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Class 9 Science Chapter 8 of 15

Chapter 8 — Motion

Overview

Chapter 8 — Motion illustration

Chapter: Motion (Class 9 Science, NCERT) — Introduction, importance, key themes, and learning outcomes. Introduction: Motion deals with change in position of an object with time. The chapter introduces basic concepts needed to describe and quantify motion along a straight line: reference frame, path, distance and displacement, speed and velocity, acceleration, and graphical representation of motion. Importance: Understanding motion is foundational for classical mechanics and many real-life applications (transport, sports, engineering). It develops skill in quantitative reasoning, interpreting graphs, and using formulae to solve problems. Key themes: (1) Distinction between scalar and vector quantities (distance vs displacement; speed vs velocity). (2) Types of motion — uniform and non-uniform, and motion with constant acceleration (including uniformly accelerated motion). (3) Definitions and calculations of average and instantaneous speed/velocity, and acceleration. (4) Equations of motion for constant acceleration: v = u + at, s = ut + 1/2 at^2, v^2 = u^2 + 2as — their meaning and application. (5) Graphical analysis: distance–time, displacement–time, velocity–time graphs; slope…

Learning Objectives

  • Define distance, displacement, speed, velocity and acceleration with their SI units and distinguish between scalar and vector quantities.
  • State the three equations of motion for uniformly accelerated motion in one dimension and list the conditions for their applicability.
  • Derive v = u + at, s = ut + 1/2·at^2 and v^2 - u^2 = 2as from basic definitions of acceleration and average velocity.
  • Calculate average speed and average velocity for given motion data and interpret the physical meaning of each in exam-style questions.
  • Apply the equations of motion to solve numerical problems involving uniform acceleration, including initial and final velocities, displacement and time.
  • Solve problems on relative velocity in one dimension for objects moving in the same and opposite directions.
  • Sketch and interpret distance–time and velocity–time graphs for uniform and uniformly accelerated motion and extract quantities (speed, velocity, acceleration) from them.
  • Interpret the slope of a distance–time graph and the area under a velocity–time graph, and use these graphical relations to solve problems.

Topics in this chapter

11 topics · tap a topic title to jump straight to it.

🏃1

Introduction to Motion

Motion is the change in position of an object with respect to a reference point over time. An object is said to be in motion if its position changes relative to a chosen frame of reference.

Reference frame: Motion is always described relative to some reference point or coordinate system (for example, ground, a car, or a lab bench).

Key terms:

  • Distance — total path length covered by an object (scalar, always positive).
  • Displacement — straight-line change in position from initial to final point (vector, can be positive, negative, or zero).
  • Speed — how fast an object moves; scalar. Instantaneous speed is the speed at a particular moment; average speed = total distance / total time.
  • Velocity — rate of change of displacement; a vector (has magnitude and direction). Average velocity = total displacement / total time.
  • Acceleration — rate of change of velocity (vector): speeding up, slowing down, or changing direction.
  • Rest — when an object’s position does not change with time relative to the chosen frame.

Types of motion (brief): Translatory (linear), rotational (about an axis), and oscillatory (back-and-forth, e.g., pendulum). In Class 9 we focus mainly on linear motion and the basic quantitative descriptions.

Uniform vs Non-uniform motion: Uniform motion: object covers equal distances in equal intervals of time (constant speed). Non-uniform motion: unequal distances in equal time intervals (speed varies).

Important conceptual points:

  • Displacement ≤ distance in magnitude. Distance measures path length; displacement is a straight-line vector from start to end.
  • Slope of a distance-time (or displacement-time) graph gives speed (or velocity). Slope of a velocity-time graph gives acceleration.
  • Area under a velocity-time graph gives displacement (for velocity as a function of time).

This introduction sets the stage for calculating and graphing motion (distance/displacement vs time, velocity vs time) and for studying uniform and non-uniform motion quantitatively.

📌 Examples
  • A car cruising at constant 60 km/h on a straight highway — approximate uniform motion.
  • A car in city traffic slowing and accelerating at signals — non-uniform motion.
  • A runner completing a 400 m track lap: distance = 400 m, displacement = 0 (start and end at same point).
  • A pendulum swinging to-and-fro — example of oscillatory motion (direction and speed change continuously).
  • Earth orbiting the Sun — nearly uniform circular motion (direction changes continuously; velocity direction changes).
  • A stationary book on a table — example of rest relative to the table.
🧮 Formulas
  1. Speed = Distance / Time (s = d / t) — scalar, SI unit m s^-1
  2. Average speed = Total distance / Total time
  3. Velocity = Displacement / Time (v = Δx / Δt) — vector, SI unit m s^-1
  4. Average velocity = Total displacement / Total time
  5. Acceleration = Change in velocity / Time (a = Δv / Δt) — vector, SI unit m s^-2
  6. Conversion: 1 km h^-1 = (5/18) m s^-1 (v m/s = v km/h × 5/18)
📊 Visual ideas
Distance‑time graph (uniform motion): plot distance on y-axis, time on x-axis. A straight line through origin with constant slope; slope = speed.
Distance‑time graph (non‑uniform motion): curved line; varying slope shows changing speed. Horizontal line indicates rest (zero speed).
Displacement‑time graph including negative region: useful when motion crosses the chosen origin. Straight line slope = constant velocity; slope sign shows direction.
Velocity‑time graph (constant velocity): horizontal line; area under curve between times t1 and t2 = displacement during that interval.
🔬2

Scalars and Vectors

Introduction
In physics, physical quantities are classified into two types: scalars and vectors. This classification depends on whether the quantity needs only magnitude or both magnitude and direction to be completely described.

Scalar
A scalar is a physical quantity that has only magnitude (size) and no direction. Examples: mass, temperature, time, speed, energy. Scalars add and subtract like ordinary numbers; their sign (if any) gives only magnitude sense, not direction.

Vector
A vector is a physical quantity that has both magnitude and direction. Examples: displacement, velocity, acceleration, force. Vectors are represented graphically by an arrow: length of the arrow = magnitude, arrowhead = direction.

Representation and Notation
Algebraically vectors are often written in bold or with an arrow above, e.g. A or \u2192A. On a plane a vector can be resolved into components: A = A_x i + A_y j, where i and j are unit vectors along x and y axes.

