Where Mindful Learning Begins
Where Mindful Learning Begins
Complete NCERT Solutions for Chapter 4 of the new Class 9 Science Exploration textbook (CBSE 2026-27) — every Think It Over, Activity, Pause & Ponder, Worked Example, Revise Reflect Refine, and Journey Beyond question on this one page, with full step-by-step working for every numerical.
This is where Class 9 Physics really begins — learning to describe motion precisely before Chapter 6 asks what causes it. You'll work through distance vs. displacement, speed vs. velocity, uniform and non-uniform acceleration, the three equations of motion, and how to read position-time and velocity-time graphs. These graph-reading skills and the kinematic equations appear again and again in board exams, so getting comfortable with them here makes every later numerical easier.
Describing Motion Around Us is Class 9's first proper physics chapter, and it's built almost entirely around numericals and graphs, unlike the biology chapters before it. It covers distance vs displacement, average speed vs average velocity, average acceleration, how to read and plot position-time and velocity-time graphs, the three kinematic equations for constant acceleration, and uniform circular motion. Every question — from the Think It Over prompts to the trickiest Revise Reflect Refine numerical — is solved here, with full step-by-step working, exactly as the textbook presents them.
Why displacement can be smaller than distance travelled, and how average speed differs from average velocity.
Reading position-time and velocity-time graphs — what the slope means, and what the area under the graph gives you.
v = u + at, s = ut + ½at², and v² = u² + 2as — used throughout for braking distance, free fall, and multi-phase journeys.
Distance vs. displacement
| Property | Distance | Displacement |
|---|---|---|
| Definition | Total path length covered | Shortest straight-line distance between initial and final position |
| Direction | Not required (scalar) | Required (vector) |
| Value | Always positive, never decreases | Can be positive, negative, or zero |
| Comparison | Distance ≥ |Displacement| always | Equal to distance only for straight-line motion in one direction |
Key formulae
Equations of motion (uniform acceleration)
where \(u\) = initial velocity, \(v\) = final velocity, \(a\) = acceleration, \(t\) = time, \(s\) = displacement.
Reading motion graphs
| Graph | Slope represents | Area under graph represents |
|---|---|---|
| Position–time graph | Velocity | — |
| Velocity–time graph | Acceleration | Distance/displacement covered |
Answer: The safe following distance depends on the stopping distance of the vehicle behind it. When a vehicle's brakes are applied, it travels some distance before stopping. This stopping distance depends on:
As derived later in the chapter (using \(v^2 = u^2 + 2as\)), for a vehicle decelerating from speed \(u\) to rest (\(v=0\)):
This means stopping distance is proportional to the square of the initial speed. A safe following distance should be at least equal to the stopping distance of your own vehicle, plus an extra margin for reaction time.
The Highway Code (and Indian traffic safety guidelines) recommend a 2-second rule at moderate speeds — always maintain a gap of at least 2 seconds' worth of travel. At higher speeds, this should increase to 4 seconds or more.
Answer: Yes, the safe following distance depends strongly on speed. Since stopping distance \(s = u^2/(2|a|)\), if speed doubles, the stopping distance quadruples. For example:
So doubling the speed increases the stopping distance four times — this is why speed limits are strictly enforced: the faster you drive, the much greater the distance required to stop safely.
| S. No. | Position | Total distance travelled from O | Displacement from O |
|---|---|---|---|
| 1 | O | 0 cm | 0 cm |
| 2 | A (40 cm) | 40 cm | 40 cm upward |
| 3 | B (140 cm) | 140 cm | 140 cm upward |
| 4 | C (80 cm, on way down) | 140 + 60 = 200 cm | 80 cm upward |
| 5 | O (returned to start) | 140 + 140 = 280 cm | 0 cm (no net displacement) |
Answer: The correct statement is its magnitude is less than or equal to the total distance travelled.
