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NEET Kinematics MCQs with answers

288 questions on Kinematics for NEET 2027, 2 of them from previous NEET papers. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

easyNEET 2023

1. A body starts from rest, acceleration 4 m/s². Distance in 3rd second is:

  1. 10 m
  2. 12 m
  3. 14 m
  4. 16 m
Show answer and explanation

Answer: (A) 10 m

Distance in the nth second: Sₙ = u + a(2n−1)/2. With u = 0, a = 4 m/s² and n = 3: Sₙ = 0 + 4(5)/2 = 10 m.

Quick tip: Sₙ = u + a(2n−1)/2. Very frequently asked in NEET.

NCERT Class 11, Chapter 3, page 44

easyNEET 2021

2. Horizontal range is maximum when angle is:

  1. 30°
  2. 45°
  3. 60°
  4. 90°
Show answer and explanation

Answer: (B) 45°

R = u²sin2θ/g. Max when sin2θ=1 → θ=45°.

Quick tip: Max range at 45°. Equal range at complementary angles (30°=60°).

NCERT Class 11, Chapter 4, page 67

easy

3. If |A⃗ + B⃗| = |A⃗ - B⃗|, then the angle between vector A⃗ and vector B⃗ is:

  1. 0°
  2. 45°
  3. 180°
  4. 90°
Show answer and explanation

Answer: (D) 90°

Squaring both sides gives A² + B² + 2A·B = A² + B² - 2A·B, which simplifies to 4(A⃗ · B⃗) = 0. Therefore, A⃗ · B⃗ = AB cos θ = 0. Since the vectors are non-zero, cos θ = 0, giving θ = 90°.

medium

4. If the dot product of two non-zero vectors A and B is zero, the vectors are:

  1. Parallel
  2. Antiparallel
  3. Equal
  4. Perpendicular to each other
Show answer and explanation

Answer: (D) Perpendicular to each other

A · B = AB cos θ. For non-zero vectors this is zero only when cos θ = 0, i.e. θ = 90°, so the vectors are perpendicular.

NCERT Class 11

hard

5. A projectile is launched from ground level with speed u at an angle θ above the horizontal. The average velocity of the projectile between its launch point and the highest point of its trajectory has a magnitude equal to:

  1. u √(4 - 3cos²θ) / 2
  2. u √(1 + cos²θ) / 2
  3. u cosθ
  4. u √(1 + 3cos²θ) / 2
Show answer and explanation

Answer: (D) u √(1 + 3cos²θ) / 2

The horizontal displacement up to the highest point is x = R/2 = (u² sin 2θ)/(2g) = (u² sin θ cos θ)/g, and the vertical displacement is y = H = (u² sin²θ)/(2g). The time taken is t = (u sin θ)/g. Therefore, the components of average velocity are v_avg,x = x/t = u cos θ and v_avg,y = y/t = (u sin θ)/2. The magnitude is v_avg = √((u cos θ)² + (u sin θ / 2)²) = (u/2) √(4cos²θ + sin²θ) = (u/2) √(3cos²θ + 1).

medium

6. A particle moves along a straight line. Its velocity-position graph is a straight line passing through (x = 0, v = 20 m/s) and (x = 40 m, v = 0). What is the acceleration of the particle when it is at x = 20 m?

  1. -5.0 m/s²
  2. +5.0 m/s²
  3. -2.5 m/s²
  4. +2.5 m/s²
Show answer and explanation

Answer: (A) -5.0 m/s²

From the given line, the equation of velocity is v = 20 - (20/40)x = 20 - 0.5x. The slope dv/dx is constant and equal to -0.5 s⁻¹. Acceleration is a = v(dv/dx). At x = 20 m, v = 20 - 0.5(20) = 10 m/s. Thus, a = (10)(-0.5) = -5.0 m/s².

easy

7. Which of the following operations between two vectors A⃗ and B⃗ results in a scalar physical quantity?

  1. Vector cross product A⃗ × B⃗
  2. Vector dot product A⃗ · B⃗
  3. Taking the component of A⃗ perpendicular to B⃗
  4. Vector subtraction A⃗ - B⃗
Show answer and explanation

Answer: (B) Vector dot product A⃗ · B⃗

The scalar (or dot) product of two vectors yields a pure scalar magnitude given by AB cos θ. Cross product, vector subtraction, and vector components along any direction retain directional character or result in vectors.

