{s}\):

\[
\begin{aligned}
x_{B/A}&=24-8.0(4.0)\\
&=-8.0\,\mathrm{m}
\end{aligned}
\]

The tracker displays the absolute value:

\[
|x_{B/A}|=8.0\,\mathrm{m}
\]

At \(t=2.0\,\mathrm{s}\), B was \(8.0\,\mathrm{m}\) east of A and approaching. At \(t=4.0\,\mathrm{s}\), B is \(8.0\,\mathrm{m}\) west of A and moving farther away.

The same unsigned separation occurs twice.

The key constraint is that the tracker measures distance along an axis, not signed position. A single distance reading loses the information needed to distinguish approaching from separating.

Question 3

Two drones move horizontally at constant velocities.

At \(t=0\):

• Drone A is at \((0,0)\,\mathrm{m}\), travelling east at \(6.0\,\mathrm{m\,s^{-1}}\).
• Drone B is at \((18,0)\,\mathrm{m}\), travelling north at \(6.0\,\mathrm{m\,s^{-1}}\).

(a) Calculate B’s velocity and speed relative to A.

(b) Determine whether the drones ever occupy the same horizontal position.

(c) Find their minimum horizontal separation and when it occurs.

(d) Explain why their relative speed does not equal the instantaneous rate at which their separation decreases at closest approach.

Solution 3

The drones never meet. Their minimum horizontal separation is approximately \(12.7\,\mathrm{m}\), reached after \(1.5\,\mathrm{s}\), even though their relative speed remains \(8.49\,\mathrm{m\,s^{-1}}\).

For (a), the ground velocities are

\[
\begin{aligned}
\vec v_{A/G}&=(6,0)\\
\vec v_{B/G}&=(0,6)
\end{aligned}
\]

with components in \(\mathrm{m\,s^{-1}}\).

Therefore,

\[
\begin{aligned}
\vec v_{B/A}&=(0,6)-(6,0)\\
&=(-6,+6)
\end{aligned}
\]

The relative speed is

\[
\begin{aligned}
s_{\mathrm{rel}}&=\sqrt{(-6)^2+6^2}\\
&=\sqrt{72}\\
&=8.49\,\mathrm{m\,s^{-1}}
\end{aligned}
\]

For (b), after \(t\) seconds, the horizontal positions in metres are

\[
\begin{aligned}
\vec r_A(t)&=(6t,0)\\
\vec r_B(t)&=(18,6t)
\end{aligned}
\]

To meet, their east and north coordinates would both have to match at the same time.

Matching east coordinates requires

\[
6t=18
\quad\Rightarrow\quad
t=3.0\,\mathrm{s}
\]

Matching north coordinates requires

\[
0=6t
\quad\Rightarrow\quad
t=0
\]

No single time satisfies both conditions. Therefore, the drones never meet.

For (c), their relative position is

\[
\vec r_{B/A}(t)=(18-6t,6t)
\]

Let \(d\) be their horizontal separation in metres, with \(t\) measured in seconds.

Then

\[
\begin{aligned}
d^2&=(18-6t)^2+(6t)^2\\
&=72(t-1.5)^2+162
\end{aligned}
\]

The squared term cannot be negative. Thus, \(d^2\) is smallest when

\[
t=1.5\,\mathrm{s}
\]

At that instant,

\[
d_{\min}=\sqrt{162}=12.7\,\mathrm{m}
\]

For (d), at \(t=1.5\,\mathrm{s}\):

\[
\begin{aligned}
\vec r_A&=(9,0)\,\mathrm{m}\\
\vec r_B&=(18,9)\,\mathrm{m}
\end{aligned}
\]

So B lies \(9\,\mathrm{m}\) east and \(9\,\mathrm{m}\) north of A.

The line joining A to B points north-east, while B’s relative velocity points north-west. These directions are perpendicular.

At that instant, the relative motion is entirely sideways to the line joining the drones.

Their distance momentarily stops decreasing, but their relative velocity is not zero.

This is why using \(18/8.49\) as a meeting time would be wrong. It assumes all the relative motion is directed along their initial separation, which it isn’t.

Question 4

Two powered boats, A and B, move through a river.

Each maintains a speed of \(5.0\,\mathrm{m\,s^{-1}}\) relative to the surrounding water. Boat A travels due east relative to the water, and boat B travels due north relative to the water.

In the first test, a uniform current flows east at \(3.0\,\mathrm{m\,s^{-1}}\) relative to the shore.

(a) Calculate the ground velocity and ground speed of each boat. Do they have equal ground speeds?

(b) Calculate B’s velocity and speed relative to A.

(c) In a second test, the current instead flows west at \(3.0\,\mathrm{m\,s^{-1}}\). Both boats maintain their original velocities relative to the water. Calculate their new ground speeds and B’s new relative velocity. Explain what changes and what remains unchanged.

