Distance and Displacement — Path vs Position Change
A surf lifesaver swims 80 m out to a struggling swimmer, then 80 m back to the beach. The training log records 160 m swum. But standing on the same patch of sand they started from, they have ended up exactly where they began. One number measured the whole path; the other measured only the change in position. That gap between path and position is what separates distance from displacement.
You walk from your front door 30 m to the letterbox, realise you forgot a parcel, walk 30 m back to the door, then 30 m to the letterbox again. A friend asks: "How far did you walk?" and "How far are you now from your door?" Sketch the trip on a line and write down both answers. Why isn't the second answer just the same as the first?
Which question is asking about displacement rather than distance?
Know
- That distance is a scalar — the total length of the path travelled
- That displacement is a vector — the straight-line change in position from start to finish
- That distance can never decrease but displacement can be zero or negative
- How to represent distance and displacement along a line using a sign convention
- The symbols and units: distance $d$ and displacement $s$ (or $\Delta x$), both in metres
Understand
- Why an out-and-back trip gives a large distance but a small (or zero) displacement
- Why displacement depends only on the start and end points, not on the route
- Why the sign of a displacement is a direction, not a smaller amount
- When distance and displacement happen to be equal (straight-line, no reversal)
Can Do
- Calculate total distance by adding every leg of a journey
- Calculate resultant displacement along a line using signed values
- Represent a journey and its displacement on a number line
- State a displacement correctly as a magnitude with a direction (or sign)
True or false: if an object's displacement for a journey is zero, then the distance it travelled must also be zero.
Core Content
Imagine tying a piece of string along the exact route an object takes — every twist, every backtrack — then pulling it straight and measuring it. That length is the distance. It is a record of the whole journey, and it cares only about how much ground was covered, never about which way.
Distance is a scalar: it has magnitude only. Its symbol is $d$ and its unit is the metre (m). To find a total distance, you simply add the length of every leg of the trip — each leg always contributes a positive amount, because you cannot "un-travel" ground you have already covered.
Distance only ever grows
If you walk 5 m forward and then 5 m back, the distance is not zero — it is $5 + 5 = 10$ m. Every metre of motion adds to the distance no matter which direction you face, so distance can never decrease as a journey continues.
Distance ($d$) is a scalar: the total length of the path travelled, in metres. It is found by adding every leg of the journey, each leg positive, so distance only ever grows and can never be negative or zero once motion has occurred.
Pause — copy the definition of distance, its symbol ($d$) and unit (m), and the rule "distance only ever grows" into your book before moving on.
A jogger runs 200 m down a path, then turns around and runs 80 m back. What total distance have they covered?
We just saw that distance records the whole path and only ever grows. That raises a question: but often we don't care about the route — we only care where you ended up compared to where you started. How do we measure that? This card answers it → with displacement, the straight-line change in position.
Forget the route for a moment. Draw a single straight arrow from where the object started to where it finished. That arrow — its length and its direction — is the displacement. It does not care how you got there; it only compares your final position with your initial one.
Displacement is a vector. Its symbol is $s$ (you will also see $\Delta x$, read "change in $x$") and its unit is the metre. It is defined as final position minus initial position:
Because it is a subtraction of positions, displacement carries a direction (a sign on a line). If you finish exactly where you began, $x_f = x_i$ and the displacement is zero — even though you may have travelled a long distance to do it.
Displacement ($s$ or $\Delta x = x_f - x_i$) is a vector: the straight-line change in position from start to finish, with direction, in metres. It depends only on the endpoints, not the route, and is zero if you finish where you started.
Pause — copy the definition of displacement, the formula $s = x_f - x_i$, and the rule "depends only on start and finish" into your book before moving on.
True or false: two cyclists take completely different roads from Town A to Town B. They must have the same displacement, even though their distances may differ.
We just saw that distance measures the path while displacement measures only the change in position. That raises a question: when do these two numbers actually disagree, and by how much? This card answers it → through the classic "there-and-back" and "around a track" trips, where the gap is largest.
The clearest place to feel the difference is a journey that reverses or loops. Whenever a path doubles back on itself, the distance keeps piling up while the displacement gets partly — or completely — cancelled.
Take the surf lifesaver from the hook. Swim 80 m out, then 80 m back, with east taken as positive:
- Distance $= 80 + 80 = 160$ m. Both legs add — distance never cancels.
- Displacement $= (+80) + (-80) = 0$ m. The return leg is the negative of the first, so they cancel exactly. You finish where you started.
The same pattern appears on a closed loop. A runner who completes exactly one lap of a 400 m oval covers a distance of 400 m, but their displacement is 0 m because the finish line and the start line are the same point.
