This MCQ module is based on: Motion, Distance and Displacement
TOPIC 11 OF 50
Motion, Distance and Displacement
🎓 Class 9
Science
CBSE
Theory
Ch 4 — Describing Motion Around Us
⏱ ~15 min
🌐 Language: [gtranslate]
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Introduction: When Do We Say Something Is Moving?
Look around for a moment. A bird flies past your window. A car rolls down the street. The hands of a clock crawl forward. Even when you sit perfectly still, the Earth is hurtling through space at roughly 30 km every second. Clearly, motion is everywhere — but how do scientists describe it precisely so that two people anywhere in the world can agree on what they mean?
This chapter answers that question with a few powerful ideas: a reference point, the difference between distance and displacement, and the difference between uniform and non-uniform motion. By the end of Part 1, you will be able to read a problem about a moving object and confidently calculate distance, displacement and average speed.
Big Idea: An object is said to be in motion if its position changes with time relative to some chosen reference point. Without a reference, motion is meaningless.
4.1 Describing Motion — The Reference Point
Imagine you are sitting inside a moving train. Your school bag on the rack appears to be at rest with respect to you, but a person standing on the platform sees both you and the bag rushing past. Who is correct? Both! The state of motion of an object is always described relative to something else — the reference point (or origin).
To completely fix the position of an object on a straight line, we need (i) a reference point and (ii) a directed distance from that reference. For example, "the post office is 5 km north of the railway station" tells us where the post office is, with the station as the reference point.
4.2 Scalars and Vectors
Some quantities in physics need only a number (with a unit) to be fully described. Others need a number and a direction. This natural split gives us two big families of quantities.
| Type | Definition | Examples |
|---|---|---|
| Scalar | Magnitude only | Distance, speed, mass, time, temperature, energy |
| Vector | Magnitude + direction | Displacement, velocity, acceleration, force, weight |
Quick test: Ask "Does the answer change if I face the other way?" If yes, it is a vector. If no, it is a scalar. "It is 30 °C outside" — same in every direction → scalar. "Push the box 5 m east" — direction matters → vector.
4.3 Distance vs Displacement
Suppose you walk 4 m east and then 3 m north. The total path you actually walked is \(4 + 3 = 7\) m. But your final position, measured straight from the start, is only \(\sqrt{4^2 + 3^2} = 5\) m (north-east). The first number (7 m) is distance; the second (5 m, north-east) is displacement.
Distance: the total length of the actual path covered by an object. It is a scalar; SI unit is the metre (m).
Displacement: the shortest straight-line distance from the initial to the final position, with direction. It is a vector; SI unit is the metre (m).
Displacement: the shortest straight-line distance from the initial to the final position, with direction. It is a vector; SI unit is the metre (m).
🚶 Path vs Straight-Line — Predict, then click each leg L2 Understand
A walker goes 4 m east, then 3 m north. Predict first: will the distance covered equal the displacement? Click the green path (the actual walk) and the orange arrow (the straight-line shortcut) to check.
Click the green path or the orange arrow above to compare distance with displacement.
Special case: If you walk 5 km north and then 5 km back south to the same starting spot, your distance is 10 km but your displacement is zero. Displacement can even be negative (when direction is opposite to the chosen positive direction); distance is always positive or zero.
Activity 4.1 — Round Trip on a Track
L3 Apply
Predict first: A jogger runs around a circular track of radius 100 m and returns to her starting point. What will be her distance and her displacement?
- Mark a circular path on the ground with a radius of 100 m (or any convenient length).
- Start at point A. Run one full lap and return to A.
- Record the path length covered (distance) and the straight-line gap between start and end (displacement).
Result: Distance covered = circumference \(= 2\pi r = 2 \times 3.14 \times 100 = 628\) m. Displacement \(= 0\) (because start and end are the same point). This shows clearly that distance and displacement are two different physical quantities.
4.4 Uniform and Non-Uniform Motion
Place yourself by the side of a highway and watch the cars. A vehicle that covers 20 m every second, never speeding up or slowing down, is in uniform motion. A car that covers 20 m in the first second, 25 m in the next, and only 10 m in the third is in non-uniform motion.
| Time interval (s) | Uniform distance (m) | Non-uniform distance (m) |
|---|---|---|
| 0–1 | 10 | 5 |
| 1–2 | 10 | 12 |
| 2–3 | 10 | 9 |
| 3–4 | 10 | 20 |
4.5 Rate of Motion — Speed
If two athletes finish a 100 m race, the one who took the shorter time is faster. The rate at which an object covers distance is called its speed.
\[\text{Speed} = \dfrac{\text{Distance travelled}}{\text{Time taken}}, \qquad v = \dfrac{s}{t}\]
The SI unit of speed is m/s. Other common units are km/h and cm/s. A useful conversion: \(1 \text{ km/h} = \dfrac{1000 \text{ m}}{3600 \text{ s}} = \dfrac{5}{18} \text{ m/s}\), and \(1 \text{ m/s} = \dfrac{18}{5} \text{ km/h}\).