Displacement is the net change in position (a straight-line distance from start to current position); distance is the total path length. When the ball moves in one direction only, they are equal (e.g., at position B, both = 140 cm). When the ball reverses direction and returns partway or fully, the total distance keeps increasing but the magnitude of displacement decreases — when the ball returns to O, distance = 280 cm but displacement = 0 cm. The magnitude of displacement can never exceed total distance — only equal it (no reversal) or fall below it (reversal of direction).
Displacement isn't always zero for intermediate positions unless the ball returns to the origin — so it can never exceed total distance, and it's only equal to it when the ball moves in one direction without turning.
Answer: Conversion: 100 km/h = 100 × 1000/3600 m/s ≈ 27.78 m/s. Average acceleration = 27.78 ÷ time (in seconds).
| Car type | Time (0 → 100 km/h) | Avg. acceleration (m/s²) |
|---|---|---|
| Maruti Suzuki Swift (budget hatchback) | ~13.7 s | 27.78/13.7 ≈ 2.03 |
| Honda City (mid-size sedan) | ~10.5 s | 27.78/10.5 ≈ 2.65 |
| Toyota Fortuner (SUV) | ~9.8 s | 27.78/9.8 ≈ 2.84 |
| Hyundai Creta (compact SUV) | ~11.5 s | 27.78/11.5 ≈ 2.42 |
| Porsche 911 Turbo S (sports car) | ~2.7 s | 27.78/2.7 ≈ 10.29 |
Formula used: \(a = (v-u)/t = (27.78 - 0)/t\). Sports cars have dramatically higher accelerations because they have more powerful engines relative to their mass. Look up current models online and substitute into the formula.
Data (Table 4.3): Time: 0, 1, 2, 3, 4, 5, 6 s | Position: 0, 20, 40, 60, 80, 100, 120 m.
Steps for plottingInterpretation: since the graph is a straight line, the vehicle moves with constant velocity = slope = (120 − 0)/(6 − 0) = 20 m/s.
Answer: From the graph, taking A at (2 s, 40 m) and B at (4 s, 80 m):
The slope BC/CA (change in position ÷ change in time) equals the average velocity. Since the graph is a straight line, the velocity is constant and equal to 20 m/s at all points.
Prediction: when the ring is lifted, the marble will no longer follow the circular path — it will continue moving in a straight line, in the direction it was moving at the instant the ring was removed.
Observation: after lifting the ring, the marble does move in a straight line (tangent to the circle at the point of release), confirming the prediction.
Explanation: the ring was providing a continuous inward (centripetal) force on the marble, forcing it to change direction continuously and follow the circular path. Once the ring is removed, there's no force to change the direction of the marble's velocity, so it continues in the direction it was moving at that instant — a straight line. This demonstrates Newton's First Law of Motion (studied later): an object continues in uniform motion in a straight line unless acted upon by an external force.
The direction of motion at the instant of release is along the tangent to the circle at that point — this is why velocity in uniform circular motion is always directed tangentially.
Answer: the athlete's displacement will be zero when they return to the starting point O. In Fig. 4.4, the athlete starts at O (0 m), reaches A (100 m), then runs back — displacement becomes zero when the athlete reaches O again. By that time:
So displacement = 0 m, but total distance = 200 m.
Answer: fuel used depends on total distance travelled. Fuel is consumed by the engine for every metre the vehicle moves, regardless of direction or whether it returns to its starting point. Displacement only measures the net change in position — it ignores the path taken. For example, if a car drives 5 km to a market and 5 km back home, displacement = 0 km, but it consumed fuel for 10 km of driving. The engine burns fuel as long as the vehicle is moving, so total distance — not displacement — determines fuel consumption.
Answer: yes, it is straight-line (linear) motion. Although the track is inclined, the ball moves along a single straight path — it never changes direction.
Can it be depicted on a horizontal number line? Yes. We can represent the inclined track as a straight-line number line (as in Fig. 4.3) by measuring distances along the incline from O — the inclination of the actual path is irrelevant for position-time analysis, since what matters is that the motion is one-dimensional.