For there-and-back or full-loop trips, distance keeps adding (e.g. 160 m, or one 400 m lap) but displacement is zero because the start and finish coincide. Distance ≥ |displacement| always; they are equal only for straight-line motion with no reversal.
Pause — copy the there-and-back example (160 m distance, 0 m displacement) and the rule "distance ≥ |displacement|" into your book before moving on.
For three of these trips the distance is greater than the magnitude of the displacement. Which one is the odd one out (distance equals displacement)?
We just saw that path and position change can disagree dramatically. That raises a question: how do we record both cleanly when an object moves back and forth along a single line? This card answers it → with a number line, an origin, and a sign convention that turns directions into + and − signs.
For motion along a line we draw a number line. Mark an origin (the zero) and choose a positive direction. Now every position is just a number, and the displacement is the difference between two positions — found by simple subtraction.
Suppose east = positive and we place the origin at the front door. An object's positions are: start at $x_i = +2$ m, finish at $x_f = +9$ m. Then:
The positive sign tells you the object finished 7 m to the east of where it began. If instead it finished at $x_f = -3$ m, the displacement would be $(-3) - (+2) = -5$ m, i.e. 5 m west. The distance, by contrast, is measured by following the path leg by leg, with every leg counted as positive regardless of direction.
On a line, mark an origin and a positive direction, then read each position as a signed number. Displacement = $x_f - x_i$ (the sign gives direction). Distance is still the leg-by-leg path total with every leg positive — the two are computed differently.
Pause — copy the number-line method, the calculation $s = x_f - x_i$, and one worked value (e.g. $(+9)-(+2)=+7$ m east) into your book before moving on.
Taking east as positive, an object starts at position $x_i = -4$ m and finishes at $x_f = +6$ m. Its displacement is:
We just saw how to read displacement off a number line as $x_f - x_i$. That raises a question: for a real multi-leg trip with several back-and-forth moves, how do we get both the distance and the displacement in one tidy method? This card answers it → a two-column routine: add path lengths for distance, add signed legs for displacement.
The safest way to handle a multi-leg journey is to compute the two quantities side by side. Distance adds the size of every leg; displacement adds the signed value of every leg. Do them in two columns and you will never confuse them.
Take the convention east = +. A delivery rider goes 12 m east, then 20 m west, then 3 m east:
- State the convention. Let east = positive (+), so west = negative (−).
- Distance — add the sizes. $d = 12 + 20 + 3 = 35$ m. (Every leg positive.)
- Displacement — add the signed legs. $s = (+12) + (-20) + (+3) = -5$ m.
- Interpret. The displacement is negative, so it is 5 m west. The trip's distance (35 m) is far larger than the magnitude of its displacement (5 m) because the rider doubled back.
Notice the check: the magnitude of the displacement (5 m) is less than the distance (35 m), exactly as the rule "distance ≥ |displacement|" demands. If they had come out equal you would suspect the path had no reversal.
For a multi-leg trip: distance = sum of leg sizes (all positive); displacement = sum of signed legs (sign = direction). e.g. 12 E, 20 W, 3 E → $d=35$ m, $s=(+12)+(-20)+(+3)=-5$ m (5 m west). Always check |displacement| ≤ distance.
Pause — copy the two-column method and the worked example ($d = 35$ m, $s = -5$ m = 5 m west) into your book before moving on.
Taking north as positive, a dog runs 9 m north, then 4 m south, then 6 m north. What are its distance and displacement?
Activities
For each scenario, state whether the quantity described is a distance or a displacement, and justify your answer in one sentence by referring to whether direction matters.
- A car's odometer reads 14 km after a round trip to the shops.
- A GPS reports you are "2.3 km north-east of home".
- A runner's watch shows "5.0 km run" after several laps of an oval.
- An aircraft's flight plan states it must end "120 km due south of the airport".
- A pedometer counts the total metres you walked around a shopping centre.
Two of these statements are true. One is a lie. Find the lie.
Take east = positive. For each trip, set up two columns: add the leg sizes for the distance, and add the signed legs for the displacement (give direction). Then check that the magnitude of the displacement is ≤ the distance.
- 15 m east, then 15 m west.
- 20 m east, then 8 m west, then 5 m east.
- 6 m west, then 6 m west, then 14 m east.
For each: explain in one line why the two answers differ (or, for any trip, why they happen to be equal).
A fresh five-question set drawn from this lesson's bank — feedback shown immediately.
Pick your answer, then rate your confidence.