Magnitude and Components
If a vector A makes an angle θ with the x-axis, its components are A_x = A cosθ and A_y = A sinθ. The magnitude is |A| = sqrt(A_x^2 + A_y^2).

Types and Properties of Vectors

  • Zero (null) vector: magnitude zero, no specific direction.
  • Unit vector: magnitude one (e.g., i, j, k).
  • Equal vectors: same magnitude and direction.
  • Opposite vectors: same magnitude but opposite direction.
  • Collinear vectors: lie on the same line (may point same or opposite).

Addition and Subtraction of Vectors
Vectors are added using geometrical rules: triangle (tip-to-tail) rule, parallelogram rule, or polygon rule for many vectors. To subtract B from A, add A and the negative of B (A - B = A + (-B)).

Key formula for resultant of two vectors
If two vectors A and B have an angle θ between them, the magnitude R of their resultant (parallelogram rule) is:
R = sqrt(A^2 + B^2 + 2AB cosθ)

Scalar multiplication
Multiplying a vector A by a scalar k changes its magnitude: |kA| = |k| |A|. If k is negative, the direction reverses.

Scalar from vectors (dot product)
Two vectors A and B can produce a scalar via dot product: A · B = |A||B| cosθ (often introduced later, but useful to connect vectors and scalars).

Why the distinction matters
In motion, for example, distance (scalar) and displacement (vector) are different: distance is total path length (no direction), displacement is shortest straight-line from initial to final position (has direction). Speed (scalar) and velocity (vector) are different in the same way; acceleration is a vector while time and temperature are scalars.

Summary
Scalars: magnitude only. Vectors: magnitude + direction; represented by arrows; added by tip-to-tail or parallelogram method; components found by trigonometry.

📌 Examples
  • Distance (scalar) vs Displacement (vector): walking 3 km east then 4 km west gives distance = 7 km but displacement = 1 km east.
  • Speed (scalar) vs Velocity (vector): a car going around a circular track at constant speed has changing velocity because its direction changes.
  • Mass (scalar) vs Weight (vector): mass is amount of matter (scalar), weight (force) acts downward and is a vector.
  • Time and Temperature: both are scalars — described by magnitude only.
  • Force (vector) vs Work (scalar): force has direction; work is scalar (force · displacement).
  • Zero vector example: a ball held stationary has zero displacement (vector) though time (scalar) passes.
🧮 Formulas
  1. |A| = sqrt(A_x^2 + A_y^2)
  2. A_x = |A| cosθ, A_y = |A| sinθ
  3. Resultant of two vectors: R = sqrt(A^2 + B^2 + 2AB cosθ)
  4. Vector addition (symbolic): R = A + B
  5. Vector subtraction: A - B = A + (-B)
  6. |kA| = |k| |A| (scalar multiplication)
📊 Visual ideas
Arrow diagram comparing a scalar (single number label, e.g., '5 km') and a vector (arrow labeled with magnitude and direction). Use different colors.
Tip-to-tail (triangle) and parallelogram diagrams showing vector addition A + B and the resultant R. Label angles and magnitudes; show formula R^2 = A^2 + B^2 + 2AB cosθ.
Component diagram: a vector A on the x–y plane with horizontal (A_x) and vertical (A_y) projections; show right triangle and the relations A_x = A cosθ, A_y = A sinθ.
Opposite vectors: two equal arrows in opposite directions showing resultant zero when added.
🔬3

Distance and Displacement

Distance and Displacement are two fundamental quantities used to describe motion.

Distance is the total length of the path travelled by an object, irrespective of the direction. It is a scalar quantity (has magnitude only). Its SI unit is metre (m).

Displacement is the shortest straight-line distance from the initial position to the final position of the object, taken along with direction. It is a vector quantity (has magnitude and direction). Displacement is usually denoted by Δx (in one dimension) or a vector →d.

Key points:

  • Distance ≥ |Displacement|. Equality holds only if motion is along a straight line in a single direction.
  • Displacement can be zero even when distance is non-zero (for example, when an object returns to its starting point).
  • For motion along a line with successive segments, distance = sum of magnitudes of each segment, while displacement = algebraic sum of signed segments.

Example in symbols for one-dimensional motion: if an object moves from xi to xf, displacement Δx = xf − xi. If it moves in segments, distance = Σ|Δxi| and displacement = ΣΔxi.

📌 Examples
  • Walking 3 m east then 2 m east: distance = 3 + 2 = 5 m; displacement = +5 m (east).
  • Walking 3 m east then 5 m west: distance = 3 + 5 = 8 m; displacement = 5 − 3 = 2 m west (or −2 m if east is positive).
  • A runner completes one full lap of a circular track (circumference 400 m): distance = 400 m; displacement = 0 (returned to start).
  • Student walks around a rectangular block 10 m, 20 m, 10 m, 20 m back to start: distance = 60 m; displacement = 0.
  • Car moves 100 km north, then 40 km south: distance = 140 km; net displacement = 60 km north.
🧮 Formulas
  1. SI unit: metre (m).
  2. Displacement (one dimension): Δx = x_f − x_i.
  3. Distance (general along path): distance = Σ |Δx_i| (sum of magnitudes of each path segment).
  4. Net displacement for multiple segments: Δx_net = Σ Δx_i (algebraic sum, sign indicates direction).
  5. Relation: distance ≥ |displacement|.
📊 Visual ideas
Position vs Time graph for one-dimensional motion: draw x (m) on vertical axis and t (s) on horizontal axis. Show a line moving up (positive displacement), then coming back down (negative displacement). Mark initial and final positions to illustrate that area/slope relate to velocity, while vertical difference gives displacement.
Trajectory diagram (top view): draw path of motion (e.g., a rectangle or circle) with arrows showing the actual path (to illustrate distance) and draw a straight arrow from start to end (to illustrate displacement). Label path length and straight-line displacement.
Segmented motion on a line: show three consecutive displacement segments with signed values (+ and −). Underneath, annotate numeric distance (sum of absolute values) and net displacement (algebraic sum) to compare the two.
Circular motion sketch: draw a circle and mark start and end points for partial and full laps. For a full lap show distance = circumference and displacement = 0; for a partial arc show displacement as the chord connecting initial and final points.
🔬4

Speed

Definition: Speed is a scalar quantity that measures how fast an object moves. It is the distance traveled per unit time.

