Detailed notes on distance and displacement for Class 9 from the NCERT Exploration textbook (Chapter 4: Describing Motion Around Us). Topics covered: definitions, the athlete example, comparison table, scalar vs vector quantities, Activity 4.1, common misconceptions and solved problems. Aligned with CBSE syllabus 2026–27.
Think about your walk from home to school. You might take a straight road, or you might take a longer route with a few turns. In both cases, you reach school — but the length of the path you walked is different. This total length of the path is what we call distance.
Distance is the total length of the path (the line or curve) described by an object moving through space. It tells you how much ground the object has covered, regardless of which direction it moved.
The odometer (distance metre) in a car measures distance — it keeps adding up every metre the car moves, regardless of direction. If you drive 10 km east and then 10 km west, the odometer reads 20 km, not zero.
Now imagine you walk from your home (point A) to a shop 500 m away, then come back home. You walked 1000 m — that is your distance. But where did you end up? Exactly where you started — at home. So your net change in position is zero. This net change in position is called displacement.
Displacement is the net change in the position of an object between two given instants of time. It is the straight-line distance between the initial position and the final position of the object, measured in the direction from initial to final position.
For motion in a straight line, we represent direction using plus (+) and minus (–) signs. Positions to the right of the reference point (origin O) are taken as positive, and positions to the left are taken as negative. This convention must remain consistent throughout a problem.
This example is directly from the NCERT Exploration textbook (Fig. 4.4, Chapter 4). Work through it carefully — it is the standard example used in board examinations.
An athlete starts running from point O (the origin, 0 m) at time t = 0 s.
Distance = total length of path covered = OA + AB
Displacement = net change in position = final position – initial position
When will the displacement of the athlete be zero? What will be the total distance travelled in that case?
| Parameter | Distance | Displacement |
|---|---|---|
| Definition | Total length of path travelled | Net change in position from start to end |
| Type of quantity | Scalar | Vector |
| Direction needed? | No | Yes |
| Can it be zero? | No (for a moving body) | Yes (when start = end position) |
| Can it be negative? | No — always positive | Yes — depends on direction chosen as positive |
| Depends on path? | Yes — different paths give different distances | No — depends only on start and end positions |
| Magnitude relation | Always ≥ magnitude of displacement | Always ≤ distance |
| SI Unit | metre (m) | metre (m) |
In physics, we deal with two broad types of physical quantities:
| Type | What It Needs | Examples |
|---|---|---|
| Scalar | Magnitude (numerical value) only — no direction needed | Distance, speed, mass, time, temperature |
| Vector | Both magnitude AND direction for a complete description | Displacement, velocity, acceleration, force |
"Physical quantities which can be specified by just their numerical value are called scalars. Physical quantities which require specifying both the direction and magnitude are called vectors. You will learn about these in higher grades."
— NCERT Exploration, Chapter 4
Activity 4.1 in the NCERT Exploration textbook asks you to analyse the motion of a ball thrown vertically upward and fill in a table with distance and displacement values. Here is the complete explanation.
A ball is thrown vertically upward from point O (the origin, at the bottom). It travels straight up to the highest point B (140 cm above O), then falls back down. Intermediate positions on the way up are marked as A (40 cm above O). On the way down, an intermediate position is C (80 cm above O). Finally, the ball returns to O.
Q. Is this a motion in a straight line?
Yes. Even though the ball goes up and comes down, it moves only along a single vertical line. It is motion in a straight line (vertical).
| S.No. | Position | Total Distance Travelled from O | Displacement from O |
|---|---|---|---|
| 1. | O (start) | 0 cm | 0 cm |
| 2. | A (40 cm above O, going up) | 40 cm | 40 cm in the upward direction |
| 3. | B (140 cm above O, highest point) | 140 cm | 140 cm in the upward direction |
| 4. | C (80 cm above O, on the way down) | 200 cm (= 140 up + 60 down) | 80 cm in the upward direction |
| 5. | O (back to start) | 280 cm (= 140 up + 140 down) | 0 cm |
From NCERT: Choose which is true for displacement.
✅ Correction: Displacement is less than or equal to distance. When the object moves in one direction without turning back, displacement = distance. The correct rule is: |displacement| ≤ distance.
✅ Correction: Displacement can be zero even if the object has moved a large distance. This happens when the final position is the same as the initial position (e.g., one complete circular trip). The object has definitely moved — it has covered a distance — but the net change in position is zero.
✅ Correction: Both distance and displacement have the same SI unit — metre (m). The difference is not in units but in the nature of the quantities (scalar vs vector).
Problem: A father goes to a shop from home located at a distance of 250 m on a straight road. On reaching there, he discovered he forgot to carry a cloth bag. He came home to take it, went to the shop again, bought provisions and came back home. How much was the total distance travelled by him? What was his displacement from home?
Journey: Home → Shop → Home → Shop → Home
Distance in each leg = 250 m
Total distance = 250 + 250 + 250 + 250 = 1000 m = 1 km
Displacement = final position – initial position = Home – Home = 0 m
The father started and ended at home, so his net displacement is zero even though he covered 1 km.
Problem: A student runs from the ground floor to the 4th floor of a school building to collect a book, and then comes down to the 2nd floor (their classroom). If the height of each floor is 3 m, find: (i) the total vertical distance travelled, and (ii) their displacement from the starting point.
Given: Height of each floor = 3 m
Ground floor to 4th floor = 4 × 3 m = 12 m (upward)
4th floor to 2nd floor = 2 × 3 m = 6 m (downward)
(i) Total vertical distance = 12 + 6 = 18 m
(ii) Displacement = 2nd floor height from ground floor = 2 × 3 = 6 m in the upward direction
Note: Distance (18 m) ≠ Displacement (6 m). The student moved upward overall, so displacement is positive.
Problem: A cyclist completes one full lap around a circular track of radius R. Find the distance covered and the displacement at the end of the lap.
The circumference of a circular track of radius R = $2\pi R$
Distance covered = $2\pi R$ (the cyclist has gone all the way around)
Displacement = 0 (the cyclist starts and ends at the same point)
This is the clearest example showing distance ≠ displacement. Even though a large distance is covered, the displacement is zero because the starting and ending positions are identical.
N1. A car travels 60 km east, then 80 km north. Find the total distance and the magnitude of displacement.
N2. A ball is thrown up and reaches a maximum height of 20 m, then falls back to the starting point. Find: (a) total distance, (b) displacement.
N3. A runner runs around a rectangular track 100 m × 50 m and completes 2.5 laps. Find: (a) total distance, (b) displacement.