Why You Subtract Velocities When Changing Reference Frames

A train travels east at \(12.0\,\mathrm{m\,s^{-1}}\). A passenger walks towards the rear of the train at \(1.5\,\mathrm{m\,s^{-1}}\) relative to the carriage.

A person standing on the platform measures the passenger’s velocity as \(10.5\,\mathrm{m\,s^{-1}}\) east.

But someone sitting inside the train measures the passenger moving at \(1.5\,\mathrm{m\,s^{-1}}\) west.

Both measurements are correct.

Before reading on, predict: how do we convert the platform observer’s measurement into the train observer’s measurement?

We subtract the train’s velocity:

\[ v_{P/T}=10.5-12.0 \]

\[ \boxed{v_{P/T}=-1.5\,\mathrm{m\,s^{-1}}} \]

The negative sign means west.

We subtract the train’s velocity because the train itself is moving. Changing reference frame means changing the coordinate system used to measure an object’s position, so we must remove the motion of the old coordinate system.

Let’s work through why this happens, how we can test it, and when subtracting speeds gives the wrong answer.

01Why do you subtract the observer’s velocity?

The key is understanding what a reference frame actually measures.

A reference frame is a coordinate system used to describe an object’s position and motion. It usually moves with an observer.

Consider our train example. We have three things to identify:

The passenger is the object. The train is the observer whose reference frame we want to use.

We write the passenger’s velocity relative to the train as:

\[ \vec v_{P/T} \]

The notation means the velocity of P as measured by an observer in T.

The platform observer measures:

\[ \vec v_{P/G} \]

The train’s velocity relative to the ground is:

\[ \vec v_{T/G} \]

The relationship between these velocities is:

\[ \boxed{ \vec v_{P/T} = \vec v_{P/G} - \vec v_{T/G} } \]

In words:

Velocity of the object relative to the observer = object velocity minus observer velocity.

Notice that both velocities on the right-hand side are measured relative to the same ground frame.

This is essential. You cannot subtract two velocities that are measured relative to different reference frames without first converting them into a common frame.

The real reason: relative position

The equation isn’t an arbitrary rule. It comes directly from how we calculate position.

Imagine measuring the passenger’s position from the ground.

The passenger has a ground coordinate \(x_{P/G}\), while the train has a ground coordinate \(x_{T/G}\).

To find the passenger’s position relative to the train, subtract the train’s position:

\[ x_{P/T}=x_{P/G}-x_{T/G} \]

Why?

Because the train itself has moved away from the ground origin. We need to account for that movement before describing where the passenger is relative to the carriage.

Now differentiate both sides with respect to time.

\[ \frac{dx_{P/T}}{dt} = \frac{dx_{P/G}}{dt} - \frac{dx_{T/G}}{dt} \]

Remember that velocity is the rate of change of position.

Therefore:

\[ \boxed{ v_{P/T}=v_{P/G}-v_{T/G} } \]

That’s the entire reason for the subtraction.

We subtract velocities because we subtract positions when changing the coordinate origin. Taking the rate of change of those positions gives us the corresponding velocity relationship.

See the subtraction using actual displacements

Return to the original train.

Over four seconds:

• The train travels \(48\,\mathrm{m}\) east.
• The passenger travels \(42\,\mathrm{m}\) east relative to the ground.

The passenger’s displacement relative to the train is:

\[ \begin{aligned} \Delta x_{P/T} &=\Delta x_{P/G}-\Delta x_{T/G}\\ &=42-48\\ &=-6\,\mathrm{m} \end{aligned} \]

The passenger has moved \(6\,\mathrm{m}\) west relative to the train.

Dividing by four seconds:

\[ v_{P/T}=\frac{-6}{4} \]

\[ \boxed{v_{P/T}=-1.5\,\mathrm{m\,s^{-1}}} \]

This matches our original calculation.

The passenger hasn’t physically reversed direction relative to the ground. The train’s coordinate origin simply moves east faster than the passenger does.