Are total distance and displacement equal at A, B, C, D? Yes, at every position, total distance and magnitude of displacement from O are equal, because the ball moves in only one direction (down the incline) and never reverses. At A: both = 10 cm; at B: both = 20 cm; at C: both = 30 cm; at D: both = 40 cm.
Given: Leg 1: 200 km north in 3 h. Leg 2: 200 km south in 2 h.
Total distance = 200 + 200 = 400 km. Total time = 3 + 2 = 5 h. Displacement = 200 km north − 200 km south = 0 km (start and end at the same point).
Conclusion: although the car was moving the whole time, average velocity is zero because the car returned to its starting point, while average speed is 80 km/h.
This happens when the object moves in one direction only, with no reversal of motion. In this case, total distance travelled equals the magnitude of displacement, so \(|\text{average velocity}| = \text{displacement}/\text{time} = \text{distance}/\text{time} = \text{average speed}\). Example: a car travelling from city A to city B in a straight line — the distance covered equals the displacement.
(ii) Average velocity zero, average speed not zeroThis happens when the net displacement is zero — the object returns to its starting point — but the object has still covered some total distance, so average speed is non-zero. Example: Sarang swimming from one end of the pool to the other and back — displacement = 0, but total distance = 50 m, so average speed = 1 m/s while average velocity = 0 m/s.
Answer: each day, the combined distance covered by both = 9 + 5 = 14 yojanas. They need to cover 210 yojanas together.
Check: in 15 days, the first postman covers 9 × 15 = 135 yojanas and the second covers 5 × 15 = 75 yojanas — total = 135 + 75 = 210 yojanas. ✓
Given: pool length = 25 m; total time = 50 s. Total distance = 25 + 25 = 50 m (one length and back). Displacement = 0 m (returns to start).
\(u = 36\text{ km/h} = 10\text{ m/s}\); \(v = 54\text{ km/h} = 15\text{ m/s}\); \(t = 10\text{ s}\).
\(u = 54\text{ km/h} = 15\text{ m/s}\); \(v = 0\text{ m/s}\); \(t = 5\text{ s}\).
The magnitude of braking deceleration (3 m/s²) is 6 times larger than the acceleration (0.5 m/s²) — brakes typically produce much larger forces than the engine does under normal conditions.
Conclusion: the average acceleration is constant at 9.8 m/s² in every interval. This is the acceleration due to gravity (\(g\)), directed downward (in the direction of motion for a falling object) — one of the most important constants in physics.
Answer: using a scale of x-axis: 5 divisions = 2 s, y-axis: 5 divisions = 5 m, plot the points (0,0), (2,1), (4,4), (6,9), (8,16), (10,25), (12,36) and join them with a smooth curve.
The graph is a curve (parabola), not a straight line, indicating that the vehicle is accelerating — velocity is not constant. The curve bends upward, showing that equal time intervals correspond to increasing distances (increasing speed), so the slope (velocity) is increasing with time.
Answer: the vehicle is stationary (at rest) at a position of 40 m from the origin. Its position does not change with time, so its velocity is 0. A horizontal line (parallel to the time axis) on a position-time graph always represents an object at rest.
Answer: object B has the higher average velocity magnitude. The slope of a position-time graph represents velocity, and B's graph has a steeper slope — for the same time interval, B covers a greater displacement than A. A steeper slope means a larger velocity.
Given: \(a = -4\text{ m/s}^2\), \(v = 0\). Using \(v^2 = u^2 + 2as \Rightarrow s = u^2/(2|a|) = u^2/8\).
(i) \(u = 54\text{ km/h} = 15\text{ m/s} \Rightarrow s = 15^2/8 = 225/8 = 28.125 \approx 28.1\text{ m}\)
(ii) \(u = 108\text{ km/h} = 30\text{ m/s} \Rightarrow s = 30^2/8 = 900/8 = 112.5\text{ m}\)
Doubling the speed quadruples the stopping distance — a critical safety insight, and the reason highway speed limits are strictly enforced and safe following distances matter so much at high speeds.