Solution 4

The boats have equal speeds relative to the water but unequal speeds relative to the shore. In the first test, A’s ground speed is \(8.0\,\mathrm{m\,s^{-1}}\), while B’s is \(5.83\,\mathrm{m\,s^{-1}}\). Their relative velocity is \((-5,+5)\,\mathrm{m\,s^{-1}}\) in both tests.

For (a), the velocity of a boat relative to the shore is the sum of its velocity through the water and the water’s velocity relative to the shore.

Let W represent the water and G the shore-fixed ground frame.

\[
\vec v_{A/G}=\vec v_{A/W}+\vec v_{W/G}
\]

Using east and north components in \(\mathrm{m\,s^{-1}}\):

\[
\begin{aligned}
\vec v_{A/G}&=(5,0)+(3,0)\\
&=(8,0)
\end{aligned}
\]

So A moves east at \(8.0\,\mathrm{m\,s^{-1}}\).

For B:

\[
\begin{aligned}
\vec v_{B/G}&=(0,5)+(3,0)\\
&=(3,5)
\end{aligned}
\]

Its ground speed is

\[
\begin{aligned}
s_{B/G}&=\sqrt{3^2+5^2}\\
&=\sqrt{34}\\
&=5.83\,\mathrm{m\,s^{-1}}
\end{aligned}
\]

The ground speeds are unequal because the current changes the magnitudes of the two ground-velocity vectors differently.

For (b), subtract A’s ground velocity from B’s:

\[
\begin{aligned}
\vec v_{B/A}&=(3,5)-(8,0)\\
&=(-5,+5)
\end{aligned}
\]

The relative speed is

\[
\begin{aligned}
s_{\mathrm{rel}}&=\sqrt{(-5)^2+5^2}\\
&=7.07\,\mathrm{m\,s^{-1}}
\end{aligned}
\]

B travels north-west at \(7.07\,\mathrm{m\,s^{-1}}\) relative to A.

For (c), the current reverses to \((-3,0)\,\mathrm{m\,s^{-1}}\).

A’s new ground velocity is

\[
\begin{aligned}
\vec v_{A/G}&=(5,0)+(-3,0)\\
&=(2,0)
\end{aligned}
\]

Its ground speed becomes \(2.0\,\mathrm{m\,s^{-1}}\).

B’s new ground velocity is

\[
\begin{aligned}
\vec v_{B/G}&=(0,5)+(-3,0)\\
&=(-3,5)
\end{aligned}
\]

Its ground speed is again

\[
\sqrt{(-3)^2+5^2}=5.83\,\mathrm{m\,s^{-1}}
\]

Now calculate B relative to A:

\[
\begin{aligned}
\vec v_{B/A}&=(-3,5)-(2,0)\\
&=(-5,+5)
\end{aligned}
\]

Its relative speed remains \(7.07\,\mathrm{m\,s^{-1}}\), north-west.

The key is that the same current contributes to both ground velocities. When we subtract those velocities, the current cancels.

This cancellation depends on the boats experiencing the same water velocity and maintaining their specified velocities relative to that water. It would not generally work if they were in different currents.

The tempting assumption is that changing the current must change every relative velocity. It changes each boat’s motion relative to shore, but a common velocity added to both boats doesn’t change their motion relative to one another.

Question 5

Two drones, A and B, travel at constant velocities in a horizontal plane.

Relative to a stationary ground observer:

• Drone A flies due east at \(10.0\,\mathrm{m\,s^{-1}}\).
• Drone B flies at \(10.0\,\mathrm{m\,s^{-1}}\), directed \(120^\circ\) anticlockwise from east.

A camera platform can move with any constant horizontal velocity relative to the ground. It measures each drone’s velocity relative to the platform.

A technician proposes choosing the platform’s velocity so that both drones appear to move at speeds no greater than \(8.0\,\mathrm{m\,s^{-1}}\).

(a) Determine whether this is possible. Justify your conclusion for every possible platform velocity, not just a few directions.

(b) Find the platform velocity that makes the larger of the two measured drone speeds as small as possible. Calculate that minimum larger speed.

(c) Another technician suggests matching A’s ground velocity exactly, making A appear stationary. Explain why this doesn’t produce the smallest possible larger speed. State B’s velocity relative to A and explain why the platform’s choice of velocity cannot change it.

Solution 5

No camera platform can make both measured speeds \(8.0\,\mathrm{m\,s^{-1}}\) or less. The smallest possible larger speed is \(8.66\,\mathrm{m\,s^{-1}}\), achieved when the platform travels at \(5.0\,\mathrm{m\,s^{-1}}\), directed \(60^\circ\) north of east.

The important constraint is that changing observers can change each drone’s measured velocity, but it cannot change the velocity of one drone relative to the other.

For (a), begin with the two ground velocities.

