The slope of the position–time (x-t) graph at any instant gives the instantaneous velocity. A straight x-t line implies uniform velocity; a curved x-t line implies non-zero acceleration.
-- NCERT Class 11 Physics, Ch. 2, p. 14Uniform Non-Uniform Motion
8 MCQs1 revision card9-step worked example
Source: NCERT KinematicsOfficial key: NTA-verifiedLast updated: 25 Sep 2026
Uniform Non-Uniform Motion, explained for NEET
NEET rarely asks "define uniform motion" as a standalone recall item — but it routinely punishes you for misclassifying motion type before choosing your formula. The real cost of this topic shows up inside other problems: you grab v = v₀ + at for a situation where acceleration is changing, and you lose 5 marks (4 lost + 1 negative) without even realising the root cause.
Uniform motion means equal displacements in equal time intervals, however small those intervals are (NCERT Class 11 Physics Chapter 2, page 14). The velocity is constant — both magnitude and direction stay fixed. The position-time graph is a straight line. No net force acts along the direction of motion.
Non-uniform motion means unequal displacements in equal time intervals. Velocity changes — in magnitude, direction, or both. This includes:
- Uniformly accelerated motion: acceleration is constant (the three kinematic equations apply here and only here).
- Non-uniformly accelerated motion: acceleration itself changes with time or position. Here you must integrate a(t) or use v dv = a dx. The standard kinematic equations are invalid.
The classification hierarchy matters:
| Motion type | Velocity | Acceleration | Kinematic eqs valid? |
|---|---|---|---|
| Uniform | Constant | Zero | Trivially (a = 0) |
| Uniformly accelerated | Changes at constant rate | Constant ≠ 0 | Yes |
| Non-uniformly accelerated | Changes at variable rate | Variable | No — integrate |
The trap NEET exploits: a problem states acceleration varies with time (e.g., a = 2t) and a distractor offers the answer you get by plugging into v = v₀ + at with a = 2t evaluated at one instant. That is wrong — you need ∫a dt. Recognising the motion type is the first checkpoint before any calculation.
Can you answer these Uniform Non-Uniform Motion MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
A car travels along a straight road covering 20 m in each successive 1-second interval. The motion of the car is:
Show answer and why every option is right or wrong
Answer: A. Equal displacements (20 m) in equal time intervals (1 s) is the definition of uniform motion (NCERT Class 11 Physics Chapter 2, page 14). Velocity is constant at 20 m/s with zero acceleration.
Why B is wrong: B is wrong because uniformly accelerated motion requires velocity to change at a constant rate. Here the displacement per second is constant, so velocity is constant and acceleration is zero.
Why C is wrong: C is wrong because non-uniformly accelerated motion requires acceleration to change. Here there is no acceleration at all — the car covers equal distances in equal times.
Why D is wrong: D is wrong because the speed is constant at 20 m/s. 'Variable speed' would require unequal distances in equal time intervals.
MCQ 2Easy RecallPractice
Which of the following is a necessary condition for uniform motion along a straight line?
Show answer and why every option is right or wrong
Answer: B. Uniform motion means constant velocity. By Newton's first law, constant velocity requires zero net force along the direction of motion. This is the defining mechanical condition for uniform straight-line motion.
Why A is wrong: A is wrong because constant non-zero acceleration means velocity is changing — that is uniformly accelerated motion, not uniform motion.
Why C is wrong: C is wrong because if velocity changes direction, the motion is not uniform. Uniform motion requires both magnitude and direction of velocity to remain constant.
Why D is wrong: D is wrong because the 1:3:5:7 ratio is the hallmark of uniformly accelerated motion from rest (Galileo's odd-number rule), not uniform motion. Uniform motion gives equal distances in equal intervals: 1:1:1:1.
MCQ 3Direct ApplicationPractice
An object moves along the x-axis with position given by x = 5t² + 3t + 2 (in SI units). The motion is:
Show answer and why every option is right or wrong
Answer: A. Velocity v = dx/dt = 10t + 3, and acceleration a = dv/dt = 10 m/s² (constant). Since acceleration is constant and non-zero, this is uniformly accelerated motion. The three kinematic equations are valid here.
Why B is wrong: B is wrong because v = 10t + 3 is not constant — it increases with time. Uniform motion requires constant velocity (zero acceleration).
