class 11 notes

Class 11 Physics Notes Chapter 3 (Chapter 3) – Examplar Problems (English) Book

Philoid

Philoid

— 9 min read

Examplar Problems (English)
Alright class, let's get straight into Chapter 3: Motion in a Straight Line from your NCERT Exemplar. This chapter forms the bedrock of mechanics, and understanding it thoroughly is crucial for many government exams that include Physics. We'll focus on the key concepts and problem-solving aspects highlighted in the Exemplar.

Chapter 3: Motion in a Straight Line - Detailed Notes

1. Introduction: Kinematics

  • Kinematics is the branch of physics that describes motion without considering its causes (forces).
  • We study motion of objects along a straight line (rectilinear motion).
  • Object is often treated as a point object if its size is much smaller than the distance it travels.

2. Position, Path Length, and Displacement

  • Position (x): Location of an object with respect to a chosen reference point (origin) on a straight line (coordinate axis). It can be positive or negative.
  • Path Length (Distance): The total length of the actual path traversed by an object between its initial and final positions.
    • It's a scalar quantity (only magnitude).
    • Always non-negative (≥ 0).
  • Displacement (Δx): The change in position of an object. It's the shortest distance between the initial (x₁) and final (x₂) positions.
    • Δx = x₂ - x₁
    • It's a vector quantity (magnitude and direction). Direction is indicated by sign (+ve or -ve along the chosen axis).
    • Can be positive, negative, or zero.
  • Key Point: Magnitude of displacement (|Δx|) is less than or equal to the path length. Equality holds only if the object moves along a straight line without changing direction.

3. Average Velocity and Average Speed

  • Average Velocity (v̄ or ): Ratio of displacement (Δx) to the time interval (Δt) over which the displacement occurred.
    • v̄ = Δx / Δt = (x₂ - x₁) / (t₂ - t₁)
    • It's a vector quantity, having the same direction (sign) as the displacement.
    • Unit: m/s (SI), km/h.
  • Average Speed: Ratio of the total path length (distance) to the total time interval.
    • Average Speed = Total Path Length / Total Time Interval
    • It's a scalar quantity.
    • Always non-negative (≥ 0).
  • Key Point: Average speed is generally greater than or equal to the magnitude of the average velocity. Equality holds only for motion in a straight line without change in direction.

4. Instantaneous Velocity and Instantaneous Speed

  • Instantaneous Velocity (v): The velocity of an object at a particular instant of time (or point in its path). It's the limit of the average velocity as the time interval Δt approaches zero.
    • v = lim (Δt→0) [Δx / Δt] = dx/dt
    • It's the first derivative of position (x) with respect to time (t).
    • Geometrically, it's the slope of the position-time (x-t) graph at that instant.
    • It's a vector quantity.
  • Instantaneous Speed: The magnitude of the instantaneous velocity at a given instant.
    • Speed = |v| = |dx/dt|
    • It's a scalar quantity.

5. Acceleration

6. Kinematic Equations for Uniformly Accelerated Motion

  • These equations are valid ONLY when acceleration 'a' is constant.
  • Let: u = initial velocity (at t=0), v = final velocity (at time t), a = constant acceleration, s = displacement (in time t).
    1. v = u + at (Velocity-time relation)
    2. s = ut + (1/2)at² (Position-time relation)
    3. v² = u² + 2as (Velocity-displacement relation)
    4. s = [(u+v)/2] * t (Displacement as average velocity × time)
    5. s<0xE2><0x82><0x99> = u + a(n - 1/2) (Displacement in the nᵗʰ second)
  • Motion under Gravity: A common case of uniformly accelerated motion (near Earth's surface, neglecting air resistance).
    • Acceleration is 'g' (acceleration due to gravity ≈ 9.8 m/s²).
    • Conventionally:
      • Upward motion: a = -g
      • Downward motion: a = +g (if downward direction is taken as positive) OR a = -g (if upward direction is taken as positive - be consistent!).