In east-north components, measured in \(\mathrm{m\,s^{-1}}\):

\[
\vec v_{A/G}=(10,0)
\]

For B, using \(\cos120^\circ=-1/2\) and \(\sin120^\circ=\sqrt3/2\):

\[
\begin{aligned}
\vec v_{B/G}&=(10\cos120^\circ,\\
&\qquad 10\sin120^\circ)\\
&=(-5,5\sqrt3)
\end{aligned}
\]

Their relative velocity is

\[
\begin{aligned}
\vec v_{B/A}&=\vec v_{B/G}-\vec v_{A/G}\\
&=(-15,5\sqrt3)
\end{aligned}
\]

Its magnitude is

\[
\begin{aligned}
s_{\mathrm{rel}}&=\sqrt{15^2+(5\sqrt3)^2}\\
&=\sqrt{300}\\
&=10\sqrt3\\
&=17.3\,\mathrm{m\,s^{-1}}
\end{aligned}
\]

Now imagine any moving camera platform C.

From that camera, the drones have velocities \(\vec v_{A/C}\) and \(\vec v_{B/C}\).

Their difference must still be

\[
\vec v_{B/A}
=\vec v_{B/C}-\vec v_{A/C}
\]

Why? Because subtracting the camera’s velocity from both ground velocities removes the same vector twice, leaving the original difference unchanged.

Suppose the camera could make both drones appear to travel at \(8.0\,\mathrm{m\,s^{-1}}\) or less.

The difference between two velocity vectors of magnitude at most \(8.0\,\mathrm{m\,s^{-1}}\) cannot have a magnitude greater than \(16.0\,\mathrm{m\,s^{-1}}\). That maximum would occur if both apparent velocity vectors had magnitude \(8.0\) and pointed in opposite directions.

But the drones’ actual relative speed is \(17.3\,\mathrm{m\,s^{-1}}\).

Therefore, the proposed camera velocity is impossible.

For (b), the smallest possible larger speed must be at least half their relative speed:

\[
\frac{10\sqrt3}{2}
=5\sqrt3
=8.66\,\mathrm{m\,s^{-1}}
\]

We can achieve this minimum by making the two apparent velocity vectors equal in magnitude and opposite in direction.

Choose the camera’s ground velocity halfway between the two drone velocity vectors:

\[
\vec v_{C/G}
=\frac{\vec v_{A/G}+\vec v_{B/G}}{2}
\]

Substituting:

\[
\begin{aligned}
\vec v_{C/G}
&=\frac{(10,0)+(-5,5\sqrt3)}{2}\\
&=(2.5,2.5\sqrt3)
\end{aligned}
\]

Its speed is

\[
\begin{aligned}
s_{C/G}
&=\sqrt{2.5^2+(2.5\sqrt3)^2}\\
&=5.0\,\mathrm{m\,s^{-1}}
\end{aligned}
\]

Its direction satisfies

\[
\tan\theta=\frac{2.5\sqrt3}{2.5}
=\sqrt3
\]

So \(\theta=60^\circ\) north of east.

Now calculate the drone velocities from the camera.

\[
\begin{aligned}
\vec v_{A/C}&=(7.5,-2.5\sqrt3)\\
\vec v_{B/C}&=(-7.5,2.5\sqrt3)
\end{aligned}
\]

These vectors are equal in magnitude and opposite in direction.

Each has speed

\[
\begin{aligned}
s&=\sqrt{7.5^2+(2.5\sqrt3)^2}\\
&=\sqrt{75}\\
&=8.66\,\mathrm{m\,s^{-1}}
\end{aligned}
\]

This achieves the theoretical minimum, so no other platform velocity can make the larger reading smaller.

For (c), if the platform copies A’s ground velocity exactly, A appears stationary:

\[
\vec v_{A/C}=(0,0)
\]

But B then appears to move with velocity

\[
\vec v_{B/C}=(-15,5\sqrt3)
\]

Its speed is \(17.3\,\mathrm{m\,s^{-1}}\).

The larger camera-measured speed would therefore be \(17.3\,\mathrm{m\,s^{-1}}\), considerably worse than the optimal \(8.66\,\mathrm{m\,s^{-1}}\).

Making one object stationary relative to the observer doesn’t necessarily minimise the speeds of both objects.

The deeper point is geometric. The platform can shift the two velocity vectors together, but it cannot change the vector joining their endpoints.

That fixed difference is B’s velocity relative to A.

08What to decide when you see a relative-motion question

Before calculating, identify three things:

1. Who is the observer? If the question asks for B relative to A, subtract A’s velocity from B’s.
2. Are you comparing velocities or just their magnitudes? Two equal speeds don’t tell you whether the velocity vectors point in the same direction.
3. Is the question asking about relative motion or changing separation? A relative speed of \(10\,\mathrm{m\,s^{-1}}\) doesn’t necessarily mean the distance between two objects decreases by \(10\,\mathrm{m}\) every second.

For equal speeds measured in one frame, the relative speed can range from zero to twice the individual speed. Direction determines where the result falls in that range.

If the question asks only for B’s velocity relative to A, vector subtraction is sufficient. If it asks whether the objects meet or how close they get, you also need their initial relative positions.

For more practice with signed velocities on a straight line, use Relative Velocity in One Dimension for HSC Physics.

To extend the same reasoning to changing separation, crossing paths, and closest approach, continue with Relative Position and Motion in Two Dimensions for HSC Physics.