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Motion in a Straight Line – Class 11 Physics Chapter 2 Notes & NCERT PDF

Class 11PhysicsChapter 2NCERT book: Physics Part I - Class 11

Notes and a simple summary of Chapter 2 of the NCERT Class 11 Physics book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 2: Motion in a Straight Line

Introduction

Everything in the universe moves — we walk, blood flows, the Earth rotates and revolves, and even the Sun moves within the Milky Way. Motion is the change in position of an object with time. In this chapter we describe rectilinear motion (motion along a straight line) using velocity and acceleration, and derive the kinematic equations for uniformly accelerated motion. We treat moving bodies as point objects, which is valid when an object's size is much smaller than the distance it moves. Studying motion without its causes is called kinematics; the causes come in Chapter 4 (Laws of Motion).

Note: the chapter outline lists "2.5 Relative velocity", but the rationalised text in this edition does not develop that section; its main content is sections 2.2–2.4.

Key Concepts

2.2 Instantaneous velocity and speed

  • Average velocity tells how fast a body moved over an interval; instantaneous velocity tells how fast it moves at an instant:

    • - v = lim(Δt→0) Δx/Δt = dx/dt.

  • Graphically, v at an instant = slope of the tangent to the position–time (x–t) graph at that instant.
  • Numerical illustration (Table 2.1): for x = 0.08 t³, shrinking Δt from 2.0 s to 0.01 s around t = 4.0 s makes Δx/Δt approach 3.84 m s⁻¹, the value of dx/dt = 0.24 t² at t = 4 s.
  • Speed is the magnitude of velocity: +24.0 m s⁻¹ and −24.0 m s⁻¹ both have speed 24.0 m s⁻¹. Over an interval, average speed ≥ |average velocity|, but instantaneous speed = |instantaneous velocity|.

    • 2.3 Acceleration

  • Galileo found that for freely falling bodies the rate of change of velocity with time is constant (not with distance) — hence acceleration is defined per unit time.
  • Average acceleration ā = (v₂ − v₁)/(t₂ − t₁) = Δv/Δt; instantaneous acceleration a = dv/dt = slope of the v–t graph. SI unit: m s⁻².
  • Acceleration can come from a change in speed, a change in direction, or both; it can be positive, negative or zero.
  • x–t graph shapes: curves upward for positive acceleration, downward for negative, a straight line for zero acceleration.
  • Area under the v–t curve = displacement over that interval (for constant velocity u, area = uT).
  • Real graphs are smooth — velocity and acceleration cannot change abruptly.

    • 2.4 Kinematic equations (constant acceleration)

    From v = v₀ + at and the area under the v–t graph:

  • v = v₀ + at
  • x = v₀t + ½at²
  • v² = v₀² + 2ax
  • x = ((v + v₀)/2) t — the average velocity is the mean of initial and final velocities (constant acceleration only).
  • If the body starts at x₀, replace x by (x − x₀): x = x₀ + v₀t + ½at².
  • The calculus derivation (Example 2.2) integrates dv = a dt and dx = v dt; the method v dv = a dx also works for non-uniform acceleration.

    • Free fall (Example 2.4)

    Taking upward as positive, a = −g = −9.8 m s⁻² and, starting from rest at y = 0:

  • v = −9.8 t m s⁻¹; y = −4.9 t² m; v² = −19.6 y m² s⁻².

    • Galileo's law of odd numbers (Example 2.5)

    Distances fallen from rest in successive equal time intervals are in the ratio 1 : 3 : 5 : 7 : 9 … because positions after τ, 2τ, 3τ … are proportional to 1, 4, 9, 16 …

    Stopping distance and reaction time

  • Stopping distance d_s = v₀²/(2a) (with braking deceleration a) — it grows as the square of the initial speed: doubling speed makes it four times longer. Measured values of 10, 20, 34 and 50 m at 11, 15, 20 and 25 m s⁻¹ roughly follow this.
  • Reaction time from a falling ruler: d = ½ g t_r² ⇒ t_r = √(2d/g). For d = 21.0 cm, t_r ≈ 0.21 s.

    • Worked Examples

  • Velocity from position (Example 2.1): x = a + bt², a = 8.5 m, b = 2.5 m s⁻². v = dx/dt = 2bt = 5.0t m s⁻¹, so v = 0 at t = 0 and v = 10 m s⁻¹ at t = 2.0 s. Average velocity from 2.0 s to 4.0 s = [x(4) − x(2)]/2.0 = 6.0 b = 15 m s⁻¹.
  • Ball thrown up from a building (Example 2.3): v₀ = 20 m s⁻¹ from 25.0 m, g = 10 m s⁻². (a) 0 = 20² − 2(10)(y − y₀) ⇒ rises 20 m. (b) 0 = 25 + 20t − 5t² ⇒ 5t² − 20t − 25 = 0 ⇒ t = 5 s. Split method: 2 s up to the top (45 m above ground) + 3 s to fall 45 m = 5 s.
  • Car stopping (Ex. 2.5): 126 km h⁻¹ = 35 m s⁻¹ stopped in 200 m. Retardation = 35²/(2 × 200) ≈ 3.06 m s⁻²; time = 35/3.06 ≈ 11.4 s.
  • Ball thrown up at 29.4 m s⁻¹ (Ex. 2.6): height = 29.4²/(2 × 9.8) = 44.1 m; time up = 3 s, so it returns after 6 s. At the top v = 0 but a = g downward.
  • Market walk (Ex. 2.10): 2.5 km at 5 km h⁻¹ and back at 7.5 km h⁻¹. 0–30 min: average velocity = average speed = 5 km h⁻¹. 0–50 min: back home, average velocity 0, average speed 5 km ÷ (5/6 h) = 6 km h⁻¹. 0–40 min: displacement 1.25 km, average velocity 1.875 km h⁻¹, average speed 5.625 km h⁻¹.