Types of speed

  • Average speed: Total distance travelled divided by total time taken. Useful when speed varies over a trip.
  • Instantaneous speed: Speed of an object at a particular instant (what you read on a speedometer). Mathematically this is the derivative of distance with respect to time.

Relation to velocity: Speed is the magnitude of velocity. Velocity is a vector (has direction); speed has no direction.

Units and conversion: SI unit is metre per second (m/s). Kilometre per hour (km/h) is common for everyday travel. To convert: 1 m/s = 3.6 km/h and 1 km/h = 1/3.6 m/s.

How to interpret graphs:

  • Distance–Time graph: slope = speed. A straight line with constant slope indicates uniform speed. A curved line means changing speed; the slope of a tangent gives instantaneous speed.
  • Speed–Time graph: height of the graph at a time gives the instantaneous speed. Area under the speed–time curve over an interval equals the distance travelled in that interval.

Measurement: Speed is measured by dividing measured distance by measured time using a stopwatch, odometer, speedometer, radar gun, etc. For instantaneous speed use instruments like speedometer or take the limit of distance/time over a very small interval.

Important notes for Class 9: Emphasise the difference between distance vs displacement and speed vs velocity. Speed does not include direction and cannot be negative; velocity can be negative depending on chosen direction.

Short worked idea: If a runner covers 100 m in 10 s, average speed = 100/10 = 10 m/s. If a car’s speedometer reads 72 km/h, convert to m/s: 72 / 3.6 = 20 m/s.

📌 Examples
  • A person walks 4 km in 1 hour → average speed = 4 km/h.
  • Runner covers 100 m in 10 s → average speed = 100 ÷ 10 = 10 m/s.
  • Car travels 180 km in 2 hours → average speed = 180 ÷ 2 = 90 km/h.
  • A speedometer showing 60 km/h gives the instantaneous speed of the vehicle at that moment.
  • Convert 72 km/h to m/s: 72 ÷ 3.6 = 20 m/s.
🧮 Formulas
  1. Speed (instantaneous) = ds/dt
  2. Average speed = (Total distance travelled) / (Total time taken) = Σd / Σt
  3. If distance = s and time = t, then speed = s / t
  4. Unit conversion: 1 m/s = 3.6 km/h ; 1 km/h = 1/3.6 m/s
📊 Visual ideas
Distance–Time graph for uniform motion: straight line through origin (or parallel if starts later). Slope = constant speed.
Distance–Time graph for non-uniform motion: curve. Slope of tangent at a point = instantaneous speed at that time.
Speed–Time graph for constant speed: horizontal line. Area under line between t1 and t2 = distance travelled.
Speed–Time graph showing acceleration and deceleration: rising section (speed increasing), falling section (speed decreasing); areas correspond to distances covered in those intervals.
🔬5

Velocity

Definition: Velocity is the rate of change of displacement of an object with respect to time. It is a vector quantity — it has both magnitude and direction. Symbol: v. SI unit: metre per second (m/s).

Average and Instantaneous Velocity:

  • Average velocity = (total displacement) / (total time) = Δx / Δt. It gives overall change in position per unit time.
  • Instantaneous velocity is the velocity of an object at a particular instant. (Conceptually it is the limit of average velocity as Δt → 0; in calculus this is dx/dt.)

Speed vs. Velocity:

  • Speed is scalar (only magnitude) and equals total distance ÷ time.
  • Velocity is vector (magnitude + direction) and equals displacement ÷ time. A body can have constant speed but changing velocity if its direction changes (e.g., uniform circular motion).

Sign and Direction: The sign of velocity (positive/negative) indicates direction along a chosen reference axis. For motion along a straight line, right/up can be taken as positive and left/down negative.

Uniform and Non-uniform Velocity:

  • Uniform velocity: both magnitude and direction remain constant (displacement-time graph is a straight line of constant slope).
  • Non-uniform velocity: magnitude and/or direction change with time.

Relation with Acceleration (for uniformly accelerated motion): If a body’s velocity changes uniformly with time, the relation v = u + at holds, where u is initial velocity, a is constant acceleration, and t is time. (Other kinematic equations like s = ut + 1/2 at^2 and v^2 = u^2 + 2as are used when acceleration is constant.)

Practical notes: To convert between units: 1 m/s = 3.6 km/h. Velocity depends on the chosen reference frame.

Graphical interpretation: On a displacement–time graph the slope at any point gives the instantaneous velocity. On a velocity–time graph the height gives velocity and the area under the curve between two times gives the displacement covered in that interval.

📌 Examples
  • A car moving east at 20 m/s: velocity = 20 m/s east (both magnitude and direction specified).
  • Runner on a circular track with constant speed 5 m/s: speed constant but velocity changing because direction changes continuously.
  • Elevator going up 2 m/s then stopping: upward velocity 2 m/s (positive), then velocity becomes 0 m/s when stopped.
  • A train reversing direction: if forward velocity is +10 m/s, backward motion could be –5 m/s (negative sign denotes opposite direction).
🧮 Formulas
  1. Average velocity: v_avg = Δx / Δt (Δx = final position − initial position)
  2. Instantaneous velocity (calculus form): v = dx/dt
  3. Uniformly accelerated motion: v = u + a t (u = initial velocity, a = acceleration, t = time)
  4. Displacement under constant acceleration: s = u t + (1/2) a t^2 (used with v = u + at)
  5. Conversion: 1 m/s = 3.6 km/h (so speed in km/h = speed in m/s × 3.6)
📊 Visual ideas
Displacement–time (x–t) graph: straight line with constant slope → constant velocity. Slope = velocity. Curved line → changing velocity; tangent gives instantaneous velocity.
Velocity–time (v–t) graph: horizontal line (nonzero) → constant velocity; height = velocity. Area under v–t curve between t1 and t2 = displacement from t1 to t2.
v–t graph with positive and negative values: shows direction (positive above time axis, negative below). Crossing the axis means change of direction.
Example classroom activity: measure a toy car on a straight track at equal time intervals, plot displacement vs time and draw the best-fit line to find its average velocity (slope).
⚖️6

Acceleration

Definition: Acceleration is the rate of change of velocity with time. It is a vector quantity — it has both magnitude and direction.

