Notes Class 9 Science Exploration Chapter 4 Describing Motion Around Us

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Introduction — Types of Motion (गति के प्रकार)

Everything in nature is in motion — from massive stars to tiny subatomic particles. A butterfly flitting, a snake slithering, a horse galloping, a falling raindrop, rising ocean tides — motion is everywhere! Scientists study complex motions by first understanding simplified, idealised types.

📏 Linear Motion

Motion in a straight line. Also called motion in one dimension. Examples: car on a straight highway, falling stone, swimming race.

⭕ Circular Motion

Motion along a circular path. Examples: merry-go-round, Earth around Sun, ceiling fan blade tip.

🔄 Oscillatory Motion

Back-and-forth motion about a fixed point. Examples: pendulum of a clock, guitar string, swing.

🛩️ Motion in a Plane

Motion in two dimensions. Examples: overtaking car, kicked football, satellite in orbit.

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Chapter Focus

In this chapter, we focus on linear motion (motion in a straight line) and uniform circular motion. We learn to describe motion using numbers, equations AND graphs!

🇮🇳

India’s Scientific Heritage — Aryabhatiya (5th Century CE)

The concept that speed = distance ÷ time was well-established in ancient India! The treatise Aryabhatiya by Aryabhata (5th century CE) contains motion problems. The Ganitakaumudi (14th century CE) by Narayana Pandita also has problems on relative speed — Example: two postmen walking towards each other!

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Position, Distance & Displacement (स्थिति, दूरी और विस्थापन)

🎯 Reference Point & Position

To describe where an object is, we first need a reference point (also called the origin O). The position of an object is described by its distance AND direction from the reference point.

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Motion vs Rest — Key Definitions

Motion:If the position of an object changes with time (relative to reference point).

Rest:If the position does NOT change with time (relative to reference point).

−40m−20mO (0m)20mB (40m)A (100m)← Negative (−)Positive (+) →

📌 Number line showing reference point O and positions of an athlete. Right of O = Positive (+), Left of O = Negative (−)

📏 Distance vs Displacement — The Most Important Difference!

📏 Distance (दूरी)

The total path length covered by an object. No direction — only a number value. Always positive (≥ 0). It is ascalarquantity. SI unit:metre (m).

↗️ Displacement (विस्थापन)

The net change in position — shortest straight line from start to end. Has both direction AND magnitude. Can be zero, positive or negative. It is avectorquantity. SI unit:metre (m).

⚠️

Critical Distinction — Never Confuse These!

Magnitude of displacement is always ≤ total distance travelled. They are equal ONLY when the object moves in one direction without turning back. If the object returns to start, displacement = 0 but distance ≠ 0!

🏃 Worked Example — Athlete on a Track

An athlete starts from O (0 m) at t = 0 s, reaches A (100 m) at t = 10 s, then runs BACK to B (40 m) at t = 16 s.

Total distance travelled = OA + AB = 100 m + 60 m = 160 m

Displacement = OB = 40 m in the positive (+) direction = +40 m

Notice: distance (160 m) ≠ magnitude of displacement (40 m) because athlete turned back!

🚗

Fuel and Distance — Not Displacement!

The fuel used by a vehicle depends on the total distance travelled — NOT displacement. This is why GPS navigation apps show “distance to destination” (displacement-like), but your fuel gauge depletes based on total path driven!

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Average Speed & Average Velocity (औसत चाल और औसत वेग)

⚡ Average Speed (औसत चाल)

Average speed tells us how fast or slow an object moves — but gives NO information about direction.

Average Speed = Total Distance Travelled ÷ Time Interval

v_avg = d / t

| SI Unit: m s⁻¹ (metres per second)

✅ Uniform Motion

Equal distances in equal time intervals. Constant speed. Position-time graph is a straight line.

📈 Non-Uniform Motion

Unequal distances in equal time intervals. Changing speed. Position-time graph is a curve.

🧭 Average Velocity (औसत वेग)

Velocity tells us both how fast AND in which direction an object moves. It is the rate of change of position.

