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Frames of reference and motion in a straight line: NEET MCQs with answers

51 questions on this topic of Kinematics (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. A ball is thrown vertically upwards from the roof of a building of height 40 m with an initial speed of 10 m/s. Taking g = 10 m/s², the time taken by the ball to reach the ground is:

  1. 4 s
  2. 3 s
  3. 2 s
  4. 5 s
Show answer and explanation

Answer: (A) 4 s

Taking the upward direction as positive and the roof as origin: y = -40 m, u = +10 m/s, a = -10 m/s². Using y = ut + (1/2)at²: -40 = 10t - 5t², which simplifies to 5t² - 10t - 40 = 0, or t² - 2t - 8 = 0. Factoring gives (t - 4)(t + 2) = 0. Since time cannot be negative, t = 4 s.

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2. Given below are two statements: Statement I: A reference frame attached to the surface of the Earth cannot strictly be an inertial frame because of the rotational and orbital motions of the Earth. Statement II: Newton's second law in the form F_net = m a can be directly applied without introducing pseudo forces in any non-inertial frame of reference. In the light of the above statements, choose the most appropriate answer from the options given below:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect

Earth has rotational motion about its axis and revolution around the Sun, both of which involve accelerations, making any frame attached to it non-inertial. Newton's laws in their standard form F_net = m a are valid only in inertial frames; to apply them in a non-inertial frame, pseudo forces must be incorporated. Hence, Statement I is correct and Statement II is incorrect.

hard

3. A particle is projected vertically upwards from ground level. If the times taken by the particle to reach half of its maximum height during its ascent and descent are t₁ and t₂ respectively, measured from the instant of projection, then the ratio t₁ : t₂ is:

  1. 1 : 3
  2. (√2 - 1) : (√2 + 1)
  3. 1 : (√2 + 1)
  4. (2 - √2) : (2 + √2)
Show answer and explanation

Answer: (B) (√2 - 1) : (√2 + 1)

Let the maximum height be H = u²/(2g) and total time of flight be T = 2u/g. At height h = H/2 = u²/(4g), the equation of motion is h = ut - (1/2)gt², so (1/2)gt² - ut + u²/(4g) = 0, or 2g²t² - 4gut + u² = 0. Solving for t: t = [4gu ± √(16g²u² - 8g²u²)] / (4g²) = (u/g)[1 ± 1/√2] = (u/g)(1 - 1/√2) and (u/g)(1 + 1/√2). Thus, t₁/t₂ = (1 - 1/√2) / (1 + 1/√2) = (√2 - 1) / (√2 + 1).

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4. A particle moves along a straight line with an initial velocity of 8 m/s and a uniform acceleration of -2 m/s². The total distance covered by the particle in the first 5 seconds of its motion is:

  1. 17 m
  2. 15 m
  3. 16 m
  4. 25 m
Show answer and explanation

Answer: (A) 17 m

The particle comes to rest momentarily when v = u + at = 8 - 2t = 0, giving t = 4 s. Distance covered in the first 4 s is s₁ = 8(4) - (1/2)(2)(4)² = 16 m. In the 5th second (from t = 4 s to t = 5 s), the distance covered in the reverse direction is s₂ = (1/2)(2)(1)² = 1 m. Hence, total distance = 16 + 1 = 17 m.

easy

5. A stone dropped from the top of a tower of height H reaches the ground in time T. The height of the stone above the ground at time t = T/2 is:

  1. H/4
  2. H/2
  3. 3H/4
  4. H/8
Show answer and explanation

Answer: (C) 3H/4

Distance fallen from rest in time t is y = (1/2)gt². In time T, H = (1/2)gT². In time T/2, distance fallen is y' = (1/2)g(T/2)² = H/4. Therefore, the height of the stone remaining above the ground is H - H/4 = 3H/4.

hard

6. A bullet loses 1/20 of its velocity after penetrating a distance of 3 cm into a wooden plank. Assuming uniform retardation, the total thickness of the plank required to bring the bullet completely to rest from its initial state is:

  1. 28.5 cm
  2. 40.0 cm
  3. 32.0 cm
  4. 30.8 cm
Show answer and explanation

Answer: (D) 30.8 cm

Let initial velocity be u. After x = 3 cm, v = u(1 - 1/20) = 19u/20. Using v² - u² = 2ax, (361/400 - 1)u² = -39u²/400 = 2a(3), so 2a = -13u²/400. For complete rest, 0² - u² = 2aX. Hence, -u² = (-13u²/400)X, giving X = 400/13 ≈ 30.77 cm ≈ 30.8 cm.