Average Speed
Real journeys speed up, slow down, and pause. To describe such a trip with a single number, we use:
\[\text{Average speed} = \dfrac{\text{Total distance covered}}{\text{Total time taken}}\]
Worked Numerical Examples
Example 1 — Distance vs Displacement (round trip)
A boy walks 600 m east from his house, then turns back and walks 400 m west. Find his total distance and displacement (taking east as positive).
Given: path AB = 600 m east; path BC = 400 m west.
Step 1. Total distance \(= 600 + 400 = 1000\) m.
Step 2. Net displacement \(= +600 - 400 = +200\) m, i.e., 200 m east of the house.
Answer: Distance \(= 1000\) m, Displacement \(= 200\) m east.
Example 2 — Unit conversion
Convert (a) 72 km/h into m/s, (b) 25 m/s into km/h.
(a) \(72 \text{ km/h} = 72 \times \dfrac{5}{18} \text{ m/s} = 20\) m/s.
(b) \(25 \text{ m/s} = 25 \times \dfrac{18}{5} \text{ km/h} = 90\) km/h.
Example 3 — Average speed of a cyclist
A cyclist covers 4.5 km in the first 15 minutes and 3.5 km in the next 15 minutes. Compute her average speed in km/h and in m/s.
Given: total distance \(= 4.5 + 3.5 = 8.0\) km; total time \(= 30\) min \(= 0.5\) h.
Step 1. \(\text{Average speed} = \dfrac{8.0 \text{ km}}{0.5 \text{ h}} = 16\) km/h.
Step 2. \(16 \text{ km/h} = 16 \times \dfrac{5}{18} \approx 4.44\) m/s.
Answer: 16 km/h or about 4.44 m/s.
Example 4 — Two-leg journey at different speeds
A car travels the first half of a 200 km journey at 50 km/h and the second half at 100 km/h. Find its average speed for the whole trip.
Given: \(s_1 = s_2 = 100\) km; \(v_1 = 50\) km/h; \(v_2 = 100\) km/h.
Step 1. \(t_1 = \dfrac{100}{50} = 2\) h, \(t_2 = \dfrac{100}{100} = 1\) h.
Step 2. Total time \(= 3\) h, total distance \(= 200\) km.
Step 3. \(\bar v = \dfrac{200}{3} \approx 66.67\) km/h.
Answer: Average speed \(\approx 66.67\) km/h. Notice it is not the simple mean (75 km/h)!
Example 5 — Athlete on a circular track
An athlete runs around a circular track of diameter 200 m and completes one full round in 40 s. Find (a) average speed, (b) magnitude of average velocity, and (c) the distance and displacement after half a round.
Given: \(d = 200\) m, so \(r = 100\) m; \(T = 40\) s.
(a) Full round: distance \(= 2\pi r = 2 \times 3.14 \times 100 = 628\) m. Average speed \(= 628/40 = 15.7\) m/s.
(b) Full round: displacement \(= 0\) (back to start), so average velocity \(= 0\) m/s.
(c) Half round: distance \(= \pi r = 314\) m; displacement \(= d = 200\) m (the diameter, straight line).
In-Text Question Check
Q. An object has moved through a distance. Can it have zero displacement? Give an example.
Yes. If the object returns to its starting point, the distance covered is non-zero but the net change in position (displacement) is zero. Example: a runner completes one full lap of a 400 m track and returns to the start — distance = 400 m, displacement = 0.
Yes. If the object returns to its starting point, the distance covered is non-zero but the net change in position (displacement) is zero. Example: a runner completes one full lap of a 400 m track and returns to the start — distance = 400 m, displacement = 0.
Competency-Based Questions
A delivery boy starts from the post office and rides 8 km north, 6 km east, and finally 8 km south, ending his trip. The total time taken for the entire ride is 0.5 h.
Q1. What is the total distance covered by the delivery boy? L2
Answer: (b) 22 km. Total path = 8 + 6 + 8 = 22 km.
Q2. The magnitude of his displacement is — L3
Answer: (c) 6 km. Net north–south = 8 − 8 = 0; net east = 6 km, so displacement is 6 km east.
Q3. His average speed in km/h is — L3
Average speed = total distance / total time = 22 / 0.5 = 44 km/h.