1. Which statement correctly describes the difference between distance and displacement?
- Distance is a vector and displacement is a scalar
- Distance and displacement are always equal
- Distance is the total path length (scalar); displacement is the change in position (vector)
- Displacement is always larger than distance
2. A cyclist rides 600 m around a circular track and returns exactly to the starting line. What are the distance and displacement?
- Distance 0 m, displacement 600 m
- Distance 600 m, displacement 0 m
- Distance 600 m, displacement 600 m
- Distance 300 m, displacement 300 m
3. Taking east as positive, a person walks 9 m east then 14 m west. What is the resultant displacement?
- −5 m (5 m west)
- +5 m (5 m east)
- +23 m (23 m east)
- −23 m (23 m west)
4. An object starts at position $x_i = +7$ m and finishes at $x_f = +2$ m on a line (east positive). Its displacement is:
- +9 m
- +5 m
- +2 m
- −5 m (5 m west)
5. For which journey is the distance travelled exactly equal to the magnitude of the displacement?
- Walking 5 m forward and 5 m back to the start
- Driving 40 km in a straight line along a road without turning back
- Running two complete laps of an athletics track
- Swimming to the far wall of a pool and back again
UnderstandBand 3(2 marks) 1. Distinguish between distance and displacement, making clear which is a scalar and which is a vector.
ApplyBand 4(3 marks) 2. Taking east as positive, a courier travels 25 m east, then 40 m west, then 5 m east. Calculate the total distance travelled and the resultant displacement (with direction).
AnalyseBand 5(4 marks) 3. A student says, "After a 400 m lap of the oval the runner's displacement must be 400 m, because that's how far they ran." Identify the error, explain it using the meaning of distance and displacement, and state the correct distance and displacement for one complete lap.
Show all answers
Multiple choice
Q1 — C. Distance is a scalar equal to the total length of the path travelled; displacement is a vector equal to the change in position from start to finish. A reverses the scalar/vector roles, B is only true for straight-line motion with no reversal, and D is false (displacement can never exceed distance).
Q2 — B. The cyclist travels a path of 600 m, so the distance is 600 m. They return to the start, so the start and finish points coincide and the displacement is 0 m. A swaps the two; C and D ignore that the trip closes back on itself.
Q3 — A. East = +, so 9 m east = +9 m and 14 m west = −14 m. Resultant displacement = (+9) + (−14) = −5 m, i.e. 5 m west. The distance (a scalar) would be 9 + 14 = 23 m — the distractor in C.
Q4 — D. Displacement = $x_f - x_i = (+2) - (+7) = -5$ m, i.e. 5 m west. The object finished west of where it started, so the sign is negative. A adds the positions, B and C drop the sign.
Q5 — B. Distance equals the magnitude of displacement only when the motion is in a single straight line with no reversal — option B. A, C and D all double back or loop, so their distance exceeds their displacement.
Short Answer — Model Answers
Q1 (2 marks): Distance is the total length of the path actually travelled; it is a scalar (magnitude only) and can never decrease. Displacement is the change in position from start to finish, measured in a straight line; it is a vector (magnitude and direction) and can be zero or negative. (1 mark for each correctly defined and classified.)
Q2 (3 marks): Taking east as positive. Distance = 25 + 40 + 5 = 70 m (every leg added as a positive length). Displacement = (+25) + (−40) + (+5) = −10 m, i.e. 10 m west. (1 mark distance, 1 mark signed working, 1 mark resultant with direction.)
Q3 (4 marks): The error is treating the distance (400 m of path) as the displacement. Distance is the total path length and is 400 m for one lap. Displacement is the change in position from start to finish; for a complete lap the runner returns to the starting line, so the start and finish points coincide and the displacement is 0 m. The 400 m the runner "ran" is the distance, not the displacement, because displacement depends only on the endpoints, not the route. (1 mark identify error; 1 mark distance = 400 m; 1 mark displacement = 0 m; 1 mark explanation referencing start/finish coinciding.)
You now have distance and displacement for motion along a line. In the next lesson these go on the top of a fraction: speed is distance ÷ time, while velocity is displacement ÷ time. Predict: for the lifesaver who swims 160 m of path but ends with 0 m displacement in 100 s, what is their average speed, and what is their average velocity? Which one is zero, and why?
The lifesaver's log read 160 m because distance is a scalar — every metre of the out-and-back swim adds up and never cancels. But their displacement was 0 m because displacement is a vector measured straight from start to finish: the 80 m out and 80 m back are equal magnitudes in opposite directions, so they cancel and the swimmer ends on the same patch of sand. Path ≠ position change the moment a journey doubles back. That same split — distance on top of speed, displacement on top of velocity — runs through every lesson that follows.