Why does adding sometimes appear instead?

Rearrange the original equation:

\[ \boxed{ \vec v_{P/G} = \vec v_{P/T} + \vec v_{T/G} } \]

This equation is useful when you know an object’s velocity relative to a moving observer and want its velocity relative to the ground.

For example, suppose a passenger walks east at \(1.5\,\mathrm{m\,s^{-1}}\) relative to a train moving east at \(12.0\,\mathrm{m\,s^{-1}}\).

The platform observer measures:

\[ v_{P/G}=1.5+12.0 \]

\[ \boxed{v_{P/G}=13.5\,\mathrm{m\,s^{-1}}\text{ east}} \]

We add because we’re combining the passenger’s movement through the train with the train’s movement through the ground.

These two equations describe the same relationship. The direction in which you are changing frames determines which one is useful.

02Testing the subtraction: an experiment with a moving trolley

Can we actually measure this relationship, rather than simply accept it as a mathematical rule?

Consider a proposed laboratory experiment using a motorised trolley and a smaller cart.

The question is straightforward: does a cart’s velocity relative to a moving trolley really equal its ground velocity minus the trolley’s ground velocity?

The original apparatus

Place a large motorised trolley on a straight horizontal track.

On its deck, place a smaller powered cart that can move independently along the trolley. The large trolley travels east, while the smaller cart travels west relative to its deck.

Install a fixed overhead camera above the track. It records the smaller cart’s position relative to the laboratory floor.

The camera can measure the cart’s ground displacement, but there is a problem.

If we only track the small cart, we don’t know how far the trolley’s coordinate origin has moved. We therefore cannot determine the cart’s motion relative to the trolley.

Revising the measurement

Add two measurements:

• A marker on the large trolley, tracked by the fixed camera.
• A second camera mounted on the trolley, looking along a ruler fixed to its deck.

Synchronise the cameras so that their position measurements refer to the same times.

Now we can measure the motion in both reference frames independently.

Predicting the result

Suppose the cameras record the following illustrative measurements.

All measurements use east as positive.

The fixed camera measures the trolley moving east by \(1.60\,\mathrm{m}\), while the small cart moves east by \(1.00\,\mathrm{m}\) relative to the floor.

However, the trolley-mounted camera measures the small cart moving \(0.60\,\mathrm{m}\) west relative to the deck.

Let’s calculate the velocities.

The small cart’s ground velocity is:

\[ v_{C/G}=\frac{1.00}{2.00} \]

\[ v_{C/G}=+0.50\,\mathrm{m\,s^{-1}} \]

The trolley’s ground velocity is:

\[ v_{T/G}=\frac{1.60}{2.00} \]

\[ v_{T/G}=+0.80\,\mathrm{m\,s^{-1}} \]

Now subtract:

\[ \begin{aligned} v_{C/T} &=v_{C/G}-v_{T/G}\\ &=0.50-0.80\\ &=-0.30\,\mathrm{m\,s^{-1}} \end{aligned} \]

The trolley-mounted camera independently measures:

\[ v_{C/T}=\frac{-0.60}{2.00} \]

\[ v_{C/T}=-0.30\,\mathrm{m\,s^{-1}} \]

The two methods agree.

What does this tell us?

The small cart moves east relative to the laboratory floor but west relative to the trolley.

The subtraction removes the movement of the trolley itself.

The important observation is that relative velocity isn’t a different physical motion happening to the cart. It’s the same motion described using a different coordinate system.

The numerical measurements above are illustrative, not reported experimental results.

03Vector subtraction versus scalar speed subtraction

There’s an important distinction to make before using this method in HSC Physics questions.

Speed and velocity aren’t interchangeable.

• Speed is a scalar. It tells us how fast something moves, without specifying direction.
• Velocity is a vector. It tells us how fast something moves and in which direction.

Relative velocity is calculated using vector subtraction.

