A ball thrown upward and caught at the starting point travels 280 cm — yet its displacement is zero. This single activity captures the entire difference between distance and displacement. From NCERT Chapter 4 (Exploration edition) Class 9 Science. Aligned with CBSE syllabus 2026-27.
Aim: To analyse the total distance travelled and the displacement of a ball at various stages of its vertical motion.
Setup: A ball is thrown vertically upward from point O (the hand). It rises to its highest point B (140 cm above O) and then falls back. Two intermediate points are marked on the vertical path: point A at 40 cm above O (on the way up) and point C at 80 cm above O (on the way down, i.e., the ball has already crossed C once while rising).
NCERT asks you to fill Table 4.1 by recording, at each labelled position, (a) the total distance the ball has travelled from O up to that point, and (b) the displacement of the ball from O at that point. Then you must choose which of four given statements is correct about the relationship between distance and displacement.
For each position, ask yourself two questions: How far has the ball moved in total from O? (distance) and How far and in which direction is the ball from O right now? (displacement).
| Position | Height above O | Total Distance from O (cm) | Displacement from O |
|---|---|---|---|
| O (start) | 0 cm | 0 | 0 |
| A (on way up) | 40 cm | 40 | 40 cm upward |
| B (highest point) | 140 cm | 140 | 140 cm upward |
| C (on way down) | 80 cm | 220 | 80 cm upward |
| O (returns) | 0 cm | 280 | 0 |
How each value is calculated:
Point A (on way up): Ball has moved 40 cm upward from O. Distance = 40 cm. Displacement = 40 cm upward. At this stage distance = displacement in magnitude, because the ball has not yet reversed direction.
Point B (highest point): Ball has moved 140 cm upward. Distance = 140 cm. Displacement = 140 cm upward. Still equal in magnitude — the ball has moved in one direction only.
Point C (on way down): The ball went up 140 cm (O→B) then came down 60 cm (B→C, since C is at 80 cm). Total distance = 140 + 60 = 220 cm. But the ball is now 80 cm above O, so displacement = 80 cm upward. Notice: distance (220) > |displacement| (80).
Back at O: Ball went up 140 cm then came back 140 cm. Total distance = 280 cm. Start and finish are the same point → displacement = 0.
NCERT gives four options based on this activity. Let us evaluate each one using the data from Table 4.1.
❌ Incorrect. This is true only at A and B (where the ball has not reversed direction). At C, distance = 220 cm but |displacement| = 80 cm. At final O, distance = 280 cm but displacement = 0. They are clearly not always equal.
❌ Incorrect. Displacement can never exceed distance — you cannot be farther from the start than the total path you walked. |Displacement| ≤ Distance always.
✅ Correct. Checking against Table 4.1:
The condition |displacement| ≤ distance holds at every point. Equality holds when motion is in one direction without reversal.
❌ Incorrect. |Displacement| is never greater than distance. It is either equal (before any reversal) or less (after reversal). It can never exceed the total path length.
Q. Why are distance and displacement equal on the way up (O to B) but not on the way down?
On the way up, the ball moves in one direction only (upward). Every centimetre added to the path length is also a centimetre added to the separation from O. The moment the ball reaches B and reverses direction, distance continues to increase while the ball is actually getting closer to O — so displacement decreases even as distance grows.
Q. This activity involves vertical motion. Is it 1D or 2D?
It is strictly one-dimensional — the ball moves along a single vertical axis. Because it is 1D, direction is described by the words "upward" or "downward" (or by + and − signs). In 2D or 3D motion, displacement would require specifying an angle or two/three components. This is why the activity is a clean illustration of the distance–displacement distinction without the complexity of 2D vectors.
Q. What is the displacement of the ball when it returns exactly to the midpoint (70 cm above O) on its way down?
Distance = 140 (up) + 70 (down from B) = 210 cm. Displacement = 70 cm upward (ball is 70 cm above O). |Displacement| (70) < distance (210) — the relationship holds.
Q1. A ball is thrown upward from O and reaches a height of 200 cm before returning to O. What are the total distance and displacement when the ball is at 60 cm height on its way down?
Q2. Can the displacement of an object ever be greater than the distance it has travelled? Justify using Activity 4.1.
Q3. At which point(s) in Activity 4.1 is the distance numerically equal to the magnitude of displacement? Why?