1-1 Velocity Vectors and Relative Velocity

Chapter 1 · Mechanics

❓ Lesson question

Why does a ferry pointed straight across the Nile land downstream — and why does falling rain appear to streak sideways on a moving bus window?

You will learn to

1Explain how two velocities combine into one resultant velocity.
2Calculate the perpendicular components of a velocity and recombine them.
3Determine the velocity of an object as seen from a moving observer.

The Nile ferry aims its bow straight at the opposite landing. The engine setting and the heading are both unchanged, yet the ferry lands well downstream of its target.

On a moving bus, rain that falls vertically in the ground frame appears as slanted streaks on a side window. Both observations follow from the same principle — and by the end of this lesson you will be able to say what that principle is.

✍️ Predict before you continue

Write your answer before reading on. A prediction you got wrong and then examined teaches you more than an answer you were handed.

(a) The ferry’s engine setting never changes. If the river current became faster, would the crossing take more time, the same time, or less time?

(b) From the moving bus, do the slanted streaks make the rain seem to come from ahead of the bus, or from behind it?

Write a line for each before this opens.

🔬 Explore — the crossing

What you’ll do: measure lengths and look for a pattern in them.

⏱️ 12 min🧰 A4 sheet · ruler · pen · pencil

1On paper: partner A pulls the sheet straight towards their chest at a slow, steady speed, while partner B slides the pen along a ruler at a slow, steady speed, drawing the whole time.
2Measure how far the pen moved along the rail (the across distance), how far the sheet was pulled (the pull distance), and the length of the slanted trace left behind.
3Repeat twice more: once pulling the sheet faster, once moving the pen faster.
4If you have no partner and no ruler right now, use the simulation below — it has the same three quantities and the same table.

🎛️ Simulation — one triangle, two situations

Boat speed (still water)4 m/s

River current3 m/s

River width60 m

Crossing time

15.0 s

Drag the current — this does not move. It is width ÷ boat speed; the current is not in it.

Downstream drift

45.0 m

Resultant speed

5.0 m/s

√(4² + 3²)

Angle to the bank

53.1°

Analyse what you collected

(a) Compare the last two columns of your table, run by run. What pattern do the numbers show?

(b) The pen moves only along the rail, while the sheet is pulled straight down. How do these two motions combine to produce the slanted trace?

✍️ Write your explanation in two lines before you read on.

In the Explore, the pen moved only along the rail — yet its trace on the sheet was slanted. The measured length of the trace matched $across^{2} + pull^{2}$ , not the simple sum of the across and pull distances. You might expect the two motions to add directly; your own results say otherwise. The model below explains why.

Velocity is a vector quantity

A velocity describes both how fast something moves and in which direction. That makes it a vector, and we draw it as an arrow. Sometimes one body carries two velocities at once: a boat is pushed by its engine while the river carries it along. Those two arrows combine tip to tail into a single resultant velocity:

v = v_{1} + v_{2}

That plus sign is doing something unfamiliar. Velocity vectors must be combined geometrically. Adding the two numbers together is only correct when the two arrows point along the same line — which is exactly what they did not do in your experiment.

Resolving a velocity into components

The same trick works backwards. Any single velocity can be split into two perpendicular components — one along each axis:

v_{x} = v cos θ v_{y} = v sin θ

and recombined with $v = v_{x}^{2} + v_{y}^{2}$ . The reason this is worth doing is the point of the whole lesson: the two axes do not talk to each other. For a river of fixed width, the component across the river sets the crossing time all by itself; the downstream drift is then simply the downstream component multiplied by that same time. Work each axis separately, then put them back together at the end.

What a moving observer sees

Now put the observer on a moving body. To find an object’s velocity relative to the bus, subtract the bus’s ground-frame velocity from the object’s ground-frame velocity:

v_{A B} = v_{B} - v_{A}

observed minus observer

Both arrows are drawn from a common origin. In the ground frame the rain falls straight down; subtracting the bus’s forward velocity adds a backward horizontal piece, and the result points downward and backward. That is the streak on the glass.