Relative speed is the magnitude of the resulting relative velocity:

\[ \boxed{ \text{Relative speed}=|\vec v_{A/B}| } \]

When the objects move in the same direction

Suppose object A travels east at \(12\,\mathrm{m\,s^{-1}}\), while observer B travels east at \(8\,\mathrm{m\,s^{-1}}\).

Taking east as positive:

\[ v_{A/B}=+12-(+8) \]

\[ v_{A/B}=+4\,\mathrm{m\,s^{-1}} \]

A appears to move east at \(4\,\mathrm{m\,s^{-1}}\) relative to B.

Here, subtracting the speed magnitudes happens to give the correct relative speed.

When the objects move in opposite directions

Now suppose A travels east at \(12\,\mathrm{m\,s^{-1}}\), while B travels west at \(8\,\mathrm{m\,s^{-1}}\).

The signed velocities are:

\[ v_{A/G}=+12\,\mathrm{m\,s^{-1}} \]

\[ v_{B/G}=-8\,\mathrm{m\,s^{-1}} \]

Subtracting gives:

\[ \begin{aligned} v_{A/B} &=12-(-8)\\ &=+20\,\mathrm{m\,s^{-1}} \end{aligned} \]

The relative speed is \(20\,\mathrm{m\,s^{-1}}\).

Simply subtracting the speed magnitudes would give \(4\,\mathrm{m\,s^{-1}}\), which is incorrect.

The negative sign on B’s velocity matters.

This is why signed velocities are so useful in one-dimensional problems. You don’t need to memorise separate addition and subtraction rules for different directions.

What changes when the motion is two-dimensional?

In two dimensions, you must subtract the velocity components separately.

Suppose:

• Object A moves at \(6\,\mathrm{m\,s^{-1}}\) east and \(8\,\mathrm{m\,s^{-1}}\) north.
• Observer B moves at \(5\,\mathrm{m\,s^{-1}}\) north.

The eastward and northward components of the relative velocity are:

\[ v_{A/B,x}=6-0=6 \]

\[ v_{A/B,y}=8-5=3 \]

Therefore:

\[ \vec v_{A/B} = (6\,\text{east},3\,\text{north}) \,\mathrm{m\,s^{-1}} \]

The relative speed is:

\[ \begin{aligned} v_{A/B} &=\sqrt{6^2+3^2}\\ &=\sqrt{45}\\ &=6.71\,\mathrm{m\,s^{-1}} \end{aligned} \]

Notice something interesting.

The speed of A is \(10\,\mathrm{m\,s^{-1}}\), and the speed of B is \(5\,\mathrm{m\,s^{-1}}\).

Subtracting those speeds gives \(5\,\mathrm{m\,s^{-1}}\), not \(6.71\,\mathrm{m\,s^{-1}}\).

The directions matter because velocity is a vector.

For more practice with vector subtraction and arrow direction, see How to Draw a Relative-Velocity Triangle Without Reversing an Arrow.

You can also explore Can Two Objects Have the Same Speed but Different Velocities? for more examples of why speed alone cannot describe relative motion.

04Worked examples

05When does this method work?

The velocity-subtraction equation is part of the classical Galilean transformation between reference frames.

For standard HSC Physics problems, we generally assume:

• The reference frames have parallel coordinate axes.
• The frames move relative to one another without rotating.
• The velocities are measured at the same time.
• The speeds are much smaller than the speed of light.

There is one subtle point worth understanding.

The simple Galilean transformation between inertial frames assumes that their relative velocity is constant.

However, the instantaneous relationship

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

still applies when the observer’s frame accelerates, provided the coordinate axes remain parallel and the measurements are made at the same instant.

The observer’s velocity simply changes over time.

This becomes important when analysing objects moving inside accelerating vehicles.

At speeds approaching the speed of light, ordinary Galilean velocity subtraction is no longer accurate. Special relativity requires a different velocity transformation. This is an extension beyond the classical calculations used here.

06A reliable method for changing reference frames

When you encounter a relative-velocity problem, use the following process.

