Motion in a Straight Line – Kinematics Equations of Uniformly and Non-uniformly: NEET MCQs with Answers

15 free questions on Motion in a Straight Line – Kinematics Equations of Uniformly and Non-uniformly for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Kinematics Equations of Uniformly and Non-uniformly Accelerated Motion: uniformly accelerated motion, velocity-time, position-time

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Question 1

Previous-year question
A particle moves along the x-axis with position x = αt⁴+βt²+γt+δ. The ratio of its initial velocity to its initial acceleration is: [RE-NEET(UG) 2024]
  1. 2α:δ
  2. γ:2δ
  3. 4α:β
  4. γ:2β
Show answer
Answer: (D) γ:2β
Explanation: v = 4αt³+2βt+γ; at t=0, v=γ. a = 12αt²+2β; at t=0, a=2β. Ratio v:a = γ:2β.

Question 2

Previous-year question
An object falls freely from height h above the ground. It travels (5/9)h in the last 1 second. Then height h is (g=10 m/s²): [NEET(UG) 2024 (Sawaimadhopur)]
  1. 25 m
  2. 45 m
  3. 58 m
  4. 5 m
Show answer
Answer: (B) 45 m
Explanation: Let total fall time be T, so h = 5T². Distance in the last second = g(2T−1)/2 = 5(2T−1), and this equals (5/9)h = (25T²)/9. Setting 5(2T−1) = 25T²/9 leads to 5T²−18T+9=0, giving T=3 s (taking the physical root). Then h = 5(3²) = 45 m.

Question 3

Previous-year question
A stone is thrown vertically downward with an initial velocity of 40 m/s from a building. It reaches the ground with velocity 60 m/s. The height of the building is (g=10 m/s²): [NEET(UG) 2022 (Overseas)]
  1. 140 m
  2. 80 m
  3. 100 m
  4. 120 m
Show answer
Answer: (C) 100 m
Explanation: v² = u² + 2gh: 60² = 40² + 2(10)h → 3600−1600 = 20h → h = 100 m.

Question 4

Previous-year question
A particle starts from the origin with zero initial velocity along the x-axis, with velocity and displacement related by v=ax⁻². The displacement after time t is: [NEET(UG) 2021 (Paper-2)]
  1. (at)^(1/2)
  2. (3at)^(1/3)
  3. (2at)^(1/2)
  4. (2at)^(1/3)
Show answer
Answer: (B) (3at)^(1/3)
Explanation: v=dx/dt=ax⁻² gives x²dx=a dt. Integrating (x=0 at t=0): x³/3=at, so x=(3at)^(1/3).

Question 5

Previous-year question
A ball is thrown vertically downward with a velocity of 20 m/s from the top of a tower. It hits the ground after some time with a velocity of 80 m/s. The height of the tower is (g = 10 m/s²) [NEET 2020]
  1. 340 m
  2. 320 m
  3. 300 m
  4. 360 m
Show answer
Answer: (C) 300 m
Explanation: Using v² = u² + 2gh, 6400 = 400 + 2(10)h, so 20h = 6000 and h = 300 m.

Question 6

Previous-year question
A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field E. Due to the force qE its velocity increases from 0 to 6 m/s in one second. At that instant the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 and 3 seconds are respectively [NEET 2018]
  1. 2 m/s, 4 m/s
  2. 1 m/s, 3 m/s
  3. 1.5 m/s, 3 m/s
  4. 1 m/s, 3.5 m/s
Show answer
Answer: (B) 1 m/s, 3 m/s
Explanation: The acceleration is +6 m/s² for the first second and −6 m/s² thereafter. The displacements in the three successive seconds are +3 m, +3 m and −3 m, giving a net displacement of 3 m and a total path length of 9 m. Hence the average velocity = 3/3 = 1 m/s and the average speed = 9/3 = 3 m/s.

Question 7

Previous-year question
If the velocity of a particle is v = At + Bt², where A and B are constants, then the distance travelled by it between 1 s and 2 s is [NEET 2016]
  1. (3/2)A + 4B
  2. 3A + 7B
  3. (3/2)A + (7/3)B
  4. A/2 + B/3
Show answer
Answer: (C) (3/2)A + (7/3)B
Explanation: The distance is ∫v dt from 1 to 2 = [At²/2 + Bt³/3] evaluated between the limits = A(4 − 1)/2 + B(8 − 1)/3 = (3/2)A + (7/3)B.
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Question 8

Previous-year question
A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x) = bx⁻²ⁿ, where b and n are constants and x is the position of the particle. The acceleration of the particle as a function of x is given by [AIPMT 2015]
  1. −2nb²x⁻⁴ⁿ⁻¹
  2. −2b²x⁻²ⁿ⁺¹
  3. −2nb²e⁻⁴ⁿ⁺¹
  4. −2nb²x⁻²ⁿ⁻¹
Show answer
Answer: (A) −2nb²x⁻⁴ⁿ⁻¹
Explanation: Acceleration a = v(dv/dx). Here dv/dx = −2nb x⁻²ⁿ⁻¹, so a = (bx⁻²ⁿ)(−2nb x⁻²ⁿ⁻¹) = −2nb²x⁻⁴ⁿ⁻¹.