Average Velocity = Displacement ÷ Time Interval

v_av = s / t
| SI Unit: m s⁻¹

Direction: Same as direction of displacement (+) or (−)

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Speed vs Velocity — Key Difference

Speed= scalar (only magnitude, no direction) | calculated from distance

Velocity= vector (magnitude + direction) | calculated from displacement

Average speed and magnitude of average velocity are equal ONLY if the object moves in ONE direction without turning back.

🏊 Worked Example — Swimming Pool

Sarang swims from one end to the other (25 m) and back in 50 seconds.

Total distance = 25 + 25 = 50 m | Displacement = 0 m (returned to start)

Average Speed = 50 m ÷ 50 s = 1 m s⁻¹

Average Velocity = 0 m ÷ 50 s = 0 m s⁻¹

Speed is NOT zero but velocity IS zero — because he came back to the start!

✏️

Exam Trick — Unit Conversion!

To convert km h⁻¹ to m s⁻¹:multiply by 5/18

To convert m s⁻¹ to km h⁻¹:multiply by 18/5

Example: 36 km h⁻¹ = 36 × 5/18 =10 m s⁻¹ | 15 m s⁻¹ = 15 × 18/5 =54 km h⁻¹

📈

Average Acceleration (औसत त्वरण)

When a vehicle speeds up from rest and you feel pushed back, or when it brakes and you lurch forward — you’re experiencing acceleration! Acceleration is the rate of change of velocity.

Average Acceleration = Change in Velocity ÷ Time Interval

a = (v − u) / (t₂ − t₁)

where u = initial velocity, v = final velocity

SI Unit: m s⁻² (metres per second squared)

🚗 Positive Acceleration

Velocity is INCREASING. Acceleration is in the SAME direction as velocity. Example: Car pressing accelerator.

🛑 Negative Acceleration (Retardation/Deceleration)

Velocity is DECREASING. Acceleration is OPPOSITE to velocity direction. Example: Car pressing brake.

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Very Common Mistake — Zero Acceleration ≠ Zero Speed!

A bus moving at constant 60 km/h on a straight highway has zero acceleration even though it’s fast! Acceleration depends on CHANGE in velocity, not on how fast the object moves. Constant velocity = zero acceleration!

🚌 Worked Example — Bus on Highway

A bus moves at 36 km h⁻¹ = 10 m s⁻¹. Driver presses accelerator for 10 s; velocity increases to 54 km h⁻¹ = 15 m s⁻¹.

(i) Acceleration while speeding up:

a = (15 − 10) / 10 = 5/10 = +0.5 m s⁻² (positive → in direction of motion)

(ii) Brakes pressed, bus stops in 5 s (u = 15 m s⁻¹, v = 0):

a = (0 − 15) / 5 = −15/5 = −3 m s⁻² (negative → opposite to motion)

🌍 Free Fall — Constant Acceleration due to Gravity

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Acceleration Due to Gravity — g = 9.8 m s⁻²

When an object falls freely from a height, its velocity increases by9.8 m s⁻¹ every second. This constant acceleration is caused by Earth’s gravitational force and is denoted byg = 9.8 m s⁻². It acts downward (in the direction of motion while falling).

Time (s)	Velocity (m s⁻¹)	Acceleration between intervals (m s⁻²)
0	0	—
1	9.8	(9.8−0)/1 = 9.8
2	19.6	(19.6−9.8)/1 = 9.8
3	29.4	(29.4−19.6)/1 = 9.8
4	39.2	(39.2−29.4)/1 = 9.8

Constant acceleration = 9.8 m s⁻² at every interval → Free fall has constant acceleration!

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Graphical Representation of Motion

Graphs give us a powerful visual picture of motion. From graphs we can identify if motion is uniform or non-uniform, and calculate velocity, acceleration and displacement without equations!

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Important Rule for All Graphs in This Chapter

All graphs here are for motion in ONE direction only. In this case: distance = displacement magnitude, and speed = velocity magnitude. Time is always on the x-axis!