Step 1: Identify the object and observer.

Decide whose motion you want to describe and whose reference frame you’re using.

Step 2: Choose a common reference frame.

Usually, this is the ground frame.

Step 3: Write both original velocities relative to that frame.

Don’t mix velocities measured relative to the ground, water, air, or another moving object.

Step 4: Subtract the observer’s velocity from the object’s velocity.

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

Step 5: Interpret the direction and magnitude.

The result is a relative velocity vector. Take its magnitude if the question asks for relative speed.

The most important habit is to keep track of the observer. If you change the observer, you change the reference frame, and therefore the velocity you measure.

07Questions and solutions

Question 1

A cyclist travels north at \(12.0\,\mathrm{m\,s^{-1}}\) relative to the road. A filming cart travels north at \(4.0\,\mathrm{m\,s^{-1}}\) relative to the road.

Calculate the cyclist’s velocity relative to the filming cart and state the relative speed.

Solution 1

The cyclist moves north at \(8.0\,\mathrm{m\,s^{-1}}\) relative to the filming cart.

Both velocities are measured relative to the same ground frame, so we can subtract them directly.

Choose north as positive.

Let C represent the cyclist and F represent the filming cart.

\[ v_{C/F}=v_{C/G}-v_{F/G} \]

Substitute:

\[ \begin{aligned} v_{C/F}&=12.0-4.0\\ &=+8.0\,\mathrm{m\,s^{-1}} \end{aligned} \]

The positive sign indicates north.

The relative speed is the magnitude:

\[ \boxed{8.0\,\mathrm{m\,s^{-1}}} \]

The cyclist appears to move north at \(8.0\,\mathrm{m\,s^{-1}}\) to someone travelling in the filming cart.

Question 2

An electric tram travels west at \(16.0\,\mathrm{m\,s^{-1}}\) relative to the road.

A passenger walks towards the front of the tram at \(1.2\,\mathrm{m\,s^{-1}}\) relative to the tram.

An inspection van travels east at \(8.0\,\mathrm{m\,s^{-1}}\) relative to the road.

Calculate:

(a) The passenger’s velocity relative to the road.

(b) The passenger’s velocity relative to the inspection van.

Take east as positive.

Solution 2

The passenger travels west at \(17.2\,\mathrm{m\,s^{-1}}\) relative to the road and west at \(25.2\,\mathrm{m\,s^{-1}}\) relative to the inspection van.

The important detail is that the passenger walks towards the front of the westbound tram. Therefore, the passenger’s walking velocity is also westwards.

Part (a): Passenger relative to the road

Choose:

• P: passenger.
• T: tram.
• G: ground.

The given velocities are:

\[ v_{T/G}=-16.0\,\mathrm{m\,s^{-1}} \]

\[ v_{P/T}=-1.2\,\mathrm{m\,s^{-1}} \]

We want the passenger’s ground velocity.

Rearrange the relative-velocity equation:

\[ v_{P/G}=v_{P/T}+v_{T/G} \]

Substitute:

\[ \begin{aligned} v_{P/G}&=-1.2+(-16.0)\\ &=-17.2\,\mathrm{m\,s^{-1}} \end{aligned} \]

Therefore:

\[ \boxed{v_{P/G}=17.2\,\mathrm{m\,s^{-1}}\text{ west}} \]

The passenger’s velocity relative to the ground includes both their walking motion and the tram’s motion.

Part (b): Passenger relative to the inspection van

Let V represent the inspection van.

Its ground velocity is:

\[ v_{V/G}=+8.0\,\mathrm{m\,s^{-1}} \]

We want the passenger’s velocity relative to the van:

\[ v_{P/V}=v_{P/G}-v_{V/G} \]

Substitute:

\[ \begin{aligned} v_{P/V}&=-17.2-(+8.0)\\ &=-25.2\,\mathrm{m\,s^{-1}} \end{aligned} \]

Therefore:

\[ \boxed{v_{P/V}=25.2\,\mathrm{m\,s^{-1}}\text{ west}} \]

The tempting mistake is to subtract \(16.0-1.2\) when calculating the passenger’s ground velocity.