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7. A body starts from rest with uniform acceleration 2 m/s². The distance it covers in the 5th second is:

  1. 10 m
  2. 5 m
  3. 9 m
  4. 25 m
Show answer and explanation

Answer: (C) 9 m

Sₙ = u + a(2n - 1)/2 = 0 + 2(10 - 1)/2 = 9 m.

NCERT Class 11

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8. Two cars P and Q start from the same point at the same time in a straight line and their positions are represented by x_P(t) = αt + βt² and x_Q(t) = ft - t². At what time do the cars have the same velocity?

  1. (α - f) / (1 + β)
  2. (f + α) / (2(β - 1))
  3. (f - α) / (2(β + 1))
  4. (f - α) / (β + 1)
Show answer and explanation

Answer: (C) (f - α) / (2(β + 1))

Velocity of car P is v_P = dx_P/dt = α + 2βt. Velocity of car Q is v_Q = dx_Q/dt = f - 2t. Equating the two velocities: α + 2βt = f - 2t, which yields 2t(β + 1) = f - α, so t = (f - α) / (2(β + 1)).

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9. A particle starts from rest at t = 0 and moves in a straight line with an acceleration varying with time as a = (4 - 2t) m/s², where t is in seconds. The maximum velocity attained by the particle is:

  1. 2 m/s
  2. 8 m/s
  3. 4 m/s
  4. 16 m/s
Show answer and explanation

Answer: (C) 4 m/s

Velocity is obtained by integrating acceleration: v(t) = ∫(4 - 2t) dt = 4t - t² (since v(0) = 0). The maximum velocity occurs when dv/dt = a = 0, which gives 4 - 2t = 0, so t = 2 s. At t = 2 s, v_max = 4(2) - (2)² = 8 - 4 = 4 m/s.

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10. Read the two statements regarding one-dimensional motion: Statement I: If the speed of a particle is constant, its acceleration along a straight line must be zero. Statement II: A particle moving in a straight line can have an instantaneous velocity of zero while having a non-zero acceleration. Choose the correct option:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct

In one dimension, direction is fixed to forward or backward; constant speed means velocity is constant, so acceleration is zero, making Statement I correct. At the highest point of vertical projection under gravity, instantaneous velocity is zero while acceleration is g downwards, so Statement II is also correct.

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11. A particle moves along the x-axis such that its position varies with time according to x = 2t³ - 9t² + 12t - 4, where x is in metres and t is in seconds. The acceleration of the particle at the instant when its velocity becomes zero for the second time is:

  1. -6 m/s²
  2. +6 m/s²
  3. -12 m/s²
  4. +12 m/s²
Show answer and explanation

Answer: (B) +6 m/s²

Velocity is v = dx/dt = 6t² - 18t + 12 = 6(t - 1)(t - 2). Thus v = 0 at t = 1 s and t = 2 s (the second time). Acceleration is a = dv/dt = 12t - 18. At t = 2 s, a = 12(2) - 18 = +6 m/s².

medium

12. Two particles A and B are projected vertically upwards from the same point on the ground with the same initial speed u = 30 m/s, but particle B is thrown 2 seconds after particle A. Neglecting air resistance and taking g = 10 m/s², at what height above the ground will the two particles collide?

  1. 25 m
  2. 35 m
  3. 45 m
  4. 40 m
Show answer and explanation

Answer: (D) 40 m

Let time t be measured from the moment A is launched. Particle A's position is y_A = 30t - 5t². Particle B is thrown at t = 2 s, so y_B = 30(t - 2) - 5(t - 2)². Setting y_A = y_B: 30t - 5t² = 30t - 60 - 5(t² - 4t + 4) = 30t - 60 - 5t² + 20t - 20, which simplifies to 0 = 20t - 80, so t = 4 s. The height at this collision time is y = 30(4) - 5(4)² = 120 - 80 = 40 m.

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