Mathematical form: For a time interval Δt during which velocity changes from u to v, average acceleration a_avg = Δv / Δt = (v - u) / (t_2 - t_1). Instantaneous acceleration a = dv/dt.

SI unit: metre per second squared (m/s2).

Types of acceleration:

  • Uniform (constant) acceleration: acceleration is same at every instant (e.g., freely falling body ignoring air resistance).
  • Non-uniform acceleration: acceleration changes with time (e.g., car accelerating unevenly).
  • Positive and negative acceleration: depends on chosen direction. If acceleration is opposite to velocity the speed decreases — often called deceleration.
  • Centripetal (directional) acceleration: when speed is constant but direction changes (e.g., uniform circular motion). Here acceleration points toward the center and magnitude a = v2/r.

Equations for uniformly accelerated (constant a) straight-line motion:

  • v = u + a t
  • s = u t + (1/2) a t2
  • v2 = u2 + 2 a s
  • For constant a, average velocity v_avg = (u + v)/2 and displacement s = v_avg · t

Physical interpretation & signs: Acceleration vector indicates how velocity changes. If velocity increases in the chosen positive direction, acceleration is positive. If the object slows down in that direction, acceleration is negative. In circular motion, acceleration is perpendicular to velocity and changes direction but not speed.

Important notes for Class 9 level: Instantaneous acceleration can be obtained from the slope of the velocity-time (v–t) graph (a = slope). For uniform acceleration, v–t graph is a straight line; position–time (s–t) graph is a parabola. Area under v–t graph between two times gives the displacement in that interval.

📌 Examples
  • A car increasing speed from 0 to 20 m/s in 10 s: average acceleration = (20 - 0)/10 = 2 m/s^2.
  • A bus braking from 15 m/s to 5 m/s in 4 s: acceleration = (5 - 15)/4 = -2.5 m/s^2 (negative means slowing down).
  • A ball dropped from rest near Earth (ignoring air resistance) accelerates downward at about g ≈ 9.8 m/s^2.
  • A cyclist pedalling to increase speed steadily — approximately uniform acceleration for short intervals.
  • A satellite moving in circular orbit has centripetal acceleration a = v^2 / r directed toward the centre, changing direction of velocity but not its magnitude.
🧮 Formulas
  1. Average acceleration: a_avg = Δv / Δt = (v - u) / (t_2 - t_1)
  2. Instantaneous acceleration: a = dv/dt
  3. First equation of motion (constant a): v = u + a t
  4. Second equation of motion: s = u t + (1/2) a t^2
  5. Third equation of motion: v^2 = u^2 + 2 a s
  6. Average velocity (constant a): v_avg = (u + v) / 2
📊 Visual ideas
Velocity–Time (v–t) graph for uniform acceleration: straight line with slope = a. Label axes: vertical 'velocity (m/s)', horizontal 'time (s)'. The slope (rise/run) gives acceleration; area under the line gives displacement.
Position–Time (s–t) graph for uniform acceleration: upward-opening parabola if acceleration is positive. Draw axes 'displacement (m)' and 'time (s)'. For zero acceleration this becomes a straight line.
Acceleration–Time (a–t) graph for constant acceleration: horizontal line at value a. Area under a–t curve between t1 and t2 equals change in velocity Δv.
v–t graph showing negative acceleration (deceleration): straight line sloping downwards. Show initial positive velocity and line crossing to lower velocity.
🏃7

Equations of Motion for Uniformly Accelerated Motion

What is uniformly accelerated motion? When an object’s velocity changes by equal amounts in equal intervals of time, the acceleration is constant. Such motion is called uniformly accelerated motion.

Symbols and units

  • u — initial velocity (m/s)
  • v — final velocity after time t (m/s)
  • a — constant acceleration (m/s²)
  • t — time interval (s)
  • s — displacement in time t (m)

Derivation and key relations

  1. Definition of acceleration: a = (v − u)/t. Rearranging gives the first equation:

    v = u + a t

  2. Average velocity when acceleration is constant is the arithmetic mean of initial and final velocities:

    v_avg = (u + v)/2

    Displacement = average velocity × time, so

    s = v_avg · t = ((u + v)/2) · t.

    Substitute v = u + at into this to get the second equation:

    s = u t + (1/2) a t²

  3. Eliminate t to get a relation between v, u and s. From v = u + a t, t = (v − u)/a. Substitute in s = ((u + v)/2)·t and simplify to obtain the third equation:

    v² = u² + 2 a s

Summary of the three standard equations (for constant a):