That would incorrectly assume the passenger walks towards the rear of the tram. In fact, both motions are westwards, so their signed velocities add when finding the ground velocity.

Question 3

A cyclist travels across a field with a ground velocity of:

• \(6.0\,\mathrm{m\,s^{-1}}\) east.
• \(8.0\,\mathrm{m\,s^{-1}}\) north.

A filming drone travels north at \(5.0\,\mathrm{m\,s^{-1}}\) relative to the ground.

Calculate:

(a) The cyclist’s velocity relative to the drone, expressed in components.

(b) The relative speed.

(c) The direction of the cyclist’s motion as measured by the drone.

Explain why simply subtracting the two ground speeds would give an incorrect answer.

Solution 3

The cyclist’s velocity relative to the drone is \(6.0\,\mathrm{m\,s^{-1}}\) east and \(3.0\,\mathrm{m\,s^{-1}}\) north.

Its relative speed is approximately \(6.71\,\mathrm{m\,s^{-1}}\), directed \(26.6^\circ\) north of east.

Part (a): Calculate the relative velocity components

We need to subtract the drone’s velocity from the cyclist’s velocity.

For the eastward component:

\[ v_{C/D,x}=6.0-0 \]

\[ v_{C/D,x}=6.0\,\mathrm{m\,s^{-1}} \]

For the northward component:

\[ v_{C/D,y}=8.0-5.0 \]

\[ v_{C/D,y}=3.0\,\mathrm{m\,s^{-1}} \]

Therefore:

\[ \boxed{ \vec v_{C/D} = (6.0\,\text{east},3.0\,\text{north}) \,\mathrm{m\,s^{-1}} } \]

Part (b): Calculate the relative speed

The relative speed is the magnitude of the relative velocity vector.

Using Pythagoras:

\[ \begin{aligned} v_{C/D} &=\sqrt{6.0^2+3.0^2}\\ &=\sqrt{45}\\ &=6.71\,\mathrm{m\,s^{-1}} \end{aligned} \]

Therefore:

\[ \boxed{v_{C/D}=6.71\,\mathrm{m\,s^{-1}}} \]

Part (c): Calculate the direction

Let \(\theta\) be the angle north of east.

\[ \tan\theta=\frac{3.0}{6.0} \]

\[ \theta=\tan^{-1}(0.5) \]

\[ \boxed{\theta=26.6^\circ} \]

The drone sees the cyclist moving \(26.6^\circ\) north of east.

Why doesn’t subtracting the speeds work?

The cyclist’s ground speed is:

\[ v_C=\sqrt{6.0^2+8.0^2} \]

\[ v_C=10.0\,\mathrm{m\,s^{-1}} \]

The drone’s ground speed is:

\[ v_D=5.0\,\mathrm{m\,s^{-1}} \]

Subtracting these magnitudes gives:

\[ 10.0-5.0=5.0\,\mathrm{m\,s^{-1}} \]

But the correct relative speed is \(6.71\,\mathrm{m\,s^{-1}}\).

The two velocity vectors aren’t parallel, so subtracting their magnitudes doesn’t produce the magnitude of their difference.

Always subtract the components first, then calculate the magnitude.

Question 4

An automated trolley carries a small puck along a straight east-west track.

A fixed camera records the positions of the puck and the trolley’s origin relative to the laboratory floor.

The following illustrative measurements are obtained:

East is positive.

(a) Calculate the average velocity of the puck relative to the trolley between \(0\) and \(2.0\,\mathrm{s}\).

(b) Calculate its average velocity relative to the trolley between \(2.0\) and \(4.0\,\mathrm{s}\).

(c) Explain how the puck can move east relative to the ground during both intervals but change its direction of motion relative to the trolley.