Point

Let $v_{1}$ be the velocity of a boat in still water and $v_{2}$ the velocity of the current. The velocity of the boat relative to the bank is $v = v_{1} + v_{2}$ — the arrows are combined, not the magnitudes. $v$ is the resultant velocity, and finding it is the addition of velocities. Going the other way — splitting $v$ back into $v_{1}$ and $v_{2}$ — is the resolution of a velocity into components.
Resolve $v$ into perpendicular $x$ - and $y$ -directions. If $θ$ is the angle between $v$ and the $x$ -axis, then $v_{x} = v cos θ$ , $v_{y} = v sin θ$ , and $v = v_{x}^{2} + v_{y}^{2}$ . A positive or negative sign on each component records its direction along that axis.
The velocity of B seen from A — the relative velocity of B with respect to A — is $v_{A B} = v_{B} - v_{A}$ : observed minus observer.

📐 Equation box

Components of a velocity

v_{x} = v cos θ v_{y} = v sin θ

Magnitude from components

v = v_{x}^{2} + v_{y}^{2}

Relative velocity of B with respect to A

v_{A B} = v_{B} - v_{A}

Key terms

Resultant velocity: — The single velocity vector with the same effect as two combined velocity vectors.
Resolution: — Splitting one velocity into two perpendicular components.
Relative velocity: — The velocity of one body as measured by an observer moving with another.
Reference frame: — The body an observer is at rest with respect to. A velocity is meaningless until you say which frame it is measured in.

🛠️ Worked example

Situation

A boat has a speed of 4.0 m/s in still water. It moves in a straight river with a current of 3.0 m/s. The boat’s bow points perpendicular to the current. Determine the speed of the boat relative to the riverbank.

Step 1 of 6 — try it yourself before you open the next one.

To a passenger on a metro train crossing an open-air section at 5.0 m/s, vertically falling rain appears to fall at 30° to the vertical. Calculate the speed of the raindrops.

A boat with a speed of 6.0 m/s in still water crosses a straight river with a current of 3.0 m/s, with its bow perpendicular to the current.
(i) Determine the angle

θ

that the resultant velocity of the boat makes with the riverbank.
(ii) If the river is 60 m wide, determine the time it takes to cross.

A cyclist travelling west at 5.0 m/s observes an apparent wind of 5.0 m/s coming from the north. Determine the direction and speed of the wind relative to the ground.

In a New Context

On a gusty afternoon at Borg El Arab Airport, the wind blows across the runway as a pilot lines up to land. If the aircraft were pointed straight down the runway, the crosswind would push it sideways off the centreline — the same effect as the ferry.

So the pilot points the aircraft slightly into the wind. The aircraft’s velocity through the air, added tip to tail to the wind’s velocity, gives a resultant directed straight along the runway. With an approach airspeed of 220 km/h and a 40 km/h crosswind, the required heading offset — the crab angle — is:

sin θ = \frac{40}{220} = 0.18 \Rightarrow θ \approx 1 0^{\circ}

➡️ Your turn: Explain why the pilot must point the aircraft’s nose away from the runway’s direction even though the aircraft must travel exactly along the runway.

The link back

The ferry, the rain and the aircraft are one problem wearing three costumes. In each case a body has a velocity of its own, something carries it sideways, and what an observer sees is the resultant. Once you can draw the triangle, you can do all three.

🧪 Nature of Science — uniform motion cannot be detected from inside the moving frame

In 1632, Galileo asked his reader to imagine a cabin below deck on a smoothly sailing ship: dripping water, fish in a bowl, drifting butterflies. His claim was that no observation made inside the cabin reveals whether the ship is at rest or moving with uniform velocity. Everything falls, swims and flies exactly as it does in harbour.

What an observer measures depends on the observer’s own motion — the same lesson as the rain on your bus window. This is why a velocity is meaningful only once the reference frame is stated. “The rain falls at 8.7 m/s” is an incomplete sentence until you say measured by whom.

Your turn: Suggest how two careful observers can record different velocities for the same raindrop without either of them being wrong.

✅ Lesson question, answered

Why does a ferry pointed straight across the Nile land downstream — and why does falling rain appear to streak sideways on a moving bus window?

The ferry. Its velocity over the ground is the tip-to-tail sum of its still-water velocity (across) and the current’s velocity (downstream). So its track slants downstream even though its bow never turns. The bow direction and the track direction are two different things, and only the first is under the pilot’s control.

