v = v₀ + a t. Final velocity equals initial velocity plus acceleration times time, for uniform acceleration in a straight line.
-- NCERT Class 11 Physics, Ch. 2, p. 16Relations Uniform Acceleration
8 MCQs3 revision cards9-step worked example
Source: NCERT KinematicsPYQ coverage: NEET 2023, 2025Official key: NTA-verifiedLast updated: 25 Sep 2026
Relations Uniform Acceleration, explained for NEET
The three kinematic equations for uniformly accelerated motion — v = v₀ + at, x − x₀ = v₀t + ½at², and v² = v₀² + 2a(x − x₀) — are among the most frequently tested tools in NEET physics. They look simple. The traps are not in the formulas themselves but in when and how you apply them.
The non-negotiable prerequisite: constant acceleration. Every one of these equations assumes acceleration is constant in both magnitude and direction (NCERT Class 11 Physics Chapter 2, pages 16–18). If a problem states that acceleration varies with time or position — for example, a = 2t or a = kx — these equations are invalid. You must integrate. A common NEET mistake is plugging a non-constant acceleration into v = v₀ + at and getting a plausible-looking but wrong answer.
Trap: dropping the initial velocity. When a problem says an object is "thrown downward" or "projected with initial speed u," that u is non-zero and must appear in your equation. Writing v² = 2gh instead of v² = u² + 2gh costs you the full 4 marks plus the −1 penalty. The word "thrown" is your signal: u ≠ 0.
Trap: linear thinking under uniform deceleration. A bullet entering a block slows from u to u/3 over distance d. How much farther to stop? Students instinctively scale distances linearly with speed. But v² is the quantity linear in distance, not v. Use v² = u² − 2as for each stage with the same deceleration a.
Trap: implicit-function kinematics. When time is given as a function of position (t = f(x) instead of x = f(t)), do not try to invert. Differentiate directly: v = dx/dt = 1/(dt/dx), then a = v·dv/dx. The chain rule is your friend here.
Galileo's odd-number rule. Distances in successive equal time intervals from rest follow 1 : 3 : 5 : 7, not 1 : 2 : 3 : 4. This follows directly from x = ½gt²: the nth-interval distance is proportional to (2n − 1).
Can you answer these Relations Uniform Acceleration MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Direct ApplicationPractice
A car moving at 20 m/s is brought to rest in 40 m by a uniform deceleration. With the same deceleration, the distance needed to stop the car from 40 m/s is:
Show answer and why every option is right or wrong
Answer: C. C is correct. From v² = u² − 2as with v = 0, the stopping distance is s = u²/2a, so it goes as the SQUARE of the speed. The first case gives a = 20²/(2 × 40) = 5 m/s²; then s = 40²/(2 × 5) = 160 m. Doubling the speed quadruples the distance, which is why speed matters so much in braking.
Why A is wrong: A is wrong because 80 m assumes stopping distance is proportional to speed. It goes as u², so doubling the speed multiplies it by four, not two.
Why B is wrong: B is wrong because 40 m assumes stopping distance does not depend on speed at all. A faster car has more kinetic energy for the same braking force to remove.
Why D is wrong: D is wrong because 320 m uses the deceleration from the first case correctly but then drops the 2 in s = u²/2a: 40²/5 = 320.
MCQ 2Easy RecallPractice
An object starts from rest and undergoes uniform acceleration. The ratio of distances covered in the 1st, 2nd, and 3rd seconds of motion is:
Show answer and why every option is right or wrong
Answer: B. For uniform acceleration from rest, distances in successive equal time intervals follow Galileo's odd-number rule: the nth interval gives distance proportional to (2n − 1). So the ratio is 1 : 3 : 5. Reference: NCERT Class 11 Physics Chapter 2, page 17 (derivable from x = ½at²).
Why A is wrong: A (1 : 2 : 3): This assumes distance increases linearly with time — the linear-intuition trap. Distance grows as t², not t (trap: free-fall linear vs quadratic).
Why C is wrong: C (1 : 4 : 9): This gives the ratio of total distances covered BY the end of 1st, 2nd, and 3rd seconds (1² : 2² : 3²), not the distances covered DURING each second.
Why D is wrong: D (1 : 2 : 4): This has no physical basis and likely results from arithmetic confusion between interval distances and cumulative distances.