Q4. Why does the magnitude of his average velocity differ from his average speed? L4
Average speed uses total distance (22 km), while average velocity uses displacement (6 km). Because the path is not a straight line, distance > |displacement|, so average speed > |average velocity| = 6/0.5 = 12 km/h.
Q5. Suggest a route between the same two end points that would make average speed equal to magnitude of average velocity. L5
A perfectly straight-line trip from start to finish (no detours). Then distance = |displacement|, so the two averages match.
Assertion–Reason Questions
Choose: (A) Both A and R are true; R is the correct explanation of A. (B) Both A and R are true; R is NOT the correct explanation of A. (C) A is true; R is false. (D) A is false; R is true.
Assertion: Displacement of an object can be zero even when distance covered is non-zero.
Reason: Displacement is the shortest straight-line distance between the initial and final positions.
(A) When an object returns to its start, the start and end positions coincide, so the shortest straight-line gap (displacement) is zero, while the path length (distance) is not.
Assertion: Speed is a scalar but velocity is a vector.
Reason: Speed depends only on distance, while velocity also depends on the direction of motion.
(A) Both true and the reason correctly explains the assertion: a quantity needing direction is a vector.
Assertion: Uniform motion always means an object is moving in a straight line.
Reason: Uniform motion requires equal distance in equal time.
(D) A is false: a body can move uniformly along a curve (e.g., uniform circular motion). R is true.
Frequently Asked Questions — Motion, Distance and Displacement
What is motion, distance and displacement in Class 9 Science (CBSE/NCERT)?
Motion, Distance and Displacement is a key topic in NCERT Class 9 Science Chapter 4 — Describing Motion Around Us. It explains concepts of motion and rest, distance vs displacement, and uniform vs non-uniform motion. Core ideas covered include motion, rest, reference point, frame of reference. Mastering this subtopic is essential for scoring well in the CBSE Class 9 Science exam and for building a strong foundation for the Class 10 board exam, because these concepts repeatedly appear in MCQs, short answers and long-answer questions. This part gives a complete, exam-ready explanation with activities, diagrams and competency-based practice aligned to NCERT.
Why is motion important in NCERT Class 9 Science?
Motion is important in NCERT Class 9 Science because it forms the foundation for understanding motion, distance and displacement in Chapter 4 — Describing Motion Around Us. Without a clear idea of motion, students cannot answer higher-order CBSE questions involving rest, reference point, frame of reference. School and competitive papers regularly include 2-mark and 3-mark questions on this concept, and competency-based questions often link motion to real-life situations. Building clarity here pays off directly in marks at Class 9 and again in the Class 10 board exam.
How is motion, distance and displacement tested in the Class 9 Science CBSE exam?
The CBSE Class 9 Science exam tests motion, distance and displacement through a mix of 1-mark MCQs, 2-mark short answers, 3-mark explanations with examples, 5-mark descriptive questions (often with diagrams or derivations) and 4-mark competency-based questions. Expect direct questions on motion, rest, reference point and application-based questions drawn from NCERT activities. Students who follow the NCERT Exploration textbook thoroughly and practise this chapter's questions consistently score in the 90%+ range.
What are the key terms to remember for motion, distance and displacement in Class 9 Science?
The key terms to remember for motion, distance and displacement in NCERT Class 9 Science Chapter 4 are: motion, rest, reference point, frame of reference, scalar quantity, vector quantity. Each of these concepts carries exam weightage and regularly appears in the CBSE Class 9 paper. Write clear one-line definitions of every term in your revision notes and revisit them before the exam. Linking these terms visually through a flowchart or concept map makes recall easier during the Class 9 Science exam.
Is Motion, Distance and Displacement included in the Class 9 Science syllabus for 2025–26 CBSE?
Yes, Motion, Distance and Displacement is part of the NCERT Class 9 Science syllabus (2025–26) prescribed by CBSE under the new NCERT Exploration textbook. It falls under Chapter 4 — Describing Motion Around Us — and is examined in the annual paper. The current syllabus retains the full treatment of motion, rest, reference point as per the NCERT textbook. Because CBSE bases every Class 9 question on NCERT, studying this part thoroughly ensures complete syllabus coverage and guarantees marks from this chapter.
How should I prepare motion, distance and displacement for the CBSE Class 9 Science exam?
Prepare motion, distance and displacement for the CBSE Class 9 Science exam in three steps. First, read this NCERT part carefully, highlighting definitions and diagrams of motion, rest, reference point. Second, solve every in-text question and end-of-chapter exercise — CBSE questions often come directly from NCERT. Third, practise competency-based and assertion-reason questions to sharpen reasoning. Write answers in the exam-style format (point-wise with diagrams) and time yourself. This method delivers confidence and full marks in the exam.
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