How do I solve NCERT Class 9 Science Chapter 4 Motion exercise questions?

Read the question, list given quantities in SI units, choose the right kinematic equation (v=u+at, s=ut+½at², or v²=u²+2as), substitute values carefully and write the final answer with units and direction. Show every step — CBSE awards step marks.

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Describing Motion Around Us — NCERT Exercises

Q: How do I solve NCERT Class 9 Science Chapter 4 (Describing Motion Around Us) exercise questions for the CBSE board exam?

Solve NCERT Chapter 4 exercise questions by reading the question carefully, writing down the given data, recalling the relevant concepts like distance, displacement, speed, and applying them step by step. CBSE Class 9 examiners award step marks even if the final answer has a small slip.

Q: What types of questions from Describing Motion Around Us are asked in the Class 9 Science exam?

The Class 9 Science paper (CBSE pattern) asks a mix: 1-mark MCQ and assertion-reason, 2-mark short answers, 3-mark explanations, 5-mark long answers with diagrams or derivations, and 4-mark competency-based / case-study questions.

Q: How many marks does Chapter 4 u2014 Describing Motion Around Us u2014 typically carry in the Class 9 Science paper?

Chapter 4 typically contributes 5u20139 marks in the annual paper. Questions are drawn from definitions, reasoning, numerical/descriptive problems and diagrams on topics like distance, displacement, speed.

Q: What is the best way to revise Describing Motion Around Us for the Class 9 Science exam?

Three-pass approach. First pass: skim the chapter and note down key terms in a one-page mind map. Second pass: solve every NCERT intext and exercise question without looking at the solution, then self-check. Third pass: attempt sample papers under timed conditions.

🎓 Class 9 Science CBSE Theory Ch 4 — Describing Motion Around Us ⏱ ~20 min

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Chapter Summary — At a Glance

Reference point

Motion is described relative to a chosen origin. Without a reference, motion is meaningless.

Distance vs Displacement

Distance is total path (scalar). Displacement is shortest straight line from start to end (vector).

Speed

\(v = s/t\). Scalar. SI unit m/s. Average speed = total distance / total time.

Velocity

Displacement per unit time. Vector. \(v_{\text{avg}} = (u+v)/2\) for uniform acceleration.

Acceleration

\(a = (v-u)/t\). Vector. Unit m/s². May be positive, negative or zero.

Equations of motion

\(v=u+at\); \(s = ut + \tfrac{1}{2}at^2\); \(v^2 = u^2 + 2as\). Valid for uniform acceleration only.

Distance–time graph

Slope = speed. Straight line ⇒ uniform speed; curve ⇒ non-uniform.

Velocity–time graph

Slope = acceleration; area under curve = displacement.

Uniform circular motion

Constant speed, changing direction → accelerated motion. \(v = 2\pi r / T\).

Key Terms

Reference pointThe fixed origin against which motion is measured.

ScalarQuantity with magnitude only.

VectorQuantity with magnitude and direction.

DistanceTotal path length (scalar).

DisplacementShortest path from initial to final position (vector).

Uniform motionEqual distance in equal time intervals.

SpeedDistance per unit time (scalar).

VelocityDisplacement per unit time (vector).

AccelerationRate of change of velocity.

RetardationNegative acceleration; slowing down.

Uniform accelerationConstant rate of change of velocity.

Centripetal accelerationAcceleration directed to centre in circular motion.

NCERT Exercise — Step-by-Step Solutions

An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?

Total time = 2 min 20 s = 140 s. Number of rounds \(= 140/40 = 3.5\).

Circumference \(= \pi d = \pi \times 200 \approx 628\) m.

Distance: 3.5 × 628 = 2198 m ≈ 2200 m.

Displacement: after 3 full rounds, athlete is at start; after 0.5 round more, he is diametrically opposite. Hence displacement = diameter = 200 m.

Joseph jogs from one end A to the other end B of a straight 300 m road in 2 min 30 s and then turns around and jogs 100 m back to point C in another 1 min. What are Joseph's average speeds and velocities in jogging (a) from A to B, (b) from A to C?

(a) A → B: distance = displacement = 300 m; time = 150 s.
Avg speed = avg velocity magnitude = 300/150 = 2 m/s.

(b) A → C: total distance = 300 + 100 = 400 m; total time = 150 + 60 = 210 s.
Displacement (A to C) = 300 − 100 = 200 m (along AB).

Avg speed = 400/210 ≈ 1.90 m/s. Avg velocity = 200/210 ≈ 0.95 m/s (towards B).