(d) Can we conclude that the puck’s instantaneous velocity relative to the trolley at exactly \(t=2.0\,\mathrm{s}\) is \(+0.10\,\mathrm{m\,s^{-1}}\)? Explain.

Solution 4

The puck’s average velocity relative to the trolley is \(-0.30\,\mathrm{m\,s^{-1}}\) during the first interval and \(+0.10\,\mathrm{m\,s^{-1}}\) during the second interval.

It moves east relative to the ground throughout both intervals, but its motion relative to the trolley changes direction.

Part (a): First time interval

Between \(0\) and \(2.0\,\mathrm{s}\):

The puck’s ground displacement is:

\[ \Delta x_{P/G}=1.00-0.00=+1.00\,\mathrm{m} \]

The trolley’s displacement is:

\[ \Delta x_{T/G}=1.60-0.00=+1.60\,\mathrm{m} \]

The relative displacement is:

\[ \begin{aligned} \Delta x_{P/T} &=\Delta x_{P/G}-\Delta x_{T/G}\\ &=1.00-1.60\\ &=-0.60\,\mathrm{m} \end{aligned} \]

Therefore, the average relative velocity is:

\[ \begin{aligned} v_{P/T} &=\frac{-0.60}{2.0}\\ &=-0.30\,\mathrm{m\,s^{-1}} \end{aligned} \]

\[ \boxed{v_{P/T}=-0.30\,\mathrm{m\,s^{-1}}} \]

The puck moves west relative to the trolley.

Part (b): Second time interval

Between \(2.0\) and \(4.0\,\mathrm{s}\):

The puck’s ground displacement is:

\[ \Delta x_{P/G}=2.40-1.00 \]

\[ \Delta x_{P/G}=+1.40\,\mathrm{m} \]

The trolley’s displacement is:

\[ \Delta x_{T/G}=2.80-1.60 \]

\[ \Delta x_{T/G}=+1.20\,\mathrm{m} \]

The relative displacement is:

\[ \begin{aligned} \Delta x_{P/T} &=1.40-1.20\\ &=+0.20\,\mathrm{m} \end{aligned} \]

Therefore:

\[ \begin{aligned} v_{P/T} &=\frac{0.20}{2.0}\\ &=+0.10\,\mathrm{m\,s^{-1}} \end{aligned} \]

\[ \boxed{v_{P/T}=+0.10\,\mathrm{m\,s^{-1}}} \]

The puck moves east relative to the trolley during this interval.

Part (c): Explain the change in relative motion

The puck’s ground position increases throughout the experiment.

Its average ground velocities are:

First interval:

\[ v_{P/G}=\frac{1.00}{2.0} =+0.50\,\mathrm{m\,s^{-1}} \]

Second interval:

\[ v_{P/G}=\frac{1.40}{2.0} =+0.70\,\mathrm{m\,s^{-1}} \]

The puck therefore moves east relative to the ground during both intervals.

However, during the first interval, the trolley moves east faster than the puck.

During the second interval, the puck moves east faster than the trolley.

The relative velocity changes sign because the motion of the trolley’s reference frame changes relative to the puck.

Part (d): Instantaneous versus average velocity

No. We cannot conclude that the instantaneous relative velocity at exactly \(t=2.0\,\mathrm{s}\) is \(+0.10\,\mathrm{m\,s^{-1}}\).

That value is the average relative velocity over the entire second interval, from \(2.0\) to \(4.0\,\mathrm{s}\).

An average over an interval doesn’t necessarily equal the instantaneous velocity at either endpoint.

We would need more detailed position measurements near \(t=2.0\,\mathrm{s}\) to determine the instantaneous velocity.

This distinction matters whenever a question gives position data over finite time intervals.

Question 5

An eastbound train travels at a constant \(20.0\,\mathrm{m\,s^{-1}}\).

At \(t=0\), a passenger throws a ball vertically upwards at \(8.0\,\mathrm{m\,s^{-1}}\) relative to the train.