Journey: Home → Shop → Home → Shop → Home = 250 + 250 + 250 + 250 = 1000 m total distance.
Displacement: he starts and ends at home, so displacement = 0 m.
(i) Total vertical distance: ground to 4th floor = 4 × 3 = 12 m (up). 4th floor to 2nd floor = 2 × 3 = 6 m (down). Total = 12 + 6 = 18 m.
(ii) Displacement: net change in position, from ground to 2nd floor = 2 × 3 = 6 m upward.
Answer: Yes, it's possible even with a constant speedometer reading. The speedometer measures speed (the magnitude of velocity). Acceleration involves a change in velocity, which includes a change in magnitude or direction (or both). If the girl is moving along a curved path (e.g., taking a turn) at constant speed, the direction of velocity is continuously changing — a change in direction means the velocity vector is changing, which means there's acceleration, even though the speed (magnitude) stays constant.
This is exactly what happens in uniform circular motion — constant speed but continuously changing direction, giving non-zero acceleration.
Given: \(u=0\), \(v=24\text{ m/s}\), \(t=6\text{ s}\).
Verification with \(v^2=u^2+2as\): \(24^2 = 0 + 2\times4\times s \Rightarrow 576=8s \Rightarrow s=72\text{ m}\) ✓
Given: \(u=28\text{ m/s}\), \(v=0\), \(s=98\text{ m}\).
The negative sign indicates deceleration (braking); magnitude = 4 m/s².
Answer: velocity is represented by the slope of a position-time graph. If A and B are both straight lines (constant velocities), they only have equal velocity if their slopes (lines) are parallel. Looking at Fig. 4.27, A's line is steeper than B's throughout, so their slopes are never equal — A and B do not have equal velocity at any point.
For curved position-time graphs (non-uniform motion), equal velocity occurs when the tangent slopes at corresponding points are equal.
Answer: (i) and (iv) are correct.
Average velocity depends only on the initial and final positions; average speed depends on the total path length.
Given: \(u=54\text{ km/h}=15\text{ m/s}\); \(v=36\text{ km/h}=10\text{ m/s}\); \(t=36\text{ s}\).
Verification — first find \(a = (10-15)/36 = -5/36\text{ m/s}^2\), then \(s = ut + \tfrac12 at^2 = 540 - 90 = 450\text{ m}\) ✓
\(u=0\), \(v=20\text{ m/s}\), \(t=5\text{ s}\) → \(s_1 = \frac{0+20}{2}\times5 = 50\text{ m}\)
Phase 2 (constant velocity)\(v=20\text{ m/s}\), \(t=10\text{ s}\) → \(s_2 = 20\times10 = 200\text{ m}\)
Phase 3 (braking)\(u=20\text{ m/s}\), \(v=0\), \(t=6\text{ s}\) → \(s_3 = \frac{20+0}{2}\times6 = 60\text{ m}\)
Given: \(u=36\text{ km/h}=10\text{ m/s}\); reaction time = 0.5 s; deceleration after braking = 2.5 m/s².
Distance during reaction time: \(d_1 = u\times t_{reaction} = 10\times0.5 = 5\text{ m}\)
Distance during braking (\(v=0\), \(u=10\text{ m/s}\)): \(d_2 = u^2/(2a) = 100/5 = 20\text{ m}\)
Since 25 m < 30 m (distance to the obstacle), the bus will stop before reaching it — with 5 m to spare.
Answer: the concept of rest and motion is always relative to a reference point (frame of reference).
So yes — the object can be considered at rest (relative to Earth) and in motion (relative to the Sun) at the same time. There is no absolute rest or absolute motion — it depends on the chosen reference point, which is the principle of relativity of motion.
Reading the graph: 0–40 s velocity increases from 0 to 3 m/s (acceleration phase); 40–80 s velocity is constant at 3 m/s (shade this area); 80–120 s velocity decreases from 3 m/s to 0 (shade in a second colour).