Abdul, while driving to school, computes the average speed for his trip to be 20 km/h. On his return trip along the same route, there is less traffic and the average speed is 30 km/h. What is the average speed for Abdul's whole trip?

Let one-way distance = \(d\). Time forward \(t_1 = d/20\); back \(t_2 = d/30\).

Total distance = 2d; total time = \(d/20 + d/30 = (3d+2d)/60 = 5d/60 = d/12\).

Avg speed = \(2d \div d/12 = 24\) km/h. Answer = 24 km/h.

A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m/s² for 8.0 s. How far does the boat travel during this time?

\(u = 0\), \(a = 3\) m/s², \(t = 8\) s.

\(s = ut + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2}(3)(64) = \mathbf{96\,\text{m}}\).

A driver of a car travelling at 52 km/h applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km/h applies brakes slowly and stops in 10 s. On the same v–t graph, plot the velocity versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

Car 1: \(u_1 = 52 \text{ km/h} \approx 14.44\) m/s, stops in 5 s.
Distance = ½ × 14.44 × 5 ≈ 36.1 m.

Car 2: \(u_2 = 3 \text{ km/h} \approx 0.833\) m/s, stops in 10 s.
Distance = ½ × 0.833 × 10 ≈ 4.17 m.

The first car (52 km/h) travels much farther after braking.

Fig 4.X shows distance–time graphs of three objects A, B and C. Study the graph and answer the following questions: (a) Which of the three is travelling the fastest? (b) Are all three ever at the same point on the road? (c) How far has C travelled when B passes A? (d) How far has B travelled by the time it passes C?

(a) The line with the steepest slope is the fastest. Object B has the steepest slope, so B is fastest.

(b) Three objects are at the same point only if their three lines meet at one common point on the graph. From the graph, no such single intersection of all three exists, so they are never simultaneously at the same point.

(c) When B passes A (intersection of A and B lines), reading C's distance at that time from its line ≈ 5.7 km (approximate from standard graph).

(d) When B passes C (intersection of B and C lines), B has covered ≈ 5 km.

A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m/s², with what velocity will it strike the ground? After what time will it strike the ground?

\(u = 0\), \(a = 10\) m/s², \(s = 20\) m.

Final velocity: \(v^2 = u^2 + 2as = 0 + 2(10)(20) = 400 \Rightarrow v = \mathbf{20\,\text{m/s}}\).

Time: \(v = u + at \Rightarrow 20 = 0 + 10t \Rightarrow t = \mathbf{2\,\text{s}}\).

The speed–time graph for a car is shown. (a) Find how far the car travels in the first 4 seconds (shade the area). (b) Which part of the graph represents uniform motion?

(a) In the standard graph, the speed rises from 0 to about 6 m/s over 0–4 s along a curve. Approximating by counting unit squares (or treating as a rough triangle ½ × 4 × 6 = 12 m). The area gives ≈ 12 m (NCERT-style answer).

(b) The horizontal portion of the graph (from \(t \approx 6\) s onward, where speed remains constant) represents uniform motion.

State which of the following situations are possible and give an example for each: (a) An object with constant acceleration but with zero velocity. (b) An object moving in a certain direction with an acceleration in the perpendicular direction.

(a) Possible. At the topmost point of an object thrown vertically up, velocity is momentarily zero, but the acceleration due to gravity (9.8 m/s² downward) is still acting.

(b) Possible. In uniform circular motion, velocity is along the tangent and the centripetal acceleration is perpendicular to it (towards the centre).

Q10

An artificial satellite is moving in a circular orbit of radius 42,250 km. Calculate its speed if it takes 24 hours to revolve around the Earth.

\(r = 42250\) km \(= 4.225 \times 10^7\) m; \(T = 24 \times 3600 = 86400\) s.

\(v = \dfrac{2\pi r}{T} = \dfrac{2 \times 3.14 \times 4.225 \times 10^7}{86400} \approx \dfrac{2.654 \times 10^8}{86400} \approx 3071\) m/s.

\(v \approx \mathbf{3.07\,\text{km/s}}\) (about 11,058 km/h).

Q11 (Additional)

A bus accelerates from 36 km/h to 72 km/h in 10 s. Find (a) acceleration, (b) distance covered.

\(u = 10\) m/s, \(v = 20\) m/s, \(t = 10\) s.

(a) \(a = (20-10)/10 = 1\) m/s².

(b) \(s = ut + \tfrac{1}{2}at^2 = 10(10) + \tfrac{1}{2}(1)(100) = 100 + 50 = \mathbf{150\,\text{m}}\).

Q12 (Additional)

A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of −0.5 m/s². Find how far the train will go before it is brought to rest.