📈 Position-Time Graph (स्थिति-समय ग्राफ)

Shows how position changes with time. Time on x-axis, Position on y-axis.

txTime (s)Position246✅ Straight line → Constant velocity (Uniform motion)

txTime (s)Position246📈 Curve → Changing velocity (Accelerated motion)

txTime (s)Position40 m➡️ Horizontal line → Object at rest (position constant)

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What Can We Get From a Position-Time Graph?

1. Position at any instant → read y-value at that time.

2.Velocity= slope of the graph = (change in position) ÷ (change in time) = BC/CA in triangle ABC.

Steeper slope = higher velocity!

📉 Velocity-Time Graph (वेग-समय ग्राफ)

Shows how velocity changes with time. Time on x-axis, Velocity on y-axis.

Time (s)VelocityvArea = Displacement➡️ Horizontal line → Constant velocity, a = 0

Time (s)VelocityArea = s↗️ Rising line → Increasing velocity, constant +a

Time (s)VelocityArea = s↘️ Falling line → Decreasing velocity, constant −a

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What Can We Get From a Velocity-Time Graph?

1. Velocity at any instant → read y-value at that time.

2.Acceleration= slope of the v-t graph = (change in velocity) ÷ (change in time)

3.Displacement= area enclosed between the line and the time axis

✏️

Exam Formula — Area under v-t graph

For constant velocity: Area =rectangle= v × t = displacement

For uniformly changing velocity: Area =trapezium= ½ × (u + v) × t = displacement

Or split into: rectangle (lower part) + triangle (upper part)

🔢

Kinematic Equations (गतिकी समीकरण)

For motion in a straight line with constant acceleration, three magical equations connect five quantities: displacement (s), time (t), initial velocity (u), final velocity (v), and acceleration (a).

🔑 Equation 1: v = u + at

(Derived from definition of acceleration)

🔑 Equation 2: s = ut + ½at²

(Derived from area under v-t graph)

🔑 Equation 3: v² = u² + 2as

(Derived by eliminating t from Eq. 1 and Eq. 2)

⚠️

Valid ONLY for Constant Acceleration!

These equations work ONLY when acceleration does not change during the motion. If acceleration is zero (constant velocity), still valid — just put a = 0. The signs of u, v, a, s indicate directions!

Equation	Quantities Involved	Use When…
v = u + at	v, u, a, t	Displacement (s) is not needed
s = ut + ½at²	s, u, a, t	Final velocity (v) is not needed
v² = u² + 2as	v, u, a, s	Time (t) is not needed

🚗 Worked Example — Car Braking

A car brakes with acceleration = −4 m s⁻², final velocity v = 0. Find stopping distance for:

(i) u = 54 km h⁻¹ = 15 m s⁻¹

Using v² = u² + 2as: 0 = (15)² + 2×(−4)×s → s = 225/8 = 28.1 m

(ii) u = 108 km h⁻¹ = 30 m s⁻¹

Using v² = u² + 2as: 0 = (30)² + 2×(−4)×s → s = 900/8 = 112.5 m

Double the speed → 4× the stopping distance! (s ∝ u²)

🛡️

Road Safety — Why Keep Safe Distance?When speed doubles, stopping distance becomes FOUR times longer (since s = u²/2|a|). This is why speed limits on highways are strictly enforced. India is also developing V2V (Vehicle-to-Vehicle) communication technology that warns drivers of possible collisions!

📝 Step-by-Step Problem Solving Strategy

List all given values — identify u, v, a, s, t from the problem
Convert units — always convert km/h to m/s (multiply by 5/18)
Identify unknown — what are you solving for?
Choose equation — pick the one that has the unknown + 3 known values
Substitute and solve — watch signs! Deceleration/braking → a is negative
Write answer with units — m, m s⁻¹, or m s⁻²

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Motion in a Plane — Uniform Circular Motion (समान वृत्तीय गति)

When an object moves in a 2D plane (two dimensions), we call it motion in a plane. Examples: a kicked football, a satellite orbiting Earth, a vehicle making a turn.