  • v = u + a t
  • s = u t + (1/2) a t²
  • v² = u² + 2 a s

Important remarks

  • These equations apply only when acceleration a is constant during the interval t.
  • Take care with signs: choose a positive direction. If acceleration is opposite to the chosen positive direction (e.g., deceleration), a is negative.
  • Area under a v–t graph between t1 and t2 equals displacement between those times; slope of v–t is acceleration.
📌 Examples
  • A car starts from rest (u = 0) and accelerates uniformly at 2 m/s² for 5 s. Find v and s: v = 0 + 2×5 = 10 m/s; s = 0×5 + 1/2×2×5² = 25 m.
  • A ball thrown upward with initial speed u = 20 m/s. Take a = −9.8 m/s² (gravity). Time to reach top (v = 0): t = (v − u)/a = (0 − 20)/(−9.8) ≈ 2.04 s. Maximum height s = u t + 1/2 a t² ≈ 20×2.04 + 0.5×(−9.8)×(2.04)² ≈ 20.4 m.
  • A cyclist moving at 15 m/s applies brakes giving acceleration a = −3 m/s². Time to stop: t = (0 − 15)/(−3) = 5 s. Stopping distance: s = (0 + 15)/2 × 5 = 37.5 m.
  • Free fall from rest near Earth’s surface (ignore air resistance): u = 0, a ≈ 9.8 m/s². Distance fallen in 3 s: s = 0 + 0.5×9.8×3² = 44.1 m.
  • An elevator speeds up uniformly from 2 m/s to 6 m/s over 4 s. Acceleration: a = (6 − 2)/4 = 1 m/s². Displacement in that time: s = ((2+6)/2)×4 = 16 m.
  • A toy car moving at 4 m/s accelerates at 0.5 m/s² for 8 s. Final speed v = 4 + 0.5×8 = 8 m/s; displacement s = 4×8 + 0.5×0.5×8² = 64 m.
🧮 Formulas
  1. v = u + a t (final velocity after time t)
  2. s = u t + (1/2) a t² (displacement in time t)
  3. v² = u² + 2 a s (relation between velocities and displacement)
  4. v_avg = (u + v)/2 (average velocity for constant acceleration)
  5. s = v_avg × t = ((u + v)/2) × t (alternate form for displacement)
📊 Visual ideas
v–t graph: plot velocity (v) on vertical axis and time (t) on horizontal axis. For uniform acceleration the graph is a straight line with slope = a. The area under the line between t1 and t2 gives displacement between those times.
s–t graph: plot displacement (s) vs time (t). For uniform acceleration the curve is a parabola (s = u t + 1/2 a t²). If u = 0 it starts at origin and opens upward for positive a.
a–t graph: plot acceleration (a) vs time (t). For uniform acceleration this is a horizontal straight line at height a. The value of a indicates whether the motion speeds up (a same sign as velocity) or slows down (opposite sign).
v² vs s graph (optional): plotting v² on vertical axis and displacement s on horizontal axis gives a straight line with slope 2a when acceleration is constant, according to v² = u² + 2 a s.
🏃8

Graphical Representation of Motion

Graphical Representation of Motion shows how position (distance or displacement), velocity and acceleration change with time using simple 2‑D graphs. Graphs convert numerical data into visual form so you can read trends (constant, increasing, decreasing), calculate rates (slopes) and totals (areas).

Key ideas

  • Axes: Time (t) is normally on the x‑axis. The y‑axis can be distance (d), displacement (s), velocity (v) or acceleration (a).
  • Slope (gradient): On a y vs t graph slope = rise/run = Δy/Δt. Interpretation depends on the y variable: on a d–t graph slope = speed (or velocity if d is displacement); on a v–t graph slope = acceleration.
  • Area under graph: On a v–t graph the area between the curve and time axis = displacement (distance with sign) for that time interval. On an a–t graph the area = change in velocity.
  • Straight lines vs curves: A straight line indicates a uniform rate (constant speed or constant acceleration). A curve indicates changing rate (non‑uniform motion).

Common graphs and how to read them

  • Distance–Time (d–t) graph:
    • Straight sloped line through origin: uniform motion. Slope = distance/time = speed.
    • Horizontal line: object at rest (speed = 0).
    • Curve (concave up): speed increasing (positive acceleration). Concave down: speed decreasing.
    • Note: distance is always non‑negative; displacement–time graphs can go negative showing direction.
  • Velocity–Time (v–t) graph:
    • Horizontal line (v = constant): uniform velocity. Area under line from t1 to t2 = v × (t2−t1) = displacement.
    • Sloped straight line: constant acceleration. Slope = (v2−v1)/(t2−t1) = acceleration.
    • If the line is below the time axis, velocity is negative (motion in the opposite direction).
  • Acceleration–Time (a–t) graph:
    • Horizontal line: constant acceleration. Area under graph = change in velocity (Δv = area).
    • Zero line: no acceleration → velocity constant.

Reading values from graphs

  • To find instantaneous speed/velocity from a d–t curve, compute the slope (tangent) at that time.
  • To find displacement from a v–t graph, calculate the area under the curve between the given times (rectangles, triangles, trapeziums for straight lines).
  • Units: distance (m), time (s), velocity (m/s), acceleration (m/s²). Slope units reflect these (e.g. slope of d–t in m/s).

Practical note: In experiments you plot measured time vs distance points and draw the best fit line or curve; then use slope/area rules above to interpret motion.

📌 Examples
  • A car traveling at constant 20 m/s: d–t graph is a straight line with slope 20; v–t graph is a horizontal line at v = 20 m/s.
  • A car accelerating uniformly from rest to 20 m/s in 5 s: on v–t graph draw a straight line from (0,0) to (5,20). Slope = (20−0)/5 = 4 m/s² (acceleration). Area under v–t from 0 to 5 s = average velocity × time = (0+20)/2 × 5 = 50 m (displacement).
  • A ball in free fall near Earth (neglect air resistance): v–t is a straight line with slope ≈ 9.8 m/s² (downwards). d–t curve is concave upward and steeper with time.
  • A bus coming to a stop from 10 m/s in 4 s with uniform deceleration: v–t is a straight line from (0,10) to (4,0). Acceleration = (0−10)/4 = −2.5 m/s². Area under v–t gives distance travelled while stopping = (10/2)×4 = 20 m.
  • A runner who rests for 5 s, runs at 4 m/s for 10 s, then stops: the v–t graph shows a horizontal line at 0 (0–5 s), then at 4 m/s (5–15 s), then 0 again. Total displacement = area under the v–t curve = 4 × 10 = 40 m.
🧮 Formulas
  1. Average speed = total distance / total time
  2. Slope (general) = rise / run = Δy / Δx
  3. For uniform acceleration: v = u + a t
  4. s (displacement) = u t + (1/2) a t²
  5. v² = u² + 2 a s
  6. \[Area under v–t graph between t1 and t2 = displacement = ∫_{t1}^{t2} v dt (use geometry for straight line segments)\]
📊 Visual ideas
Distance–Time (uniform motion): x‑axis = time (s), y‑axis = distance (m). Draw a straight line through origin with constant positive slope. Label slope = speed (m/s).
Distance–Time (object at rest): x‑axis = time, y‑axis = distance. Draw a horizontal line. Speed = 0.
Distance–Time (non‑uniform): x‑axis = time, y‑axis = distance. Draw a curve getting steeper with time (concave up) to show increasing speed. To find instantaneous speed, draw tangent and compute its slope.
Velocity–Time (constant velocity): x‑axis = time (s), y‑axis = velocity (m/s). Draw horizontal line at v = constant. Shade rectangle under this line between t1 and t2; area = v×(t2−t1) = displacement.
🏃9

Uniform Circular Motion

Definition: Uniform circular motion (UCM) is the motion of a particle along a circular path with constant speed. Although speed is constant, the velocity is not constant because its direction changes continuously.

