Class 11 Physics – Chapter 3 Motion in a Straight Line (Solved Numericals)

Class 11 Physics – Chapter 3: Motion in a Straight Line

Numericals Book with Detailed Solutions

Q1. A car starts from rest and moves with a uniform acceleration of 2 m/s². Find:

  • (a) Its velocity after 5 s
  • (b) The distance covered in this time

Given:

  • Initial velocity, u = 0 m/s (starts from rest)
  • Acceleration, a = 2 m/s²
  • Time, t = 5 s

To Find:

  • (a) Final velocity, v
  • (b) Distance covered, s

Concept & Formula:

For uniformly accelerated motion in a straight line, we use:

  • v = u + at
  • s = ut + ½at²

Solution:

(a) Final velocity v

  1. Use the formula: v = u + at
  2. Substitute values: v = 0 + (2)(5)
  3. So, v = 10 m/s

Answer (a): The velocity after 5 s is 10 m/s.

(b) Distance covered s

  1. Use the formula: s = ut + ½at²
  2. Since u = 0, the first term becomes zero: s = 0 × t + ½ × 2 × (5)²
  3. So, s = 1 × 25 = 25 m

Answer (b): The distance covered in 5 s is 25 m.

Q2. A stone is dropped from a height of 80 m. Find:

  • (a) The time taken to reach the ground
  • (b) The velocity just before hitting the ground

Given:

  • Initial velocity, u = 0 m/s (dropped, not thrown)
  • Displacement, s = 80 m (downwards)
  • Acceleration, a = g = 9.8 m/s² (downwards)

To Find:

  • (a) Time taken, t
  • (b) Final velocity, v

Concept & Formula:

  • For vertical motion under gravity: s = ut + ½at²
  • And: v² = u² + 2as

Solution:

(a) Time taken t

  1. Use: s = ut + ½at²
  2. Substitute: 80 = 0 × t + ½ × 9.8 × t²
  3. So, 80 = 4.9 t²
  4. t² = 80 / 4.9 ≈ 16.33
  5. t ≈ √16.33 ≈ 4.04 s

Answer (a): Time taken ≈ 4.0 s (approx).

(b) Final velocity v

  1. Use: v² = u² + 2as
  2. Substitute: v² = 0 + 2 × 9.8 × 80
  3. v² = 1568
  4. v = √1568 ≈ 39.6 m/s

Answer (b): Velocity just before hitting the ground ≈ 40 m/s downward.

Q3. A particle moves along a straight line with velocity v = 3t² m/s, where t is in seconds. Find the displacement in the first 2 seconds.

Given:

  • Velocity as a function of time: v(t) = 3t²
  • Time interval: from t = 0 to t = 2 s

To Find:

  • Displacement, s, in first 2 s

Concept & Formula:

Velocity is the rate of change of displacement:

v = ds/dt ⇒ ds = v dt

To find displacement, integrate velocity over time:

s = ∫ v dt

Solution:

  1. We have v(t) = 3t²
  2. So, s = ∫ from 0 to 2 of 3t² dt
  3. Integrate: ∫ 3t² dt = 3 × (t³/3) = t³
  4. Evaluate from 0 to 2: s = (2³) − (0³) = 8 − 0 = 8 m

Answer: Displacement in first 2 s is 8 m.

Q4. A train moves with a uniform velocity of 54 km/h for 10 s and then comes to rest in the next 20 s with uniform retardation. Find:

  • (a) The acceleration during retardation
  • (b) The distance covered during retardation

Given:

  • Initial uniform speed: 54 km/h
  • Convert to m/s: 54 × (1000/3600) = 15 m/s
  • Final velocity after retardation: v = 0 m/s
  • Time of retardation: t = 20 s

To Find:

  • (a) Acceleration a (actually retardation, so negative)
  • (b) Distance s during retardation

Concept & Formula:

  • v = u + at
  • s = ut + ½at²

Solution:

(a) Acceleration a

  1. Use: v = u + at
  2. 0 = 15 + a × 20
  3. a × 20 = −15
  4. a = −15 / 20 = −0.75 m/s²

Answer (a): Acceleration (retardation) = −0.75 m/s².

(b) Distance s during retardation

  1. Use: s = ut + ½at²
  2. Substitute: s = 15 × 20 + ½ × (−0.75) × (20)²
  3. First term: 15 × 20 = 300
  4. Second term: ½ × (−0.75) × 400 = −0.375 × 400 = −150
  5. So, s = 300 − 150 = 150 m

Answer (b): Distance covered during retardation = 150 m.

Q5. Two cars A and B are moving along a straight road in the same direction. Car A is ahead of car B by 50 m. Car A moves with a constant speed of 15 m/s, while car B moves with 10 m/s and then accelerates uniformly at 1 m/s². Find the time after which car B will overtake car A.

Given:

  • Initial separation: 50 m (A ahead of B)
  • Speed of A: vA = 15 m/s (constant)
  • Initial speed of B: uB = 10 m/s
  • Acceleration of B: aB = 1 m/s²

To Find:

  • Time t when B overtakes A (i.e., their positions become equal)

Concept & Setup:

Let us choose the position of car B at t = 0 as x = 0.

  • Then car B’s position: xB(t) = uB t + ½ aB t²
  • Car A is 50 m ahead at t = 0, so its initial position is x = 50 m.
  • Car A moves with constant speed: xA(t) = 50 + vA t

At the instant of overtaking: xA(t) = xB(t)

Solution:

  1. Write positions:
    • xA(t) = 50 + 15t
    • xB(t) = 10t + ½ × 1 × t² = 10t + 0.5t²
  2. Set them equal:

    50 + 15t = 10t + 0.5t²

  3. Rearrange:

    0.5t² + 10t − 15t − 50 = 0

    0.5t² − 5t − 50 = 0

  4. Multiply by 2 to remove decimal:

    t² − 10t − 100 = 0

  5. Solve quadratic: t² − 10t − 100 = 0
  6. Use quadratic formula:

    t = [10 ± √(10² − 4 × 1 × (−100))] / 2

    t = [10 ± √(100 + 400)] / 2

    t = [10 ± √500] / 2

    √500 ≈ 22.36

    So, t = (10 + 22.36)/2 or (10 − 22.36)/2

    t₁ ≈ 32.36/2 ≈ 16.18 s

    t₂ ≈ −12.36/2 ≈ −6.18 s (reject, negative time)

Answer: Car B overtakes car A after approximately 16.2 s.

End of Chapter 3 – Motion in a Straight Line (Detailed Numericals)

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