7. Graphical Interpretation of Motion

  • Position-Time (x-t) Graph:
    • Slope at any point = Instantaneous velocity.
    • Straight line graph => Uniform velocity (constant slope).
    • Curved graph => Non-uniform velocity.
    • Horizontal line (slope=0) => Object at rest.
    • Line parallel to position axis is not possible (infinite velocity).
  • Velocity-Time (v-t) Graph:
    • Slope at any point = Instantaneous acceleration.
    • Straight line graph => Uniform acceleration.
    • Horizontal line (slope=0) => Uniform velocity (zero acceleration).
    • Area under the v-t graph (between time t₁ and t₂) = Displacement (Δx) during that interval. (Area above time axis is positive displacement, below is negative).
    • Total area under the v-t graph (treating all areas as positive) = Total path length (distance).

8. Relative Velocity

  • Velocity of object A relative to object B (v<0xE2><0x82><0x91><0xE2><0x82><0x82>) is the velocity with which A appears to move when observed from B.
  • For motion in one dimension:
    • v<0xE2><0x82><0x91><0xE2><0x82><0x82> = v<0xE2><0x82><0x91> - v<0xE2><0x82><0x82>
    • v<0xE2><0x82><0x82><0xE2><0x82><0x91> = v<0xE2><0x82><0x82> - v<0xE2><0x82><0x91> = -v<0xE2><0x82><0x91><0xE2><0x82><0x82>
  • Remember to assign signs to v<0xE2><0x82><0x91> and v<0xE2><0x82><0x82> according to the chosen coordinate system.
  • Useful for problems involving objects moving towards/away from each other, overtaking, etc.

Exemplar Specific Focus Points:

  • Problems often involve interpreting graphs carefully, especially non-uniform motion graphs.
  • Questions might test the subtle difference between average speed and magnitude of average velocity.
  • Application of calculus (differentiation and integration) for non-uniform acceleration (though less common in basic government exams, understanding dx/dt = v and dv/dt = a is essential).
  • Relative velocity problems can be slightly more complex.

Multiple Choice Questions (MCQs)

Here are 10 MCQs based on the concepts discussed, suitable for government exam preparation:

  1. A particle moves from point P(2, 3) to Q(4, 1) in the x-y plane. What is its displacement vector?
    a) 2i + 2j
    b) 2i - 2j
    c) -2i + 2j
    d) -2i - 2j
    (Note: Although the chapter is 'Motion in a Straight Line', sometimes basic vector concepts related to displacement are tested.)

  2. A car travels half the distance with constant velocity 40 km/h and the other half with constant velocity 60 km/h along a straight line. The average velocity of the car is:
    a) 48 km/h
    b) 50 km/h
    c) 0 km/h
    d) 100 km/h
    (Hint: Think about average speed vs average velocity here. Since it's along a straight line without turning back, the magnitude of average velocity equals average speed.)

  3. The slope of the velocity-time graph for motion with uniform velocity is:
    a) Positive
    b) Negative
    c) Zero
    d) Variable

  4. A ball is thrown vertically upwards. Which of the following graphs best represents the velocity-time graph of the ball during its flight (neglecting air resistance)?
    a) A parabola opening upwards
    b) A straight line with positive slope
    c) A straight line with negative slope
    d) A horizontal line

  5. The displacement of a particle is given by x = (t - 2)² where x is in metres and t in seconds. The distance covered by the particle in the first 4 seconds is:
    a) 4 m
    b) 8 m
    c) 0 m
    d) 2 m

  6. Two trains A and B are moving on parallel tracks in opposite directions with speeds 36 km/h and 72 km/h respectively. The velocity of train B relative to train A is:
    a) 108 km/h
    b) 36 km/h
    c) -108 km/h
    d) -36 km/h
    (Hint: Choose a direction as positive, convert speeds to m/s if needed, and apply the relative velocity formula.)