At \(t=0.30\,\mathrm{s}\), the train begins braking with a constant acceleration of \(2.0\,\mathrm{m\,s^{-2}}\) westwards. It continues braking until \(t=1.00\,\mathrm{s}\).

Assume:

• The ball experiences no horizontal force after release.
• Air resistance is negligible.
• Gravitational acceleration is \(9.8\,\mathrm{m\,s^{-2}}\) downwards.
• The train continues travelling east throughout the stated interval.

At \(t=1.00\,\mathrm{s}\), calculate:

(a) The train’s velocity relative to the ground.

(b) The ball’s velocity relative to the ground.

(c) The ball’s velocity relative to the train.

(d) The ball’s horizontal displacement relative to the train since release.

(e) The ball’s vertical displacement from its release point, and whether it is still in the air.

Finally, evaluate this statement:

“The ball started with the same horizontal velocity as the train, so it must remain directly above its release point, even while the train brakes.”

Solution 5

At \(t=1.00\,\mathrm{s}\), the train travels east at \(18.6\,\mathrm{m\,s^{-1}}\), while the ball travels east at \(20.0\,\mathrm{m\,s^{-1}}\) and downwards at \(1.8\,\mathrm{m\,s^{-1}}\).

Relative to the train, the ball moves east at \(1.4\,\mathrm{m\,s^{-1}}\) and downwards at \(1.8\,\mathrm{m\,s^{-1}}\).

It is \(0.49\,\mathrm{m}\) ahead of its original horizontal position relative to the train and remains \(3.1\,\mathrm{m}\) above its release point.

The statement is incorrect because the ball retains its original horizontal velocity while the train slows down.

Part (a): Train velocity relative to the ground

The train initially travels east at:

\[ u_T=+20.0\,\mathrm{m\,s^{-1}} \]

It begins braking at \(t=0.30\,\mathrm{s}\).

The braking duration is:

\[ \Delta t=1.00-0.30 \]

\[ \Delta t=0.70\,\mathrm{s} \]

Taking east as positive, its acceleration is:

\[ a_T=-2.0\,\mathrm{m\,s^{-2}} \]

Using:

\[ v=u+at \]

\[ \begin{aligned} v_{T/G} &=20.0+(-2.0)(0.70)\\ &=20.0-1.4\\ &=18.6\,\mathrm{m\,s^{-1}} \end{aligned} \]

Therefore:

\[ \boxed{v_{T/G}=18.6\,\mathrm{m\,s^{-1}}\text{ east}} \]

Part (b): Ball velocity relative to the ground

The ball has no horizontal force acting on it after release.

Therefore, its horizontal velocity remains constant:

\[ v_{B/G,x}=+20.0\,\mathrm{m\,s^{-1}} \]

Its initial vertical velocity is:

\[ u_y=+8.0\,\mathrm{m\,s^{-1}} \]

Its vertical acceleration is:

\[ a_y=-9.8\,\mathrm{m\,s^{-2}} \]

After \(1.00\,\mathrm{s}\):

\[ \begin{aligned} v_{B/G,y} &=u_y+a_yt\\ &=8.0+(-9.8)(1.00)\\ &=-1.8\,\mathrm{m\,s^{-1}} \end{aligned} \]

Therefore:

\[ \boxed{ \vec v_{B/G} = (20.0\,\text{east},1.8\,\text{down}) \,\mathrm{m\,s^{-1}} } \]

The ball continues moving horizontally at its original speed, even though the train has started braking.

Part (c): Ball velocity relative to the train

Now apply the relative-velocity equation at \(t=1.00\,\mathrm{s}\).

\[ \vec v_{B/T} = \vec v_{B/G} - \vec v_{T/G} \]

For the horizontal component:

\[ \begin{aligned} v_{B/T,x} &=20.0-18.6\\ &=+1.4\,\mathrm{m\,s^{-1}} \end{aligned} \]

For the vertical component:

\[ v_{B/T,y}=-1.8-0 \]

\[ v_{B/T,y}=-1.8\,\mathrm{m\,s^{-1}} \]

Therefore:

\[ \boxed{ \vec v_{B/T} = (1.4\,\text{east},1.8\,\text{down}) \,\mathrm{m\,s^{-1}} } \]

An observer inside the train sees the ball moving forwards and downwards.