MCQ 3CalculationPractice
The time t (in seconds) at which a particle is at position x (in metres) is given by t = x² + x. The acceleration of the particle at x = 1 m is:
Show answer and why every option is right or wrong
Answer: A. A is correct. The relation gives t as a function of x, so differentiate it that way and take the reciprocal: dt/dx = 2x + 1, so v = dx/dt = 1/(2x + 1), which is 1/3 m/s at x = 1. Acceleration needs the chain rule, a = v·(dv/dx), because v is known as a function of x rather than of t: dv/dx = −2/(2x + 1)² = −2/9 at x = 1, so a = (1/3)(−2/9) = −2/27 m/s². The sign says the particle is slowing as it advances. (trap: implicit differentiation kinematics)
Why B is wrong: B is wrong because −6/27 is −2/9, which is dv/dx on its own. That is the rate of change of velocity with POSITION, not with time; converting it to an acceleration needs the extra factor of v = 1/3.
Why C is wrong: C is wrong because 2/27 has the right magnitude but the wrong sign. dv/dx = −2/(2x + 1)² is negative for every x, so the acceleration here cannot be positive.
Why D is wrong: D is wrong because 1/3 m/s is the velocity at x = 1, reported as the acceleration. (trap: implicit differentiation kinematics)
MCQ 4CalculationPractice
A bullet travelling at 360 m/s penetrates a wooden block and its speed reduces to 120 m/s after passing through 20 cm of wood. Assuming uniform deceleration, the additional thickness of wood needed to stop the bullet completely is:
Show answer and why every option is right or wrong
Answer: A. Using v² = u² − 2as for the first stage: 120² = 360² − 2a(0.20). So 2a = (360² − 120²)/0.20 = (129600 − 14400)/0.20 = 576000 m/s². For the remaining stage (120 m/s to 0): 0 = 120² − 2a·d₂ → d₂ = 14400/576000 = 0.025 m = 2.5 cm. Reference: NCERT Class 11 Physics Chapter 2, page 18 (third kinematic equation).
Why B is wrong: B (2.2 cm): This scales the 20 cm by (120/360)² = 1/9, giving 20/9 ≈ 2.2 cm. The ratio needed is the remaining v² over the v² already lost, 120²/(360² − 120²) = 1/8, not over the whole initial v².
Why C is wrong: C (5.0 cm): This squares the ratio of remaining speed to speed lost, (120/240)² = 1/4, giving 20/4 = 5.0 cm. Distance goes with the change in v², 120²/(360² − 120²) = 1/8, not with the square of the change in v (trap: speed vs distance ratio linear).
Why D is wrong: D (10.0 cm): This treats distance as proportional to speed lost: 240 m/s was lost in 20 cm, so the remaining 120 m/s is taken to need 10 cm — the linear speed-distance trap (trap: speed vs distance ratio linear).
MCQ 5Direct ApplicationPractice
A car accelerates uniformly from rest to 20 m/s in 10 s. The distance covered by the car during this time is:
Show answer and why every option is right or wrong
Answer: D. From v = v₀ + at: a = (20 − 0)/10 = 2 m/s². Using x = v₀t + ½at² = 0 + ½(2)(100) = 100 m. Alternatively, x = (v₀ + v)t/2 = (0 + 20)(10)/2 = 100 m. Reference: NCERT Class 11 Physics Chapter 2, pages 16–17.
Why A is wrong: A (200 m): This results from using x = vt = 20 × 10, treating the final velocity as if it were constant throughout — ignoring that the car started from rest.
Why B is wrong: B (400 m): This results from using x = v × t with incorrect doubling, such as x = 2 × 20 × 10 = 400 m, a dimensional-analysis failure.
Why C is wrong: C (50 m): This results from using x = ½at² but computing a = 20/10 = 2 m/s² and then ½ × 2 × 10 = 10, squaring t but misplacing a factor somewhere, or using ½vt with v = 10 (mean speed error).
MCQ 6Direct ApplicationPractice
A particle has velocity v₀ at t = 0 and decelerates at a constant rate of magnitude a. The distance it travels before coming to rest is:
Show answer and why every option is right or wrong
Answer: B. Using v² = v₀² + 2(−a)s with v = 0: 0 = v₀² − 2as → s = v₀²/(2a). Reference: NCERT Class 11 Physics Chapter 2, page 18 (third kinematic equation).
Why A is wrong: A (v₀/(2a)): This confuses the stopping distance formula with the stopping time formula t = v₀/a, then halves it without justification.