Key ideas:

  • Path: a circle of radius r.
  • Speed: constant magnitude v.
  • Velocity: tangent to the circle at each point (direction changes with time).
  • Acceleration: always directed towards the centre of the circle (radially inward). This acceleration is called centripetal acceleration and arises because the direction of velocity changes.

Angular motion description: Let θ(t) be the angular position measured from a fixed direction. For uniform motion, angular speed (angular velocity magnitude) ω is constant and

θ(t) = ωt + θ₀

Position coordinates (choosing origin at centre):

x(t) = r cos(ωt + θ₀), y(t) = r sin(ωt + θ₀)

Velocity (by differentiating):

v_x = −rω sin(ωt + θ₀), v_y = rω cos(ωt + θ₀)

|v| = v = rω (constant)

Centripetal acceleration (by differentiating velocity):

a_x = −rω² cos(ωt + θ₀), a_y = −rω² sin(ωt + θ₀)

a⃗ = −ω² r⃗ (points to centre), |a| = a_c = ω² r = v² / r

Physical interpretation: A continuous inward force (centripetal force) is required to change the direction of velocity and keep the object moving in a circle. If that force is removed, the object will move off tangentially.

📌 Examples
  • A stone tied to a string and whirled in a horizontal circle (string provides centripetal force).
  • Blades of a ceiling fan or rotor of an electric motor (approximate UCM for blade tips).
  • The tip of a second hand on an analogue clock (roughly uniform angular speed).
  • A car moving at constant speed around a circular roundabout (friction between tyres and road supplies centripetal force).
  • A Ferris wheel or merry-go-round where seats move approximately with uniform speed along a circle.
🧮 Formulas
  1. Angular speed: ω = dθ/dt (for uniform motion ω = constant)
  2. Relation between ω, period T and frequency f: ω = 2π/T = 2πf, f = 1/T
  3. Linear speed: v = ω r = 2π r / T
  4. Centripetal acceleration (magnitude): a_c = v² / r = ω² r
  5. Centripetal force: F_c = m a_c = m v² / r = m ω² r
📊 Visual ideas
x(t) = r cos(ωt) and y(t) = r sin(ωt): plot x vs t and y vs t — sinusoidal (cosine and sine) waves of same frequency, phase-shifted by 90°.
θ vs t: a straight line with slope ω (shows uniform angular increase).
v vs t (speed magnitude): a horizontal line (constant speed).
a_c vs t (magnitude): a horizontal line (constant magnitude), but a vector-field diagram on the circle showing acceleration arrows pointing towards centre and rotating with the particle.
🏃10

Motion under Gravity (Free Fall)

Definition: Free fall (motion under gravity) is the motion of an object when gravity is the only force acting on it (air resistance neglected). All objects in free fall near Earth's surface accelerate downward with the same constant acceleration called acceleration due to gravity, g.

Value and sign convention: g ≈ 9.8 m/s² (often approximated as 10 m/s² in Class 9 problems). Choose a sign convention: e.g. downward positive gives acceleration +g; upward positive gives acceleration −g.

Key points:

  • Acceleration is constant (magnitude g) and is independent of the mass of the object.
  • Free-fall motion includes objects dropped from rest, objects thrown downward, and objects thrown upward (which rise, stop momentarily at the highest point, and then fall).
  • In real life air resistance can be important (feathers fall slower than stones in air). In vacuum, all objects fall identically.

Equations of motion (use with chosen sign for g): The kinematic equations for uniformly accelerated motion apply with a = ±g:

  • v = u + g t
  • s = u t + (1/2) g t²
  • v² = u² + 2 g s

Here u = initial velocity, v = velocity after time t, s = displacement (positive in the direction chosen as positive), and g = 9.8 m/s². For an object dropped from rest, u = 0 and s = (1/2) g t², v = g t, v² = 2 g s. For an object thrown upward with initial speed u (upward positive), time to reach maximum height t_up = u/g, and maximum height h = u²/(2 g). Total time of flight (if it lands at launch level) = 2u/g.

📌 Examples
  • Dropped from rest: A ball is dropped from 20.0 m. Using s = (1/2) g t² with g = 9.8 m/s², t = sqrt(2s/g) = sqrt(40/9.8) ≈ 2.02 s. Speed on impact v = g t ≈ 9.8 × 2.02 ≈ 19.8 m/s.
  • Thrown upward: A ball is thrown vertically up with u = 15 m/s. Time to reach top t_up = u/g ≈ 15/9.8 ≈ 1.53 s. Maximum height h = u²/(2g) = 225/(19.6) ≈ 11.48 m. Total time of flight ≈ 3.06 s (up + down).
  • Mass independence (Galileo idea): In the absence of air resistance, a heavy stone and a light stone dropped together from the same height reach the ground simultaneously because their acceleration is the same (g).
  • Real-life non-ideal case — skydiver: A skydiver eventually reaches a terminal velocity when air resistance equals weight; this is not free fall (because other forces matter).
🧮 Formulas
  1. Acceleration due to gravity: g ≈ 9.8 m/s² (approx. 10 m/s² for simple problems).
  2. v = u + g t (use sign of g consistent with chosen positive direction)
  3. s = u t + (1/2) g t²
  4. v² = u² + 2 g s
  5. For free fall from rest (u = 0): s = (1/2) g t², v = g t, v² = 2 g s
  6. For upward throw: time to top t_up = u/g, maximum height h = u²/(2 g), total time (return to launch level) = 2u/g
📊 Visual ideas
Position vs Time (s–t): Parabola. If object is dropped from origin, s(t) = (1/2) g t² — upward concave plot rising (if downward is positive) or opening downward if upward positive and object thrown up (shows rise to a maximum then fall). Label axes: s (m) vertical, t (s) horizontal. Mark apex for thrown-up case.
Velocity vs Time (v–t): Straight line with slope = g. For downward positive, v increases linearly from initial u; for upward positive and object thrown up, v starts positive, decreases linearly, crosses zero at top (t = u/g), then becomes negative. Label axes: v (m/s), t (s); mark v = 0 at peak.
Acceleration vs Time (a–t): Horizontal line at a = g (or a = −g depending on sign convention). Shows acceleration is constant throughout free fall.
v² vs s: Straight line from equation v² = u² + 2 g s. Useful to show linear relation between v² and displacement; slope = 2 g.
🔬11

Problem-solving and Numerical Applications

What this topic covers: 'Problem-solving and Numerical Applications' applies formulas and reasoning from the chapter Motion to solve numerical questions about speed, velocity, displacement and acceleration. It emphasizes choosing the right equation, careful unit conversion, sign convention and interpreting graphs (s–t, v–t, a–t).

