KBAT: Vectors in Navigation Problems

Q: Why is the displacement magnitude 5 km, not 10 km?

The 101010 km h−1^{-1}−1 is a speed; the boat travels for only half an hour, so its distance is 10×0.5=510\times0.5=510×0.5=5 km. Equivalently, the displacement vector is 3i+4j3\mathbf{i}+4\mathbf{j}3i+4j with magnitude 32+42=5\sqrt{3^2+4^2}=532+42=5. Add Math Paper 2 is 2 hours 30 minutes and 100 marks with analytic marking, so keeping speed, time and displacement distinct protects every method mark.

What makes this a KBAT question

A routine vectors question gives you two vectors and asks for their sum or magnitude. A navigation KBAT question hides the vectors inside a situation, a boat steered one way while a current pushes another, and asks you to build them, add them, and then interpret the resultant as a speed, a bearing, and a landing point.

That is Kemahiran Berfikir Aras Tinggi, higher-order thinking: writing $\mathbf{i}$ and $\mathbf{j}$ components and using $\sqrt{x^2+y^2}$ are ordinary Form 4 skills, but choosing east–north as your directions, realising that the time to cross a river depends only on the velocity component across it, and that downstream drift depends only on the component along it, are insights the question leaves to you. Reading a bearing correctly from the components adds one more layer.

The situation, not the arithmetic, is what makes it hard.

One worked problem, in the style of Paper 2

This is an original question written in the style of SPM Paper 2. Try it yourself before reading the solution.

Q1[8 marks]

Take $\mathbf{i}$ as a unit vector due east and $\mathbf{j}$ as a unit vector due north. In still water a boat travels at $8$ km h $^{-1}$ due north.

A straight river flows at $6$ km h $^{-1}$ due east, and the boat's engine keeps it pointing due north throughout. (a) Express the boat's resultant velocity in terms of $\mathbf{i}$ and $\mathbf{j}$ , and find the resultant speed.

(b) Find the bearing on which the boat actually moves. (c) The river is $4$ km wide from the south bank to the north bank.

Find the time taken to cross, the distance the boat is carried downstream, and the magnitude of its resultant displacement.

Show worked solution

Understand. Two velocities act at once: the boat's $8$ km h $^{-1}$ due north, written $8\mathbf{j}$ , and the current's $6$ km h $^{-1}$ due east, written $6\mathbf{i}$ .

The boat's true motion is their vector sum. The banks run east–west, so 'across the river' is the north ( $\mathbf{j}$ ) direction and 'downstream' is the east ( $\mathbf{i}$ ) direction.

Plan. Add the two velocity vectors for the resultant, use $\sqrt{x^2+y^2}$ for the speed, and $\tan\theta=\tfrac{\text{east}}{\text{north}}$ for the bearing.

For the crossing, divide the $4$ km width by the north component, then multiply that time by the east component for the drift.

Execute. (a) Add the velocities:

\mathbf{v} = 6\mathbf{i} + 8\mathbf{j}

The resultant speed is the magnitude:

|\mathbf{v}| = \sqrt{6^2 + 8^2} = \sqrt{36+64} = \sqrt{100} = 10 \text{ km h}^{-1}

(b) The resultant points into the north-east quadrant. Measuring the angle $\theta$ east of due north:

\tan\theta = \frac{\text{east component}}{\text{north component}} = \frac{6}{8} = 0.75

\theta = \tan^{-1}(0.75) = 36.87^\circ

A bearing is measured clockwise from north, so the boat moves on a bearing of $036.87^\circ$ , which is about $037^\circ$ .

t = \frac{4}{8} = 0.5 \text{ hours} = 30 \text{ minutes}

During that time the east component $6$ km h $^{-1}$ carries it downstream:

\text{drift} = 6 \times 0.5 = 3 \text{ km}

The resultant displacement is $\mathbf{s}=\mathbf{v}\,t=(6\mathbf{i}+8\mathbf{j})(0.5)=3\mathbf{i}+4\mathbf{j}$ , with magnitude

|\mathbf{s}| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5 \text{ km}

So the boat crosses in 30 minutes, lands $3$ km downstream, and its straight-line displacement from the start is $5$ km.

Check. The displacement $3\mathbf{i}+4\mathbf{j}$ points the same way as the velocity $6\mathbf{i}+8\mathbf{j}$ , both simplify to the direction $3:4$ , as it must, since the boat moves in a straight line at constant velocity.

The north part of the displacement, $4$ km, equals the river width, confirming the crossing is complete.

Finding a sensible first step

When motion is described in words and directions, the dependable first step is to fix a pair of perpendicular directions and write every velocity in components. Here east ( $\mathbf{i}$ ) and north ( $\mathbf{j}$ ) are natural, because the river flows east and the boat is steered north.

Translate each phrase into a component: ' $8$ km h $^{-1}$ due north' becomes $8\mathbf{j}$ ; ' $6$ km h $^{-1}$ due east' becomes $6\mathbf{i}$ . With both in component form, the resultant is a single addition, and the two later questions separate cleanly: distance across the river uses only the north component, drift downstream uses only the east one.

Committing to components first, rather than trying to reason about the slanted path directly, is what turns a wordy navigation scene into three short calculations.

What markers reward

Marking is analytic, so method marks are awarded line by line. On a navigation vectors question a marker looks for:

Each velocity written in components, $8\mathbf{j}$ for the boat and $6\mathbf{i}$ for the current.
The resultant found by adding, $\mathbf{v}=6\mathbf{i}+8\mathbf{j}$ .
The speed as a magnitude, $\sqrt{6^2+8^2}=10$ km h $^{-1}$ .
A bearing measured clockwise from north, $\tan\theta=\tfrac{6}{8}$ giving about $037^\circ$ .
Crossing time from the across-river component only, $t=\tfrac{4}{8}=0.5$ h.
Downstream drift from the along-river component, $6\times0.5=3$ km, and the displacement magnitude $5$ km.

How a teacher helps

Navigation questions reward students who resolve into components cleanly and then interpret each one, and both habits grow with feedback. In a one-to-one lesson our teachers ask you to fix east and north first, to write each velocity as $\mathbf{i}$ and $\mathbf{j}$ components before adding, and to say which component controls the crossing time and which controls the drift.

We are careful with the bearing, measured clockwise from north, because that is where marks are quietly lost. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

Message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.

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Frequently asked questions

Why does the crossing time use only the northward speed?

The river is crossed in the north direction, so only the north component of the boat's velocity moves it from bank to bank. Here that component is $8$ km h $^{-1}$ , and the width is $4$ km, so the time is $\tfrac{4}{8}=0.5$ h.

The eastward current does not help or hinder the crossing itself, it only carries the boat downstream during that same time.

How do I turn components into a bearing?

A bearing is measured clockwise from north. With an east component and a north component, find the angle east of north from $\tan\theta=\tfrac{\text{east}}{\text{north}}$ .

Here $\tan\theta=\tfrac{6}{8}=0.75$ , so $\theta=36.87^\circ$ and the bearing is about $037^\circ$ . Always check which quadrant the resultant points into before writing the three-figure bearing.

Why is the displacement magnitude 5 km, not 10 km?

The $10$ km h $^{-1}$ is a speed; the boat travels for only half an hour, so its distance is $10\times0.5=5$ km. Equivalently, the displacement vector is $3\mathbf{i}+4\mathbf{j}$ with magnitude $\sqrt{3^2+4^2}=5$ .

Add Math Paper 2 is 2 hours 30 minutes and 100 marks with analytic marking, so keeping speed, time and displacement distinct protects every method mark.

Source:SRC-DSKP-EN SRC-FORMAT

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026