Why C is wrong: C is wrong because the acceleration a = d²x/dt² = 10 m/s² is constant, not variable. Non-uniformly accelerated motion would require a to depend on t or x.
Why D is wrong: D is wrong because the position function x(t) already contains all the information needed. Differentiating gives velocity and acceleration directly — no additional initial conditions are required.
MCQ 4Direct ApplicationPractice
The acceleration of a particle moving along a straight line is given by a = 4t m/s². The kinematic equation v = v₀ + at:
Show answer and why every option is right or wrong
Answer: C. The equation v = v₀ + at requires constant acceleration. Here a = 4t varies with time, so the motion is non-uniformly accelerated. The correct approach is to integrate: v = v₀ + ∫₀ᵗ 4t′ dt′ = v₀ + 2t².
Why A is wrong: A is wrong because substituting a = 4t into v = v₀ + at gives v = v₀ + 4t² — this is incorrect. The formula v = v₀ + at was derived assuming 'a' is constant throughout the motion. When a varies, you must integrate a(t) with respect to time (trap: applying constant-acceleration formulas to variable-acceleration problems).
Why B is wrong: B is wrong because there is no valid 'average time' substitution that rescues this formula. The correct method is direct integration of a(t).
Why D is wrong: D is wrong because at t = 0 the equation trivially gives v = v₀ regardless of acceleration — this tells you nothing useful about the motion and is not a meaningful 'application' of the kinematic equation.
MCQ 5Easy RecallPractice
A body moves along a straight line. Its velocity-time graph is a straight line with positive slope. The motion is:
Show answer and why every option is right or wrong
Answer: C. A straight-line v-t graph with positive slope means velocity increases linearly with time. The slope of the v-t graph is acceleration, and a constant slope means constant acceleration. This is the definition of uniformly accelerated motion.
Why A is wrong: A is wrong because uniform motion has constant velocity — its v-t graph is a horizontal line (zero slope), not a line with positive slope.
Why B is wrong: B is wrong because a straight-line v-t graph has constant slope, meaning constant acceleration. Non-uniformly accelerated motion would show a curved v-t graph (changing slope).
Why D is wrong: D is wrong because 'uniform motion' by definition means constant velocity. 'Uniform motion with increasing speed' is a contradiction in terms.
MCQ 6Direct ApplicationPractice
A particle has position x = 3t³ − 2t (SI units). At t = 1 s, the acceleration is:
Show answer and why every option is right or wrong
Answer: B. v = dx/dt = 9t² − 2; a = dv/dt = 18t. At t = 1 s: a = 18 × 1 = 18 m/s². Note that since a = 18t depends on time, this is non-uniformly accelerated motion — the standard kinematic equations (which assume constant a) would not apply here.
Why A is wrong: A is wrong — this comes from evaluating 9t² at t = 1, which gives the coefficient of the velocity expression, not the acceleration. Acceleration requires a second derivative: a = d²x/dt² = 18t.
Why C is wrong: C is wrong — this comes from evaluating v = 9t² − 2 at t = 1 (giving 7 m/s), which is the velocity, not the acceleration. Velocity and acceleration are different derivatives of position.
Why D is wrong: D is wrong — 16 m/s² comes from differentiating v = 9t² − 2 but keeping the −2 (a = 18t − 2). The derivative of a constant is zero, so a = 18t = 18 m/s² at t = 1 s.
MCQ 7Concept TrapPractice
For which of the following motions can the equation s = v₀t + ½at² be correctly used?
Show answer and why every option is right or wrong
Answer: B. Free fall near Earth's surface (air resistance neglected) has constant acceleration g ≈ 9.8 m/s² downward. The equation s = v₀t + ½at² requires constant acceleration, which is satisfied only by option B among the choices.
Why A is wrong: A is wrong because friction increasing with distance means the decelerating force changes, so acceleration is not constant. The kinematic equation requires constant acceleration throughout the motion.
Why C is wrong: C is wrong because alternating acceleration and braking means the acceleration changes discontinuously. The equation applies within each phase separately (if each phase has constant a), but not over the entire motion as a single equation.
Why D is wrong: D is wrong because a non-uniform electric field produces a force that varies with position, giving non-constant acceleration. The kinematic equations are invalid here — you would need to integrate the position-dependent force.