Displacement during acceleration phase (0–40 s): \(s = \tfrac12\times3\times40 = 60\text{ m}\)
Displacement during constant velocity (40–80 s): \(s = v\times t = 3\times40 = 120\text{ m}\)
Displacement during decreasing velocity (80–120 s): \(s = \frac{3+0}{2}\times40 = 60\text{ m}\)
Average acceleration over 120 s: \(a = (v_{final}-v_{initial})/t = (0-0)/120 = 0\text{ m/s}^2\)
Average acceleration over the full 120 s is zero because the initial and final velocities are both zero — but the cyclist was accelerating and decelerating within that interval.
Answer: the area under a velocity-time graph gives displacement (= distance, for one-direction motion). Reading Fig. 4.31 approximately:
The actual values depend on careful reading of the graph — use graph paper to estimate areas as accurately as possible.
6 m/s for 2 min = 120 s → rectangle on the graph. \(s_1 = 6\times120 = 720\text{ m}\)
Phase 2 — acceleration\(u=6\text{ m/s}\), \(a=1\text{ m/s}^2\), \(t=6\text{ s}\) → \(v = u+at = 6+6 = 12\text{ m/s}\), so \(s_2 = \frac{6+12}{2}\times6 = 54\text{ m}\)
Graph description: plot velocity (y-axis) vs time (x-axis) — a horizontal line at 6 m/s from 0 to 120 s (rectangle, area 720 m), then a line rising from 6 to 12 m/s from 120 to 126 s (trapezium, area 54 m).
Car A: \(a_A = 5/5 = 1\text{ m/s}^2\). Velocities: 0 (0 s), 1 (1 s), 2 (2 s), 3 (3 s), 4 (4 s), 5 (5 s) — a straight line from the origin.
Car B: \(a_B = 3/10 = 0.3\text{ m/s}^2\). Velocities: 0 (0 s), 0.3 (1 s), … 3 (10 s) — a shallower straight line.
Displacement of Car A in 5 s: \(s_A = \tfrac12\times5\times5 = 12.5\text{ m}\)
Displacement of Car B in 10 s: \(s_B = \tfrac12\times10\times3 = 15\text{ m}\)
On the same graph, Car A's line is steeper (greater acceleration) and Car B's is less steep — both start at the origin, and the area under each line gives its displacement.
Time interval: 6:00 PM to 7:30 PM = 1.5 hours = 90 minutes → 1.5 complete revolutions.
(i) Distance: each revolution = circumference = \(2\pi r = 2\pi\times7 = 44\text{ cm}\). Total distance = 1.5 × 44 ≈ 66 cm.
(ii) Displacement: after 1 complete revolution, the tip returns to start (displacement = 0). After the extra 0.5 revolution (half circle), the tip is diametrically opposite. Displacement = diameter = 2r = 14 cm (from the 12 o'clock position to the 6 o'clock position).
(iii) Speed: total time = 90 min = 5400 s. Speed = 66 cm ÷ 5400 s ≈ 1.22 × 10⁻² cm/s.
(iv) Velocity: magnitude = 14 cm ÷ 5400 s ≈ 2.59 × 10⁻³ cm/s, directed from the 12 o'clock to the 6 o'clock position (straight downward on the clock face).
Answer: No, the speeds are different. Even though both sets complete one revolution in the same time (same angular speed/time period \(T\)), they're at different distances from the centre.
Numbers at \(r=7\text{ cm}\): \(v_{numbers} = 14\pi/T\) cm per revolution. Letters at \(r=4\text{ cm}\): \(v_{letters} = 8\pi/T\) cm per revolution.
The numbers move faster (nearly 1.75× faster than the letters) because they're farther from the centre. When an object moves fast, our eyes cannot resolve it clearly — it appears as a blur. The numbers blur and seem to disappear, while the slower-moving letters (closer to the centre) remain visible for longer.