Part (d): Horizontal displacement relative to the train

Before braking begins, the train and ball have identical horizontal velocities.

Therefore, during the first \(0.30\,\mathrm{s}\), their horizontal separation remains unchanged.

Once braking begins, their horizontal accelerations differ.

The ball’s horizontal acceleration is zero, while the train’s is:

\[ a_T=-2.0\,\mathrm{m\,s^{-2}} \]

The ball’s horizontal acceleration relative to the train is:

\[ a_{B/T,x}=a_{B/G,x}-a_{T/G,x} \]

\[ a_{B/T,x}=0-(-2.0) \]

\[ a_{B/T,x}=+2.0\,\mathrm{m\,s^{-2}} \]

At the instant braking starts, their relative horizontal velocity is zero.

During the remaining \(0.70\,\mathrm{s}\), use:

\[ s=ut+\frac12 at^2 \]

Substitute:

\[ \begin{aligned} \Delta x_{B/T} &=0(0.70)\\ &\quad+\frac12(2.0)(0.70)^2\\ &=0.49\,\mathrm{m} \end{aligned} \]

Therefore:

\[ \boxed{\Delta x_{B/T}=0.49\,\mathrm{m}\text{ east}} \]

The ball moves \(0.49\,\mathrm{m}\) forwards relative to the train during braking.

Part (e): Vertical displacement and whether the ball is still airborne

The ball’s vertical motion is unaffected by the train’s horizontal braking.

Use:

\[ s_y=u_yt+\frac12a_yt^2 \]

Substitute:

\[ \begin{aligned} s_y &=8.0(1.00)\\ &\quad+\frac12(-9.8)(1.00)^2\\ &=8.0-4.9\\ &=+3.1\,\mathrm{m} \end{aligned} \]

Therefore:

\[ \boxed{s_y=3.1\,\mathrm{m}\text{ upwards}} \]

The ball is still above its release point.

We can also check its total flight time:

\[ t=\frac{2u_y}{g} \]

\[ t=\frac{2(8.0)}{9.8} \]

\[ t=1.63\,\mathrm{s} \]

Since \(1.00\,\mathrm{s}\) is less than \(1.63\,\mathrm{s}\), the ball is still airborne.

Evaluate the statement

The statement is incorrect.

The ball initially shares the train’s horizontal velocity because it was moving with the train before release.

However, once released, the ball experiences no horizontal force. Its horizontal velocity therefore remains \(20.0\,\mathrm{m\,s^{-1}}\).

The train, meanwhile, slows to \(18.6\,\mathrm{m\,s^{-1}}\).

The two horizontal velocities are no longer equal, so the ball moves forwards relative to the train.

There is also an important distinction here: subtracting velocities remains valid even when the observer’s frame accelerates. What changes is the observer’s velocity over time.

The instantaneous relative-velocity equation still works, but you must use the train’s current velocity rather than its original velocity.

This is the conceptual difference between changing reference frames at constant velocity and changing reference frames while accelerating.

08What to remember

When changing reference frames, subtract the observer’s velocity from the object’s velocity:

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

The reason is simple: relative position is the object’s position minus the observer’s position. Differentiating that relationship gives relative velocity.

Three things will keep your calculations reliable:

• Identify the object and observer before writing an equation.
• Make sure both original velocities are measured relative to the same reference frame.
• Subtract velocity vectors, not simply their speed magnitudes.

Once you understand this relationship, you can move comfortably between ground, vehicle, aircraft, and other moving reference frames.

The next useful step is applying the same vector relationships to situations involving wind, currents, and changing directions, where choosing the correct reference frame becomes just as important as performing the calculation.