Warm-up
Three quick ones from earlier work. Try each in your head or on paper before you open the answer.
- From lesson 4: find the magnitude of (5, 12).
Answer
13. - From lesson 5: work out (3, 2) + 2 × (1, −4).
Answer
(5, −6). - From Year 10: in a right-angled triangle with hypotenuse 10 and an angle of 30 degrees, how long is the side opposite the angle?
Answer
5. 10 × sin 30.
Words used in this lesson
- bearing
- an angle clockwise from north. For example: 090 for east.
- component
- the part of a vector in one direction. For example: the east part.
- resultant
- the single vector that has the same effect as several. For example: the actual path.
- heading
- the direction a vehicle points. For example: the way the bow faces.
- current
- the steady movement of water or air. For example: a river flowing east.
Which way should the ferry captain steer to reach the point opposite?
A river ferry near Chiang Mai crosses a river that flows towards the east at 3 kilometres per hour. The captain wants to land at the point directly north of the jetty, across the river. If the boat's own speed through the water is 10 kilometres per hour, should the bow point due north, a little west of north, or a little east of north?
A bearing is an angle measured clockwise from north, written with three digits, so east is 090 and south-west is 225. A velocity or a force is a vector, and the effect of two together is their sum, the resultant.
Predict first: will the boat that points due north still land directly north of the jetty? Then use the figure to see how the current shifts the resultant.
Boat 10 km/h on bearing 060 and a current of 2 km/h towards the east: the resultant is 11.8 km/h on bearing 64.9 degrees.
Show me the steps
- Boat 10 km/h on bearing 000 with no current: east 0, north 10, so the resultant is 10 km/h on bearing 0.
- Boat 10 km/h on bearing 000 with a current of 4 km/h east: east 4, north 10, so the speed is the square root of 116 which is 10.8 km/h, on bearing 21.8 degrees.
- Boat 10 km/h on bearing 270 (west) with a current of 4 km/h east: east −6, north 0, so the resultant is 6 km/h on bearing 270, which the current has slowed.
- Any boat and current: east = s sin t + c, north = s cos t, speed = the square root of east squared plus north squared, bearing from the inverse tangent of east over north.
Components from a bearing. A vector of size r on bearing t has an east component r sin t
and a north component r cos t, because the angle is measured from north. As a column (east over north) it is
r sin t
r cos t.
Resultant. Add the components of every vector. The resultant's size is the square root of the east component squared plus the north component squared, and its bearing is the inverse tangent of east over north, adjusted for the quadrant. To steer across a current, choose a heading whose component against the current cancels it.
Two common slips. Using the angle from east, as in a graph, so that sine and cosine are swapped: for a bearing the north component uses the cosine. And forgetting to add the current to the east component: the boat's own east component is s sin t, and the current adds to it.
Worked examples
1. A ship sails 20 km on bearing 060. Find its east and north components to one decimal place.
- East = 20 sin 60 = 17.3 and north = 20 cos 60 = 10.0.
- So the ship has moved 17.3 km east and 10.0 km north.
2. A boat steers north at 12 km/h in a current of 5 km/h towards the east. Find its speed and bearing.
- The components are east 5 and north 12, so the speed is the square root of 25 + 144 = 13.
- The bearing is tan−1(5 ÷ 12) = 22.6 degrees, so 13 km/h on bearing 022.6.
3. A force of 7 N acts due north and a force of 24 N acts due east. Find the resultant.
- Magnitude: the square root of 49 + 576 = 25, so 25 N.
- Bearing: tan−1(24 ÷ 7) = 73.7 degrees.
Your go. The first step is given. An aircraft flies at 200 km/h on bearing 090. A wind adds 50 km/h towards the north. Find the aircraft's speed over the ground and its bearing, to one decimal place.
- Write the components: east 200 and north 50.
- Find the speed from the components, then the bearing from tan−1(east ÷ north).
Check the answer
The speed is the square root of 40 000 + 2 500 = 206.2 km/h, and the bearing is tan−1(200 ÷ 50) = 76.0 degrees, so 206.2 km/h on bearing 076.0.
With a calculator
Keep the calculator in degree mode. A bearing is measured clockwise from north, so for a vector with east component E and north component N the angle from north is tan−1(E ÷ N), with E on the top. This is the other way round from the angle in a graph, where it is the up entry over the across entry.
Take care with the quadrant: if the vector points south or west, the calculator's acute angle needs to be added to 90, 180 or 270 degrees to give a bearing between 0 and 360. Sketch the vector first. Round only the final answer, to one decimal place unless the question says otherwise.
Check before you go on
1. Two forces act at a point: 9 N due north and 40 N due east. What is the bearing of the resultant, to one decimal place?
Your turn
Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.
Level 1 · Getting started
2. An aircraft travels 120 km east and then 50 km north. How far is it from its starting point, in kilometres?
3. Two forces act on a hook: 9 N due east and 12 N due north. What is the magnitude of the resultant force, in newtons?
Level 2 · Building
4. A walker in the Scottish hills goes 15 km east and 8 km north of a hut. What is the bearing of the walker from the hut, to one decimal place?
5. A boat steers due north at 8 km/h in a current of 5 km/h towards the east. What is its speed over the ground, to one decimal place?
Level 3 · Stretch
6. A boat steers on bearing 060 at 9 km/h through the water, in a current of 3 km/h towards the east. What is the bearing of its path over the ground, to one decimal place?
7. A current of 4 km/h flows towards the east. A boat that can travel at 12 km/h through the water must go straight north across the river. At what angle west of north should it steer, to one decimal place?
Something to think about
A boat can travel at 10 km/h through still water and the current is 10 km/h. Can the boat go straight across the river if the current is at right angles to the bank? What is the largest current for which it can?
One way to think about it
No. To go straight across, the boat must cancel the current by pointing upstream with a component of 10 km/h against it, which uses all of its speed, so nothing is left to cross. It can go straight across only if the current is less than its own speed.
Exit ticket
8. A boat steers due east at 12 km/h across a current of 5 km/h flowing due south. Which describes its path over the ground?