Step-by-step problem-solving strategy

  1. Read the problem carefully; list known quantities (with units) and what is asked.
  2. Choose a sign convention (e.g., right or up = positive) and state it.
  3. Convert all quantities to SI units (metre, second) if needed — e.g., km/h to m/s multiply by 5/18.
  4. Identify whether motion is uniform, non-uniform or uniformly accelerated; pick the suitable formula(s).
  5. Solve algebraically for the unknown, substitute numbers, keep track of units, compute numerical answer with reasonable significant figures.
  6. Check result for physical sense (units, sign, magnitude) and, if possible, check using a second method (e.g., use another kinematic equation).

Common types of numerical problems: finding speed from distance and time, converting speeds (km/h ↔ m/s), finding final velocity and distance under uniform acceleration (using the three standard kinematic equations), calculating average speed vs average velocity for multi-stage journeys, and interpreting areas/slopes on motion graphs.

Important points to remember

  • Speed is scalar (distance/time); velocity is vector (displacement/time). Average speed = total distance / total time; average velocity = total displacement / total time.
  • For uniformly accelerated motion use: v = u + at, s = ut + (1/2)at², v² = u² + 2as. Here u = initial velocity, v = final velocity, a = acceleration, s = displacement, t = time.
  • Slope of s–t graph = instantaneous velocity. Slope of v–t graph = acceleration. Area under v–t graph = displacement.

This combination of conceptual understanding, careful unit handling, and equation selection is what CBSE problems test. Practise a variety of problems (constant speed, changing speed, round trips) and always interpret answers physically.

📌 Examples
  • Example 1 (uniform speed): A cyclist travels at 18 km/h for 30 minutes. How far does she go?\nGivens: speed = 18 km/h = 18 × (5/18) = 5 m/s, time = 30 min = 1800 s. Distance = speed × time = 5 × 1800 = 9000 m = 9 km. (Or do in km: 18 km/h × 0.5 h = 9 km.)
  • Example 2 (constant acceleration — find final speed and distance): A car starts from rest and accelerates uniformly at 2 m/s² for 10 s. Find the final velocity and distance covered.\nGivens: u = 0, a = 2 m/s², t = 10 s. Use v = u + at ⇒ v = 0 + 2×10 = 20 m/s. Use s = ut + 1/2 at² ⇒ s = 0 + 0.5×2×(10)² = 1×100 = 100 m.
  • Example 3 (use v² = u² + 2as): A ball moving at 20 m/s is brought to rest by a constant negative acceleration over 50 m. What is the acceleration?\nGivens: u = 20 m/s, v = 0, s = 50 m. Use v² = u² + 2as ⇒ 0 = 400 + 2a(50) ⇒ 0 = 400 + 100a ⇒ a = -4 m/s². (Negative sign indicates deceleration.)
  • Example 4 (average speed vs average velocity — two-leg journey): A student walks 200 m east in 4 min and then 100 m west in 2 min. (a) Average speed = total distance / total time = (200+100) m / (6×60 s) = 300 / 360 = 0.833 m/s. (b) Average velocity = displacement / time = (200-100) m / 360 s = 100 / 360 ≈ 0.278 m/s east.
🧮 Formulas
  1. Speed (average) = distance / time (s = d / t)
  2. Velocity (average) = displacement / time (v_avg = Δx / t)
  3. Uniform acceleration: a = (v - u) / t
  4. v = u + at
  5. s = ut + (1/2)at²
  6. v² = u² + 2as
📊 Visual ideas
Distance–Time (s–t) for uniform motion: straight line with constant slope (slope = speed). Label axes 'time (s)' and 'distance (m)'. Example: s = vt gives a straight line through origin if motion starts at s=0.
Distance–Time for uniformly accelerated motion: a parabola (s = ut + 1/2 at²). If u=0 it’s a simple upward-opening parabola; slope at a point gives instantaneous velocity.
Velocity–Time (v–t) for constant velocity: horizontal straight line (slope = 0 ⇒ acceleration = 0). Area under the line between t1 and t2 = displacement.
Velocity–Time for uniform acceleration: straight line with constant slope (slope = a). The vertical intercept is u, the final point gives v; area under this line = displacement (use trapezoid area or base×average height).

Key Concepts

Motion
Change in position of an object with respect to a reference point over time.
Rest
State in which an object does not change its position with respect to a reference point.
Reference point
A fixed point or object used to determine whether another object is in motion or at rest.
Distance
Total length of the path travelled by an object; scalar quantity.
Displacement
Shortest straight-line distance from initial to final position with direction; vector quantity.
Speed
Rate of change of distance with time; scalar (speed = distance/time).
Velocity
Rate of change of displacement with time; vector (velocity = displacement/time).
Acceleration
Rate of change of velocity with time (can be positive or negative).
Uniform motion
Motion in which an object covers equal distances in equal intervals of time (constant speed).
Non-uniform motion
Motion in which an object covers unequal distances in equal intervals of time (variable speed).
Average speed
Total distance travelled divided by total time taken for the entire journey.
Instantaneous speed
Speed of an object at a particular instant of time.
Uniform acceleration
Acceleration that remains constant in magnitude and direction.
Retardation (Deceleration)
Negative acceleration; velocity decreases with time.
Scalar quantity
Physical quantity described by magnitude only, no direction.
Vector quantity
Physical quantity described by both magnitude and direction.
Relative motion
Description of motion of an object as observed from a particular reference frame.
Distance–time graph
Graph plotting distance versus time; slope gives speed for uniform motion.
Velocity–time graph
Graph plotting velocity versus time; slope gives acceleration and area under curve gives displacement.
Equations of motion
Set of formulae relating displacement (s), initial velocity (u), final velocity (v), acceleration (a) and time (t): v = u + at; s = ut + 1/2 at²; v² = u² + 2as.