🎠 Circular Motion Basics

When an object moves along a circular path, its motion is called circular motion. For one complete revolution around a circle of radius R:

📏 Distance in One Revolution

= Circumference =2πR

(This is the actual path length travelled)

↗️ Displacement in One Revolution

=Zero!

(Object returns to same starting point)

Average Speed in Circular Motion (one revolution):

v_av = 2πR / T

where T = time period (time for one revolution), R = radius

Average Velocity for one complete revolution = 0 (displacement = 0)

🔄 Uniform Circular Motion (समान वृत्तीय गति)

When an object moves in a circular path at constant speed, it is called uniform circular motion.

⚠️

Tricky Point — Uniform Circular Motion Has Acceleration!

In uniform circular motion, the speed is constant but the direction of velocity changes continuously. Since velocity (direction) changes, there IS acceleration — even though speed doesn’t change! This is a very common exam question.

✅ What is CONSTANT

Speed (magnitude of velocity) is constant throughout. Same value at every point on the circle.

🔄 What CHANGES

Direction of velocity changes continuously. Velocity is always tangent to the circle at that point. Therefore acceleration ≠ 0!

OAvBvCvRUniform circular motion: speed |v| is constant, but direction of v changes at every point → there is acceleration!

🏃

Marble in a Ring — Key Observation!

If a marble is moving in a circular ring and the ring is suddenly lifted, the marble immediately moves in a STRAIGHT LINE (tangent to the circle at that point). This shows that circular motion requires a continuous force — without it, the object continues in a straight line. You’ll learn WHY in the next chapter (Newton’s Laws)!

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Real-World Uniform Circular Motion Examples!

Earth revolving around the Sun (approximately), a satellite in a circular orbit, the tip of a clock hand, a stone being whirled on a string, a car taking a circular turn at constant speed — all are examples of (approximately) uniform circular motion!

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Quick Revision Summary

📍 Position & MotionPosition = distance + direction from reference point. Motion = position changes with time. Rest = position unchanged.

📏 Distance vs DisplacementDistance = total path length (scalar). Displacement = net change in position (vector). |displacement| ≤ distance always.

🚀 Speed vs VelocitySpeed = d/t (scalar). Velocity = s/t (vector, has direction). SI unit: m s⁻¹. Equal only if object moves in ONE direction.

📈 Accelerationa = (v−u)/t. SI unit: m s⁻². +a: speeding up. −a: slowing down. g = 9.8 m s⁻² (free fall, constant).

📊 Position-Time GraphStraight line = constant velocity. Curve = acceleration. Horizontal line = at rest. Slope = velocity.

📉 Velocity-Time GraphHorizontal = constant v, a=0. Rising = +a. Falling = −a. Slope = acceleration. Area under graph = displacement.

🔢 Kinematic Equationsv = u+at | s = ut+½at² | v² = u²+2as. Valid ONLY for constant acceleration!

⭕ Uniform Circular MotionConstant speed, changing direction. v = 2πR/T. Displacement after 1 revolution = 0. Average velocity = 0. Has acceleration!

🔄 Unit Conversionkm h⁻¹ to m s⁻¹: × 5/18. m s⁻¹ to km h⁻¹: × 18/5. Example: 72 km/h = 20 m/s.

⚠️ Common TrapsConstant high speed ≠ acceleration. Displacement can be zero while distance ≠ 0. Circular motion has acceleration even at constant speed!

📝

Important Exam Questions with Answers

Q1. Distinguish between distance and displacement with an example. When are they equal? (CBSE Pattern / 3 Marks)

Ans: Distance is the total length of path covered by an object. It is a scalar quantity (no direction). It is always positive. Displacement is the net change in position — the shortest straight-line distance from initial to final position, with direction. It is a vector quantity. Example: An athlete runs 100 m forward then 60 m back — distance = 160 m, displacement = 40 m forward. Distance and displacement are equal when the object moves in one direction only, without turning back.