\(u = 90\) km/h \(= 25\) m/s, \(v = 0\), \(a = -0.5\) m/s².

\(v^2 = u^2 + 2as \Rightarrow 0 = 625 + 2(-0.5)s \Rightarrow s = 625\) m.

Distance = 625 m.

Q13 (Additional)

A trolley, while going down an inclined plane, has an acceleration of 2 cm/s². What will be its velocity 3 s after the start?

\(u = 0\), \(a = 2\) cm/s², \(t = 3\) s.

\(v = u + at = 0 + 2 \times 3 = \mathbf{6\,\text{cm/s}}\).

Q14 (Additional)

A racing car has a uniform acceleration of 4 m/s². What distance will it cover in 10 s after start?

\(s = ut + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2}(4)(100) = \mathbf{200\,\text{m}}\).

Q15 (Additional)

A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s², find the maximum height reached and the total time it remains in the air.

At max height \(v=0\). Use \(v^2 = u^2 - 2gh\): \(0 = 1600 - 20h \Rightarrow h = \mathbf{80\,\text{m}}\).

Time up: \(t_{up} = u/g = 40/10 = 4\) s. Total time of flight \(= 2 t_{up} = \mathbf{8\,\text{s}}\).

Exam Tip: Always (i) write down "Given" in SI units, (ii) pick the right kinematic equation, (iii) solve algebraically before substituting numbers, and (iv) state the unit clearly with the final answer.

Frequently Asked Questions — NCERT Exercises & Intext Questions

How do I solve NCERT Class 9 Science Chapter 4 (Describing Motion Around Us) exercise questions for the CBSE board exam?

Solve NCERT Chapter 4 — Describing Motion Around Us — exercise questions by first reading the question carefully, writing down the given data, recalling the relevant concepts like distance, displacement, speed, and applying them step by step. This Part 4 covers every intext and end-of-chapter exercise from the NCERT textbook. Write balanced equations, label diagrams clearly and show each step — CBSE Class 9 examiners award step marks even if the final answer has a small slip. Practising these solutions strengthens conceptual clarity and builds speed for both the school exam and the upcoming Class 10 board exam.

Are the NCERT intext questions from Describing Motion Around Us important for the Class 9 Science exam?

Yes, NCERT intext questions for Chapter 4 Describing Motion Around Us are highly important for the CBSE Class 9 Science exam. Many questions in school and competitive papers are directly lifted or only slightly modified from these intext questions, and they test the foundational concepts — distance, displacement, speed — that chapter-end questions and the Class 10 board build on. Attempt every intext question first, then move on to the exercises. This practice ensures complete NCERT coverage, which is the CBSE syllabus's primary source.

What types of questions from Describing Motion Around Us are asked in the Class 9 Science exam?

The Class 9 Science paper (CBSE pattern) asks a mix of question types from Describing Motion Around Us: 1-mark MCQ and assertion-reason, 2-mark short answers, 3-mark explanations, 5-mark long answers with diagrams or derivations, and 4-mark competency-based / case-study questions. These test understanding of distance, displacement, speed, velocity. Practising every NCERT exercise and intext question prepares you to answer all of these formats with confidence.

How many marks does Chapter 4 — Describing Motion Around Us — typically carry in the Class 9 Science paper?

Chapter 4 — Describing Motion Around Us — is part of the CBSE Class 9 Science syllabus and typically contributes 5–9 marks in the annual paper, depending on the year's weightage. Questions are drawn from definitions, reasoning, numerical/descriptive problems and diagrams on topics like distance, displacement, speed. Solving the NCERT exercises in this part is essential because CBSE directly references the NCERT Exploration textbook for question design.

Where can I find step-by-step NCERT solutions for Chapter 4 Describing Motion Around Us Class 9 Science?

You can find complete, step-by-step NCERT solutions for Chapter 4 Describing Motion Around Us Class 9 Science on MyAiSchool. Every intext and end-of-chapter exercise question is solved with full working, labelled diagrams and CBSE-aligned mark distribution. Solutions highlight key points about distance, displacement, speed that examiners look for. This makes revision quick and exam-focused for Class 9 CBSE students.

What is the best way to revise Describing Motion Around Us for the Class 9 Science exam?

The best way to revise Describing Motion Around Us for the CBSE Class 9 Science exam is a three-pass approach. First pass: skim the chapter and note down key terms like distance, displacement, speed in a one-page mind map. Second pass: solve every NCERT intext and exercise question without looking at the solution, then self-check. Third pass: attempt sample papers and competency-based questions under timed conditions. This structured revision secures full marks for this chapter.

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