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Distance, displacement and vectors
Describing position and motion, scalar and vector quantities, the difference between distance and displacement, finding displacement with Pythagoras and bearings, and adding vectors head to tail.
An athlete who runs a perfect 400 m race finishes exactly where they started. They have run 400 metres, yet in one important sense they have gone nowhere. Physics needs two different words for "how far": one for the path you followed, and one for where you ended up.
Describing where things are
To describe motion you first need a reference point (an origin) and a direction. "The bus stop is 300 m" tells you very little. "The bus stop is 300 m north of the school gate" tells you exactly where to go. Motion is a change of position over time, measured relative to that reference point.
Physicists sort quantities into two kinds.
- A scalar has only a size (magnitude) and a unit: distance (5 m), speed (12 m/s), time (3 s), mass (60 kg), energy (200 J), temperature.
- A vector has a size and a direction: displacement (5 m north), velocity (12 m/s east), acceleration (9.8 m/s² downwards), force (300 N backwards), momentum.
Vectors are drawn as arrows. The length of the arrow shows the size, and the arrowhead shows the direction. In one dimension (along a straight line) we often show direction with a sign: choose one direction as positive, so the other direction is negative.
Distance and displacement
Distance is the total length of the path an object travels. It is a scalar, measured in metres (m), and it can never go down: every step adds to it.
Displacement (symbol s) is the straight-line change in position from the start to the finish, with a direction. It is a vector, also measured in metres. If you finish where you started, your displacement is zero, however far you walked.
The size of a displacement can never be bigger than the distance travelled. The two are equal only when the object moves in a straight line without turning back.
Finding displacement on a map
When a journey is made of straight sections at right angles, you can find the displacement in three steps: find the net movement north–south and east–west, use Pythagoras' theorem for the size, and use trigonometry for the direction.
A student walks 5 m north, then 4 m west, then 2 m south. The distance is simply the total: 5 + 4 + 2 = 11 m.
North–south: 5 m north − 2 m south = 3 m north. East–west: 4 m west.
These two movements are at right angles, so by Pythagoras: s = √(3² + 4²) = √25 = 5 m.
The angle west of north is θ, where tan θ = opposite ÷ adjacent = 4 ÷ 3, so θ = 53°. The direction is N53°W, which is a true bearing of 360° − 53° = 307°.
Distance 11 m; displacement 5 m on a bearing of 307°.
A true bearing is measured clockwise from north and written with three digits: east is 090°, south is 180° and west is 270°.
- Walk 5 m North, then 4 m West, then 2 m South. Check that the distance is 11 m and the displacement is 5 m on a bearing of 307°.
- Press Start again. Walk 6 m East, then 8 m North. Predict the displacement before you look (hint: 6, 8, ...). What bearing does the sim give?
- Make a walk that brings you back to the start. What happens to the displacement? Does the distance go back to zero too?
- Try to make a walk where the displacement is bigger than the distance. Can you? Explain why not.
Adding vectors
Two vectors are added by drawing them head to tail: draw the first arrow, start the second arrow where the first one ends, and draw the resultant from the very first tail to the very last head. The resultant has the same effect as the two vectors together.
- Vectors in the same direction add: 3 m/s east + 4 m/s east = 7 m/s east.
- Vectors in opposite directions subtract: 3 m/s east + 4 m/s west = 1 m/s west.
- Vectors at right angles combine by Pythagoras: 3 m/s north + 4 m/s east = 5 m/s.
This is why a pilot or a kayaker must aim upstream or into the wind to travel in a straight line to their destination: the velocity they produce and the velocity of the air or water add as vectors.
Test yourself
Which of these is a vector quantity?
Displacement has a size and a direction. The others have size only.
A swimmer swims 4 lengths of a 50 m pool, finishing at the starting end. Their distance and displacement are
The path length is 4 × 50 = 200 m, but they finish where they started, so the displacement is zero.
A hiker walks 9 km east then 12 km north. The size of the displacement is
√(9² + 12²) = √225 = 15 km.
A true bearing of 270° points
Bearings are measured clockwise from north: 090° east, 180° south, 270° west.
A plane flies north at 250 km/h relative to the air while a wind blows it south at 50 km/h. Its velocity over the ground is
Vectors in opposite directions subtract: 250 − 50 = 200 km/h, in the direction of the larger one.
Label the walk.
Sort these quantities.
A vector has both and . The total length of the path travelled is the . The straight-line change in position is the . Vectors are added by joining them head to .
A delivery rider travels 600 m south, then 800 m east, then 1200 m north. Calculate the distance travelled and the displacement (size and bearing), showing your working, and explain why the two answers are different.
Distance = 600 + 800 + 1200 = 2600 m. For displacement, the north–south movement is 1200 m north − 600 m south = 600 m north, and the east–west movement is 800 m east. These are at right angles, so the size is √(600² + 800²) = √1 000 000 = 1000 m. The direction is east of north by θ, where tan θ = 800 ÷ 600, so θ ≈ 53°, a bearing of 053°. The displacement is 1000 m on a bearing of 053°. The answers differ because distance is a scalar that adds up the whole path, including the 600 m south that was later cancelled out, while displacement is a vector that only measures the straight line from start to finish.