  7. If the velocity of a particle is v = At + Bt², where A and B are constants, then the distance travelled by it between 1s and 2s is:
    a) (3/2)A + (7/3)B
    b) A/2 + B/3
    c) (3/2)A + 4B
    d) 3A + 7B

  8. A particle starts from rest and accelerates uniformly. The ratio of the distance covered in the nᵗʰ second to the distance covered in n seconds is:
    a) 2/n - 1/n²
    b) 1/n² - 1/n
    c) 2/n² - 1/n
    d) (2n-1)/n²

  9. The position-time (x-t) graph for a particle in one-dimensional motion is shown. When is the particle's instantaneous velocity zero?
    ![Sample x-t graph: Parabola opening downwards, vertex at t=T] (Assume a graph showing x increasing, reaching a maximum, then decreasing)
    a) At the start (t=0)
    b) At the point where the graph is steepest
    c) At the point where the position x is maximum
    d) Never

  10. Which of the following statements is FALSE for a particle moving in a straight line?
    a) If the velocity is zero at any instant, the acceleration must also be zero at that instant.
    b) If the velocity is zero for a time interval, the acceleration is zero at every instant within that interval.
    c) If speed is constant, acceleration must be zero.
    d) If velocity is constant, acceleration must be zero.

Answer Key for MCQs:

  1. b) 2i - 2j (Displacement = Final Position Vector - Initial Position Vector = (4i + 1j) - (2i + 3j) = 2i - 2j)
  2. a) 48 km/h (Let total distance be 2D. Time t1 = D/40, Time t2 = D/60. Total time T = t1+t2 = D/40 + D/60 = (3D+2D)/120 = 5D/120 = D/24. Average speed = Total Distance / Total Time = 2D / (D/24) = 48 km/h. Since motion is in a straight line without reversal, magnitude of average velocity = average speed).
  3. c) Zero (Uniform velocity means velocity is constant, hence its rate of change (slope of v-t graph, which is acceleration) is zero).
  4. c) A straight line with negative slope (Velocity decreases linearly due to constant downward acceleration 'g'. Taking upward as positive, v = u - gt. This is a straight line equation with negative slope -g).
  5. b) 8 m (Velocity v = dx/dt = 2(t-2). Velocity is zero at t=2s. Particle moves in -ve direction for t<2s and +ve direction for t>2s.
    x(0) = (-2)² = 4m.
    x(2) = (2-2)² = 0m.
    x(4) = (4-2)² = 4m.
    Distance in 0 to 2s = |x(2) - x(0)| = |0 - 4| = 4m.
    Distance in 2 to 4s = |x(4) - x(2)| = |4 - 0| = 4m.
    Total distance = 4m + 4m = 8m).
  6. c) -108 km/h (Let direction of A be positive. v<0xE2><0x82><0x91> = +36 km/h. v<0xE2><0x82><0x82> = -72 km/h. v<0xE2><0x82><0x82><0xE2><0x82><0x91> = v<0xE2><0x82><0x82> - v<0xE2><0x82><0x91> = -72 - 36 = -108 km/h. The magnitude is 108 km/h, the negative sign indicates the direction relative to A).
  7. a) (3/2)A + (7/3)B (Distance s = ∫v dt = ∫(At + Bt²) dt = (1/2)At² + (1/3)Bt³ + C.
    Distance travelled between t=1 and t=2 is s(2) - s(1).
    s(2) = (1/2)A(2)² + (1/3)B(2)³ = 2A + (8/3)B.
    s(1) = (1/2)A(1)² + (1/3)B(1)³ = (1/2)A + (1/3)B.
    s(2) - s(1) = (2A - A/2) + (8B/3 - B/3) = (3/2)A + (7/3)B).
  8. d) (2n-1)/n² (Distance in nᵗʰ second, s<0xE2><0x82><0x99> = u + a(n - 1/2). Since u=0, s<0xE2><0x82><0x99> = a(n - 1/2). Distance in n seconds, s = ut + (1/2)at² = 0 + (1/2)an². Ratio = s<0xE2><0x82><0x99> / s = [a(n - 1/2)] / [(1/2)an²] = [a(2n-1)/2] / [(1/2)an²] = (2n-1)/n²).
  9. c) At the point where the position x is maximum (Instantaneous velocity is the slope of the x-t graph. At the maximum position (peak of the curve), the tangent is horizontal, so the slope is zero).
  10. a) If the velocity is zero at any instant, the acceleration must also be zero at that instant. (This is false. Example: A ball thrown upwards has zero velocity at its highest point, but its acceleration is still 'g' downwards).

Make sure you understand the reasoning behind each answer. Go through these concepts thoroughly, practice problems from the Exemplar book, and you'll be well-prepared. Let me know if any part needs further clarification.

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