hard

8. Two cars A and B are traveling in the same direction on a straight highway with speeds 25 m/s and 15 m/s, respectively. When car A is 200 m behind car B, driver A applies brakes producing a constant deceleration of 2 m/s². The minimum distance of approach between the two cars is:

  1. 25 m
  2. 150 m
  3. 175 m
  4. 200 m
Show answer and explanation

Answer: (C) 175 m

Relative initial velocity u_rel = 25 - 15 = 10 m/s. Relative acceleration a_rel = -2 m/s². Relative velocity becomes zero when v_rel² = u_rel² + 2 a_rel s_rel, so 0 = 10² - 2(2)s_rel, giving s_rel = 25 m. Since the initial separation was 200 m, the closest approach is 200 - 25 = 175 m.

medium

9. The displacement x of a particle moving in a straight line is related to time t by the equation t = 2x² + 3x. The acceleration of the particle as a function of its velocity v is:

  1. -4v²
  2. -3v²
  3. -2v³
  4. -4v³
Show answer and explanation

Answer: (D) -4v³

Differentiating t = 2x² + 3x with respect to x gives dt/dx = 4x + 3. Since velocity v = dx/dt, we have v = 1/(4x + 3). Acceleration is a = v(dv/dx). We find dv/dx = -4/(4x + 3)² = -4v². Therefore, a = v(-4v²) = -4v³.

medium

10. A ball is projected with speed 10 m/s at 30° to the horizontal. Taking g = 10 m/s², the maximum height it reaches is:

  1. 0.625 m
  2. 5 m
  3. 1.25 m
  4. 2.5 m
Show answer and explanation

Answer: (C) 1.25 m

H = u² sin²θ / 2g = 10² × (sin 30°)² / 20 = 1.25 m.

easy

11. A body starts from rest and moves with a uniform acceleration of 4 m/s². The distance it covers in the first 9 s is:

  1. 162 m
  2. 18 m
  3. 324 m
  4. 36 m
Show answer and explanation

Answer: (A) 162 m

Starting from rest, u = 0, so s = ut + ½at² = 0 + ½ × 4 × 9² = 162 m.

medium

12. A balloon is rising vertically with a constant velocity of 12 m/s. When it is at a height of 65 m above the ground, a packet is released from it. Taking g = 10 m/s², the time taken by the packet to hit the ground is:

  1. 3 s
  2. 4 s
  3. 5 s
  4. 6.5 s
Show answer and explanation

Answer: (C) 5 s

The initial velocity of the packet is u = +12 m/s upwards, and its displacement upon reaching ground is s = -65 m. Using s = ut - (1/2)gt², we get -65 = 12t - 5t², which rearranges to 5t² - 12t - 65 = 0. Factoring gives (5t + 13)(t - 5) = 0. Since t > 0, t = 5 s.

easy

13. A vector of magnitude 10 units makes an angle of 30° with the x-axis. Its x-component is:

  1. 10√3 units
  2. 5√3 units
  3. 5 units
  4. 10 units
Show answer and explanation

Answer: (B) 5√3 units

The x-component is A cos θ = 10 cos 30° = 10 × (√3/2) = 5√3 units. The y-component is A sin θ = 5 units.

NCERT Class 11

medium

14. A vector A⃗ is rotated through an angle θ about an axis perpendicular to the plane containing A⃗ without changing its magnitude. The magnitude of the change in the vector, |ΔA⃗|, is equal to:

  1. |A⃗| (1 - cos θ)
  2. 2 |A⃗| cos(θ/2)
  3. |A⃗| sin θ
  4. 2 |A⃗| sin(θ/2)
Show answer and explanation

Answer: (D) 2 |A⃗| sin(θ/2)

The change in vector is ΔA⃗ = A⃗_final - A⃗_initial. Since |A⃗_final| = |A⃗_initial| = A and the angle between them is θ, |ΔA⃗| = √(A² + A² - 2A² cos θ) = √(2A²(1 - cos θ)) = √(4A² sin²(θ/2)) = 2A sin(θ/2).

easy

15. For uniformly accelerated motion, the correct relation is:

  1. v² = u² + 2as
  2. v = u + 2at
  3. s = ut + at²
  4. v² = u + 2as
Show answer and explanation

Answer: (A) v² = u² + 2as

The kinematic equations for constant acceleration are v = u + at, s = ut + ½at² and v² = u² + 2as.

NCERT Class 11

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