End-of-Chapter Trial Paper & Test Questions

Topic-wise questions to test your understanding of every concept in this chapter.

  1. A car starts from rest and accelerates uniformly at 3 m/s² for 6 seconds. What is its final velocity? (a) 9 m/s (b) 18 m/s (c) 12 m/s (d) 6 m/s एक कार विराम से शुरू होकर 6 सेकंड तक 3 m/s² से एकसमान त्वरण लेती है। इसका अंतिम वेग क्या होगा? (a) 9 m/s (b) 18 m/s (c) 12 m/s (d) 6 m/s
    Show answer

    (b) 18 m/s — Using v = u + at = 0 + 3×6 = 18 m/s. The first equation of motion applies since acceleration is constant. (b) 18 m/s — v = u + at = 0 + 3×6 = 18 m/s का उपयोग करते हैं। त्वरण स्थिर होने पर गति का पहला समीकरण लागू होता है।

  2. A runner completes one full lap of a 400 m track and returns to starting point. Which is correct? (a) Distance=400 m, Displacement=400 m (b) Distance=0 m, Displacement=400 m (c) Distance=400 m, Displacement=0 m (d) Distance=0 m, Displacement=0 m एक धावक 400 m के ट्रैक का एक पूरा चक्कर लगाकर शुरुआती बिंदु पर वापस आता है। कौन सा सही है? (a) दूरी=400 m, विस्थापन=400 m (b) दूरी=0 m, विस्थापन=400 m (c) दूरी=400 m, विस्थापन=0 m (d) दूरी=0 m, विस्थापन=0 m
    Show answer

    (c) Distance=400 m, Displacement=0 m — Distance is the total path length (400 m), but displacement is zero because initial and final positions are the same. (c) दूरी=400 m, विस्थापन=0 m — दूरी कुल मार्ग की लंबाई (400 m) है, लेकिन विस्थापन शून्य है क्योंकि प्रारंभिक और अंतिम स्थिति एक ही है।

  3. On a velocity–time graph, what does the area under the curve between two time instants represent? (a) Acceleration (b) Speed (c) Displacement (d) Force वेग-समय ग्राफ पर दो समय बिंदुओं के बीच वक्र के नीचे का क्षेत्रफल क्या दर्शाता है? (a) त्वरण (b) चाल (c) विस्थापन (d) बल
    Show answer

    (c) Displacement — The area under a velocity-time graph gives the displacement in that time interval. (c) विस्थापन — वेग-समय ग्राफ के नीचे का क्षेत्रफल उस समय अंतराल में तय किया गया विस्थापन देता है।

  4. The slope of a displacement–time graph gives ________. विस्थापन-समय ग्राफ की ढाल ________ देती है।
    Show answer

    Velocity. A steeper slope means higher velocity; a horizontal line means zero velocity (object at rest). वेग। अधिक ढाल का अर्थ है अधिक वेग; क्षैतिज रेखा का अर्थ है शून्य वेग (वस्तु विराम में)।

  5. A body moving at 20 m/s is brought to rest in 50 m. The acceleration is ________ m/s². 20 m/s की चाल से गतिमान एक वस्तु 50 m में रुक जाती है। त्वरण ________ m/s² है।
    Show answer

    –4 m/s² — Using v² = u² + 2as: 0 = 400 + 2a(50), so a = –4 m/s² (deceleration). –4 m/s² — v² = u² + 2as से: 0 = 400 + 2a(50), अतः a = –4 m/s² (मंदन)।

  6. True or False: An object can have zero velocity but non-zero acceleration. सत्य या असत्य: किसी वस्तु का वेग शून्य हो सकता है परंतु त्वरण शून्य नहीं।
    Show answer

    True — At the highest point of a vertically thrown object, velocity is zero but gravitational acceleration (9.8 m/s²) still acts downward. सत्य — ऊर्ध्वाधर रूप से फेंकी गई वस्तु के उच्चतम बिंदु पर वेग शून्य होता है, लेकिन गुरुत्वीय त्वरण (9.8 m/s²) अभी भी नीचे की ओर कार्य करता है।

  7. Distinguish between uniform motion and non-uniform motion with one example each. एकसमान गति और असमान गति में अंतर बताइए, प्रत्येक का एक उदाहरण दीजिए।
    Show answer

    Uniform motion: equal distances in equal time intervals — e.g., a car at constant 60 km/h on a highway. Non-uniform motion: unequal distances in equal intervals — e.g., a car in city traffic that keeps stopping and accelerating. एकसमान गति: समान समय अंतरालों में समान दूरियाँ — जैसे हाईवे पर 60 km/h की स्थिर गति से चलती कार। असमान गति: समान समय में असमान दूरियाँ — जैसे शहरी ट्रैफिक में बार-बार रुकती और त्वरित होती कार।

  8. A ball is thrown vertically upward with initial velocity 20 m/s. Taking g = 10 m/s², find (i) time to reach the highest point and (ii) maximum height reached. एक गेंद को 20 m/s के प्रारंभिक वेग से ऊर्ध्वाधर ऊपर फेंका जाता है। g = 10 m/s² लेते हुए (i) उच्चतम बिंदु तक पहुँचने में लगा समय और (ii) अधिकतम ऊँचाई ज्ञात कीजिए।
    Show answer

    (i) t = u/g = 20/10 = 2 s. (ii) h = u²/(2g) = 400/20 = 20 m. At the top, v = 0; using v = u – gt gives t = 2 s, and h = ut – ½gt² = 40 – 20 = 20 m. (i) t = u/g = 20/10 = 2 s। (ii) h = u²/(2g) = 400/20 = 20 m। शीर्ष पर v = 0; v = u – gt से t = 2 s और h = ut – ½gt² = 40 – 20 = 20 m।

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