The rain. The velocity of the rain relative to the bus equals the rain’s ground-frame velocity minus the bus’s ground-frame velocity. Rain that falls vertically in the ground frame therefore appears, to an observer on the bus, to fall at an angle — downward and backward.

One principle, stated once: perpendicular components are independent, and what you observe depends on the frame you observe from.

A boat with a speed of 2.0 m/s in still water moves in a straight river that has a current of 1.5 m/s, with its bow pointing perpendicular to the riverbank. Determine the speed of the boat relative to the riverbank.

An airplane is climbing at a constant speed of 50 m/s at an angle of 30° to the horizontal. Determine the height gained by the airplane in 10 s.

To an observer looking out of the window of train A, which is moving on a horizontal track at a speed of 10 m/s, vertically falling rain appears to fall at an angle of 45° to the vertical.
(i) Determine the speed of the raindrops.
(ii) To an observer looking out of the window of train B passing train A, the rain appears to be falling at an angle of 60° to the vertical. Determine the speed of train B.

Car A travels eastward at 20 m/s and car B travels northward at 20 m/s. Determine the direction and magnitude of the relative velocity of B with respect to A.

How the real paper is built

The ministry sets 50% multiple choice and 50% written answer. They test different things: the MCQ half rewards eliminating fast and spotting a distractor built from a plausible wrong method; the written half asks you to show that you understand why. Practise both halves — being fluent at one and weak at the other costs you exactly half the paper.

Part 1 — multiple choice (50% of the paper)

Every wrong option here is built from a real mistake, not a random number. If you pick one, the explanation tells you which mistake you made.

A boat crosses a river with its bow perpendicular to a current. Its speed in still water is 8.0 m/s and the current is 6.0 m/s. Its speed relative to the bank is:

A ferry crosses a river pointing straight across. If the current speed doubles, the time to cross:

An aeroplane flies at 50 m/s at 30° above the horizontal. Its vertical velocity component is:

Car A travels east at 20 m/s; car B travels east at 30 m/s. The velocity of B relative to A is:

Rain falls vertically. To a passenger on a bus moving forward, the rain appears to come from:

A swimmer heads directly across a river 60 m wide at 1.5 m/s while the current is 2.0 m/s. How far downstream do they land?

Part 2 — written answer (the other 50%)

Here the method is the marks. Work down the levels; if Level 1 is not automatic yet, stay there before moving on. Use the notebook button on each question so your working is saved on its own page and you can compare it with the model answer afterwards.

Level 1 — direct substitution

One step. You are given the formula; put the numbers in.

1.1

A motorboat heads straight across a canal at 3.0 m/s while the current flows at 4.0 m/s. Determine the boat’s speed relative to the bank.

1.2

A drone flies at 20 m/s at 37° above the horizontal. Determine its horizontal and vertical velocity components.

Level 2 — application in a new context

Two steps, or you must choose the right component before you can start.

2.1

A ferry crosses a 240 m stretch of the Nile. Its engine gives it 3.0 m/s in still water and its bow points perpendicular to the bank; the current is 2.0 m/s.
(i) How long does the crossing take? (ii) How far downstream does it land? (iii) What is its speed over the ground?

2.2

Rain falls vertically at 6.0 m/s while a car drives at 6.0 m/s. Determine the angle to the vertical at which the streaks appear on the side window, and the speed at which the rain appears to travel.

Level 3 — exam challenge

Multi-step, or a case the textbook does not set and examiners do.

3.1

A swimmer whose speed in still water is 1.5 m/s wants to land directly opposite on the far bank of a 90 m wide river that flows at 0.9 m/s.
(i) At what angle must they aim? (ii) What is their speed relative to the bank? (iii) How long does the crossing take?

3.2

An aircraft must fly due north. Its airspeed is 200 km/h and a wind blows from the west at 50 km/h.
(i) In which direction must the nose point? (ii) What is the ground speed?

📝 Exam-style question

A river flows at 1.2 m/s parallel to its banks. A swimmer whose speed in still water is 0.9 m/s heads directly across the river. Determine the magnitude of the swimmer’s velocity relative to the bank.

Keep reading — this lesson marks itself done automatically.