This is why the outer edge of a spinning wheel or fan blade appears to vanish while the hub remains visible — the same principle applies to helicopter rotors and turbine blades.
(i) Phone on outstretched palm: the readings fluctuate, showing small non-zero values. This is because even a steady hand has involuntary micro-tremors — tiny, unconscious movements that the accelerometer can detect.
(ii) Phone on the floor: the readings are very close to zero (nearly flat). The floor is rigid and stationary, so there's no acceleration — the tiny residual readings come from electronic noise in the sensor.
What this shows: even when we think we're perfectly still, our body is constantly making tiny adjustments — motion and acceleration are present even in "stillness." This is studied in medical research for conditions like Parkinson's disease and essential tremor, where involuntary movements differ from healthy micro-tremors.
From \(v=u+at \Rightarrow u = v-at\). Substituting into \(s=ut+\tfrac12at^2\):
From \(at = v-u \Rightarrow a = (v-u)/t\). Substituting into \(s=ut+\tfrac12at^2\):
Trapezium method: the velocity-time graph for uniformly accelerating motion forms a trapezium with parallel sides \(u\) and \(v\) and height (width) \(t\). Area of a trapezium = ½ × (sum of parallel sides) × height = \(\tfrac12(u+v)t\) = displacement. ✓
Answer: the shape of the curve (parabolic) stays the same regardless of scale, but the visual appearance changes:
Same Table 4.4 data, same underlying parabola — but the two scale choices create very different visual impressions of how "fast" the vehicle appears to accelerate.
As a rule of thumb, a good scale is chosen by first finding the full range of values on each axis, then dividing that range into a convenient number of equal intervals (round numbers like 1, 2, 5, 10 rather than awkward fractions) so the plotted graph fills most of the available space without any part of the curve being cramped or cut off.
Create a poster comparing stopping distances at 40, 60, 80, 100 km/h on dry vs wet roads, using a visual scale such as number of car lengths.
Motion can be described precisely with just five linked quantities — displacement, time, initial velocity, final velocity, and acceleration — and the three kinematic equations that connect them let you predict exactly where an object will be and how fast it will be moving, whether it's a braking car, a falling ball, or a child on a merry-go-round.
Here are extra practice questions for Class 9 Chapter 4, Describing Motion Around Us. These are numerical problems on uniform acceleration — apart from the NCERT questions — to make you confident in your understanding of the chapter. Attempt the questions yourself first, and then cross-check your answers below.
Given: \(u = 0\), \(a = 0.1\text{ m/s}^2\), \(t = 2\text{ min} = 120\text{ s}\).
(a) Speed acquired:
(b) Distance travelled:
The bus reaches a speed of 12 m/s and covers a distance of 720 m.
Given: \(u = 90\text{ km/h} = 25\text{ m/s}\), \(v = 0\) (brought to rest), \(a = -0.5\text{ m/s}^2\).
Using \(v^2 = u^2 + 2as\):
The train will travel 625 m before it comes to rest.
Given: Starting "from the start" means the car begins from rest, so \(u = 0\), \(a = 4\text{ m/s}^2\), \(t = 10\text{ s}\).
Using \(s = ut + \tfrac{1}{2}at^2\):
The racing car covers a distance of 200 m in the first 10 seconds.
Given: \(u = 5\text{ m/s}\) (upward), \(a = -10\text{ m/s}^2\) (deceleration due to gravity acting downward, opposing the upward motion). At the highest point, the stone's velocity becomes zero: \(v = 0\).
Height attained, using \(v^2 = u^2 + 2as\):
Time taken, using \(v = u + at\):
The stone reaches a maximum height of 1.25 m, taking 0.5 s to get there.
Pair these solutions with our free Class 9 Science Notes PDF — quick, chapter-wise revision notes covering every unit, perfect for last-minute recall before a test.
📘 Get Class 9 Science Notes →Next up is Chapter 5 — Exploring Mixtures and their Separation. Or explore the full chapter list, browse the Class 9 Science hub, or book a free demo class for personalised coaching.
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