Question 9

Previous-year question
From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle to hit the ground is n times that taken by it to reach the highest point of its path. The relation between H, u and n is [AIPMT 2014]
  1. 2gH = n²u²
  2. gH = (n − 2)²u²
  3. 2gH = nu²(n − 2)
  4. gH = (n − 2)u²
Show answer
Answer: (C) 2gH = nu²(n − 2)
Explanation: Time to the top is u/g, so the total time is nu/g. Taking upward as positive, −H = u(nu/g) − ½g(nu/g)², which gives −H = nu²/g − n²u²/(2g). Rearranging yields 2gH = nu²(n − 2).

Question 10

Previous-year question
A car is moving along a straight road with a uniform acceleration. It passes through two points P and Q, separated by a distance, with velocities of 30 km/h and 40 km/h respectively. The velocity of the car midway between P and Q is [AIPMT 1988]
  1. 33.3 km/h
  2. 20√2 km/h
  3. 25√2 km/h
  4. 35 km/h
Show answer
Answer: (C) 25√2 km/h
Explanation: Let PQ = x and let R be the midpoint, so PR = x/2. For the whole stretch PQ: (40)² − (30)² = 2ax ⇒ 1600 − 900 = 2ax ⇒ 2ax = 700, i.e. ax = 350 For the first half PR: v² − (30)² = 2a(x/2) = ax = 350 v² = 900 + 350 = 1250 v = √1250 = 25√2 km/h ≈ 35.36 km/h Note carefully that the midpoint velocity is the root-mean-square of the two velocities, √[(30² + 40²)/2], and not their arithmetic mean 35 km/h — that value is the velocity at the mid-time, not the mid-point.

Question 11

Previous-year question
What will be the ratio of the distance moved by a freely falling body from rest in the 4th and 5th second of its journey? [AIPMT 1989]
  1. 4 : 5
  2. 7 : 9
  3. 16 : 25
  4. 1 : 1
Show answer
Answer: (B) 7 : 9
Explanation: For a body falling freely from rest, the distance covered in the nth second is sₙ = ½g(2n − 1) s₄ = ½g(2 × 4 − 1) = (7/2)g s₅ = ½g(2 × 5 − 1) = (9/2)g Hence s₄ : s₅ = 7 : 9. The distractor 16 : 25 is the ratio of the total distances fallen in the first 4 s and the first 5 s, since total distance varies as t².

Question 12

Previous-year question
A body dropped from the top of a tower falls through 40 m during the last two seconds of its fall. Taking g = 10 m/s², the height of the tower is [AIPMT 1991]
  1. 60 m
  2. 45 m
  3. 80 m
  4. 50 m
Show answer
Answer: (B) 45 m
Explanation: Let t be the total time of fall. The distance covered in the last 2 s equals the total distance minus the distance covered in the first (t − 2) s: 40 = ½gt² − ½g(t − 2)² 40 = ½(10)[t² − (t − 2)²] 40 = 5[t² − (t² − 4t + 4)] 40 = 5(4t − 4) = 20t − 20 20t = 60 ⇒ t = 3 s Height of the tower: h = ½gt² = ½(10)(3)² = 45 m.

Question 13

Previous-year question
A body starts from rest. What is the ratio of the distance travelled by the body during the 4th and the 3rd second? [AIPMT 1993]
  1. 7/5
  2. 5/7
  3. 7/3
  4. 3/7
Show answer
Answer: (A) 7/5
Explanation: The distance covered in the nth second, for a body starting from rest, is sₙ = u + (a/2)(2n − 1) = (a/2)(2n − 1) s₄ = (a/2)(2 × 4 − 1) = (a/2)(7) s₃ = (a/2)(2 × 3 − 1) = (a/2)(5) Dividing, the factor a/2 cancels: s₄/s₃ = 7/5. The result is independent of the value of the acceleration.

Question 14

Previous-year question
The velocity of a train increases uniformly from 20 km/h to 60 km/h in 4 h. The distance travelled by the train during this period is [AIPMT 1994]
  1. 160 km
  2. 180 km
  3. 100 km
  4. 120 km
Show answer
Answer: (A) 160 km
Explanation: Because the acceleration is uniform, the average velocity is simply the mean of the initial and final velocities: v_avg = (20 + 60)/2 = 40 km/h Distance = v_avg × t = 40 × 4 = 160 km. Equivalently, a = (60 − 20)/4 = 10 km/h² and s = ut + ½at² = 20(4) + ½(10)(16) = 80 + 80 = 160 km. ✔

Question 15

Previous-year question
A stone released with zero velocity from the top of a tower reaches the ground in 4 s. Taking g = 10 m/s², the height of the tower is [AIPMT 1995]
  1. 20 m
  2. 40 m
  3. 80 m
  4. 160 m
Show answer
Answer: (C) 80 m
Explanation: The stone is released from rest, so u = 0 and the acceleration is g downwards. h = ut + ½gt² = 0 + ½(10)(4)² = 5 × 16 = 80 m.

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