Q2. A swimmer completes one length of a 50 m pool in 40 seconds, then swims back. What is the average speed and average velocity for the entire round trip? (CBSE Pattern / 3 Marks)

Ans: Total distance = 50 + 50 = 100 m. Total displacement = 0 m (returned to start). Total time = 40 + 40 = 80 s. Average speed = 100/80 = 1.25 m s⁻¹. Average velocity = 0/80 = 0 m s⁻¹. The average velocity is zero because the swimmer starts and ends at the same point, so net displacement is zero.

Q3. A car starts from rest and reaches 20 m s⁻¹ in 4 seconds. Find the acceleration and distance covered. (CBSE Pattern / 3 Marks)

Ans: Given: u = 0 m s⁻¹, v = 20 m s⁻¹, t = 4 s. Using v = u + at: a = (20 − 0)/4 = 5 m s⁻². Using s = ut + ½at²: s = 0×4 + ½×5×(4)² = 0 + ½×5×16 = 40 m. (Or using v² = u² + 2as: 400 = 0 + 2×5×s → s = 40 m ✓)

Q4. What does the slope of a position-time graph and velocity-time graph represent? (CBSE Pattern / 2 Marks)

Ans: The slope of a position-time graph represents the velocity of the object. A steeper slope means higher velocity. The slope of a velocity-time graph represents the acceleration of the object. A positive slope = positive acceleration; a negative slope = negative acceleration (deceleration). A horizontal line (slope = 0) on the velocity-time graph means zero acceleration (constant velocity).

Q5. In uniform circular motion, is the object accelerating? Explain. (CBSE Pattern / 2 Marks)

Ans: Yes, an object in uniform circular motion IS accelerating, even though its speed is constant. This is because acceleration depends on the change in velocity, not just speed. In uniform circular motion, the direction of velocity continuously changes (it is always along the tangent to the circle). Since velocity (which is a vector, having both magnitude and direction) is continuously changing, there is acceleration. This is called centripetal acceleration.

Q6. A motorbike moving at 28 m s⁻¹ stops after travelling 98 m. Find the acceleration and time to stop. (CBSE Pattern / 3 Marks)

Ans: Given: u = 28 m s⁻¹, v = 0 m s⁻¹, s = 98 m. Using v² = u² + 2as: 0 = (28)² + 2×a×98 → 0 = 784 + 196a → a = −784/196 = −4 m s⁻² (deceleration). Using v = u + at: 0 = 28 + (−4)×t → t = 28/4 = 7 seconds.

Q7. A car travels at constant 72 km h⁻¹. (a) What is its speed in m s⁻¹? (b) If radius of circular turn = 50 m, what is the average speed for one complete turn? Is velocity constant? (CBSE Pattern / 3 Marks)

Ans: (a) 72 km h⁻¹ = 72 × 5/18 = 20 m s⁻¹. (b) For uniform circular motion, speed is constant at 20 m s⁻¹ everywhere — the average speed is still 20 m s⁻¹. However, velocity is NOT constant because the direction of motion keeps changing at every point on the circular path. This is the key feature of circular motion — constant speed but changing velocity (and hence non-zero acceleration).

Q8. A bus is at 36 km h⁻¹ when the driver sees an obstacle 30 m ahead. Reaction time = 0.5 s, braking deceleration = 2.5 m s⁻². Will the bus stop in time? (CBSE Higher Order / 5 Marks)

Ans: u = 36 km h⁻¹ = 10 m s⁻¹, a = −2.5 m s⁻², reaction time = 0.5 s. Distance during reaction time (constant speed): s₁ = 10 × 0.5 = 5 m. After brakes applied (u = 10, v = 0, a = −2.5): using v² = u² + 2as: 0 = 100 + 2×(−2.5)×s₂ → s₂ = 100/5 = 20 m. Total stopping distance = 5 + 20 = 25 m. Since 25 m < 30 m, the bus will stop before the obstacle with 5 m to spare!

Notes Class 9 Science Exploration Chapter 4 Describing Motion Around Us

Amresh Academy

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