    • Formula / Key-Facts Sheet

  • v = dx/dt; a = dv/dt
  • ā = Δv/Δt; average speed ≥ |average velocity|
  • v = v₀ + at; x − x₀ = v₀t + ½at²; v² = v₀² + 2a(x − x₀)
  • x = ½(v + v₀)t (constant a)
  • Area under v–t graph = displacement; slope of x–t = velocity; slope of v–t = acceleration
  • Free fall from rest: v = −gt, y = −½gt², v² = −2gy
  • Stopping distance d_s = v₀²/2a; reaction time t_r = √(2d/g)

    • Exam Focus

  • Graph reading (Ex. 2.2): from x–t graphs decide who lives closer, starts earlier, walks faster, and how many times one overtakes the other. Outline: compare intercepts, slopes and crossings.
  • Statements with reasons (Ex. 2.7): a particle with zero speed may have non-zero acceleration (top of a vertical throw) ✔; constant speed in one dimension means zero acceleration ✔; positive acceleration need not mean speeding up (it depends on the sign of velocity).
  • Distance vs displacement, average speed vs average velocity (Ex. 2.9): path length ≥ |displacement|, equality only when the body moves in one direction without turning back.
  • Numericals with the three equations — always fix the origin and positive direction first and use signs consistently.

    • Key Terms

  • Rectilinear motion (सरल रेखीय गति)
  • Point object (बिंदु वस्तु)
  • Kinematics (शुद्धगतिकी)
  • Displacement (विस्थापन)
  • Instantaneous velocity (तात्क्षणिक वेग)
  • Speed (चाल)
  • Average speed (औसत चाल)
  • Acceleration (त्वरण)
  • Retardation (मंदन)
  • Uniformly accelerated motion (एकसमान त्वरित गति)
  • Free fall (मुक्त पतन)
  • Position–time graph (स्थिति-समय ग्राफ)
  • Velocity–time graph (वेग-समय ग्राफ)
  • Stopping distance (अवरोधन दूरी)
  • Reaction time (प्रतिक्रिया काल)

    • Common Mistakes

  • Assuming zero velocity means zero acceleration — at the top of a throw v = 0 but a = g.
  • Thinking negative acceleration always means slowing down; its sign depends on the chosen positive direction.
  • Using the kinematic equations when acceleration is not constant — they hold only for constant acceleration.
  • Mixing up the sign of g: if upward is positive, a = −g even when the body is falling.
  • Reporting average speed as zero for a round trip; that is the average velocity — average speed is total path ÷ time.
  • Taking the slope of a v–t graph as displacement instead of acceleration (displacement is the area).

    • 💡 Key Learning Points

    • ✓Instantaneous velocity v = dx/dt is the slope of the x–t graph; speed is its magnitude
    • ✓Acceleration a = dv/dt is the slope of the v–t graph; the area under the v–t graph is the displacement
    • ✓For constant acceleration: v = v₀ + at, x = v₀t + ½at², v² = v₀² + 2ax
    • ✓In free fall (upward positive) a = −g; distances in successive equal intervals are in the ratio 1 : 3 : 5 : 7 …
    • ✓Stopping distance ∝ v₀² and reaction time t_r = √(2d/g) apply kinematics to road safety

    👨‍🏫 Teaching Tips

    • →Recreate Table 2.1 on a spreadsheet with x = 0.08t³ and let students watch Δx/Δt converge to 3.84 m s⁻¹
    • →Do the ruler-drop reaction-time experiment in pairs and compute t_r = √(2d/g) from each student's catch distance
    • →Sketch the four v–t cases of Fig. 2.3 and have students describe the motion in words before revealing captions
    • →Solve Example 2.3 both ways (split path vs single equation) to show why a single equation with signs is safer
    • →Use the car braking data (10, 20, 34, 50 m at 11, 15, 20, 25 m s⁻¹) to test d ∝ v² and discuss school-zone speed limits
    • →Insist on an explicit 'origin and positive direction' line at the top of every numerical solution

    📋 Assessment Questions

    1. Define instantaneous velocity and explain how to find it from an x–t graph
    2. Derive v² = v₀² + 2ax from the other kinematic equations
    3. A ball thrown up at 29.4 m s⁻¹: find the maximum height and time of return
    4. Show that distances in free fall over equal intervals follow 1 : 3 : 5 : 7
    5. Can a particle have zero speed and non-zero acceleration? Give an example

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    Keep learning

    All chapters in Physics Part I - Class 11

    1. 1. Units and Measurement
    2. 2. Motion in a Straight Line
    3. 3. Motion in a Plane
    4. 4. Laws of Motion
    5. 5. Work, Energy and Power
    6. 6. System of Particles and Rotational Motion
    7. 7. Gravitation