MCQ 8Concept TrapPractice
An object moves with velocity v = (6 − 2t) m/s along a straight line. The motion is:
Show answer and why every option is right or wrong
Answer: D. a = dv/dt = −2 m/s², which is constant. The motion is uniformly accelerated (with negative acceleration, i.e., deceleration when v > 0). The object slows down, stops at t = 3 s (v = 0), then moves in the negative direction with increasing speed. Throughout all of this, the acceleration remains −2 m/s² — the motion is uniformly accelerated the entire time, not just during the deceleration phase.
Why A is wrong: A is wrong because v = 6 − 2t changes with time. Uniform motion requires constant velocity.
Why B is wrong: B is wrong because 'deceleration' means the object is slowing down. After t = 3 s, the object reverses direction and speeds up in the negative direction — it is accelerating (speeding up), not decelerating, even though a = −2 m/s² is unchanged. The acceleration is constant throughout; the label 'decelerated throughout' incorrectly describes the motion after reversal.
Why C is wrong: C is wrong because the motion is not uniform for t < 3 s — velocity is decreasing during that interval. The object does reverse at t = 3 s, but the first half is not uniform motion.
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Uniform Non-Uniform Motion: quick recall before you leave
How do you solve a Uniform Non-Uniform Motion question? A worked example
- 1
Given
Position: x = 2t² + t (x in metres, t in seconds).
- 2
Required
(a) Type of motion. (b) v at t = 2.0 s. (c) Displacement from t = 1.0 s to t = 3.0 s.
- 3
Concept
Differentiate x(t) to get v(t), then differentiate again to get a(t). If a is constant, the motion is uniformly accelerated and kinematic equations are valid. If a varies with t, they are not.
- 4
Formula
v = dx/dt; a = dv/dt. For displacement under constant acceleration: Δx = x(t₂) − x(t₁), or equivalently s = v₀t + ½at² within the interval.
- 5
Substitution
v = d(2t² + t)/dt = 4t + 1
a = d(4t + 1)/dt = 4 m/s²
Since a = 4 m/s² is constant, this is uniformly accelerated motion. The kinematic equations apply.
(b) v(2.0) = 4(2.0) + 1 = 9.0 m/s
(c) x(1.0) = 2(1)² + 1 = 3.0 m; x(3.0) = 2(9) + 3 = 21.0 m
Δx = 21.0 − 3.0 = 18.0 m - 6
Calculation
All arithmetic is shown above. Cross-check part (c) using kinematic equation: at t = 1.0 s, v₀ = 4(1) + 1 = 5.0 m/s, a = 4.0 m/s², Δt = 2.0 s.
s = v₀Δt + ½a(Δt)² = 5.0 × 2.0 + ½ × 4.0 × 4.0 = 10.0 + 8.0 = 18.0 m ✓
Note on exact values: The coefficients 2 and 1 in x = 2t² + t, and the time values 1.0, 2.0, 3.0 s, are problem-defined exact quantities. They do not limit significant figures — the answers are exact within the problem's framework. - 7
Final answer
(a) Uniformly accelerated motion (constant a = 4.0 m/s²).
(b) v = 9.0 m/s at t = 2.0 s.
(c) Displacement = 18.0 m. - 8
Common trap
If this problem had given x = 2t³ + t instead, then a = d²x/dt² = 12t — acceleration depends on time, making it non-uniformly accelerated. A student who blindly plugs a = 12t at some instant into s = v₀t + ½at² gets the wrong answer. Always classify the motion before choosing your equation.
- 9
Similar NEET-style question
A particle moves along a line with x = t³ − 6t² + 9t + 4 (SI units). Find the acceleration at t = 2 s and classify the motion.
Approach: v = 3t² − 12t + 9, a = 6t − 12. At t = 2 s, a = 0. But since a depends on t, the motion is non-uniformly accelerated overall. The kinematic equations v = v₀ + at cannot be used across the entire trajectory.
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What to remember before solving Uniform Non-Uniform Motion questions
More in Kinematics: 14 exam traps and mistakes · 6 formulas · 10 question patterns from its other lessons.
Uniform Non-Uniform Motion questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
NCERT refs: Class 11 Physics Chapter 2, p.14
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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