Why C is wrong: C (v₀²/a): This forgets the factor of 2 in the denominator of v² = v₀² − 2as, effectively doubling the correct answer.
Why D is wrong: D (v₀/a): This is the time to stop, not the distance — it comes from v = v₀ − at with v = 0, giving t = v₀/a.
MCQ 7Direct ApplicationPractice
A body starts from rest and moves with uniform acceleration. If it covers 5 m in the first second, the distance it covers in the 4th second is:
Show answer and why every option is right or wrong
Answer: A. Distance in 1st second from rest: s₁ = ½a(1)² = ½a = 5 m → a = 10 m/s². Distance in nth second from rest = ½a(2n − 1). For n = 4: ½(10)(7) = 35 m. Reference: Galileo's odd-number rule derived from NCERT Class 11 Physics Chapter 2, page 17.
Why B is wrong: B (25 m): This is the distance covered in the 3rd second, ½(10)(2 × 3 − 1) = 25 m — an off-by-one in n. The question asks for the 4th second.
Why C is wrong: C (20 m): This results from using 4 × s₁ = 4 × 5 = 20 m, assuming distance grows linearly with the second number (trap: free-fall linear vs quadratic).
Why D is wrong: D (40 m): This computes ½a(2n) instead of ½a(2n − 1), using 2(4) = 8 instead of 2(4) − 1 = 7, giving ½(10)(8) = 40 m.
MCQ 8Easy RecallPractice
The kinematic equations v = v₀ + at and s = v₀t + ½at² are valid when:
Show answer and why every option is right or wrong
Answer: C. These equations are derived under the assumption that acceleration a is constant (uniform) in both magnitude and direction. If acceleration varies, integration of a(t) or a(x) is required instead. Reference: NCERT Class 11 Physics Chapter 2, page 16 (derivation premise).
Why A is wrong: A: If velocity is constant, then acceleration is zero and the equations reduce to s = v₀t (trivially true but not the condition for validity — the equations are valid for any constant acceleration, not just a = 0).
Why B is wrong: B: If a ∝ t, acceleration is not constant and these equations cannot be used. You would need to integrate a(t) = kt to get v(t) and x(t) (mistake: kinematic equations applied to non-uniform acceleration).
Why D is wrong: D: If a ∝ x, this is simple harmonic-type or exponential motion, and the standard kinematic equations do not apply. Integration or energy methods are needed.
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Relations Uniform Acceleration: quick recall before you leave
How do you solve a Relations Uniform Acceleration question? A worked example
Pattern: Multi-stage deceleration (bullet-block) — NEET pattern: multi stage deceleration bullet block
- 1
Given
• Initial speed: u = 400 m/s• Speed after first stage: v₁ = 200 m/s• Distance of first stage: d₁ = 10 cm = 0.10 m• Deceleration is uniform (constant a throughout)
- 2
Required
Additional penetration distance d₂ from v₁ = 200 m/s to rest (v₂ = 0).
- 3
Concept
Under constant deceleration, v² is linear in displacement. The third kinematic equation connects speeds to distances without needing time.
- 4
Formula
v² = u² − 2a·d (taking deceleration as positive magnitude with minus sign explicit)
- 5
Substitution
First stage: 200² = 400² − 2a(0.10)
Second stage: 0² = 200² − 2a·d₂ - 6
Calculation
First stage: 40000 = 160000 − 0.20a → 0.20a = 120000 → a = 600000 m/s²
Second stage: 0 = 40000 − 2(600000)d₂ → d₂ = 40000/1200000 = 1/30 m ≈ 0.0333 m = 3.33 cm
Note on exact constants: The numbers 400, 200, and 10 cm are given problem values and are treated as exact. The factor 2 in the kinematic equation is a mathematical constant. Neither constrains significant figures in this calculation. - 7
Final answer
d₂ = 10/3 cm ≈ 3.33 cm
The bullet penetrates an additional 3.33 cm before stopping. The total penetration is 10 + 3.33 = 13.33 cm. - 8
Common trap
The temptation is to reason: "Speed halved from 400 to 200 in 10 cm, so it should halve again in another 10 cm and reach 100, then another 10 cm to reach 50..." — this linear-speed reasoning is wrong because v² (not v) decreases linearly with distance. The remaining KE at 200 m/s is only (200/400)² = 1/4 of the original, so only 1/4 of the remaining-distance budget is left after the initial 10 cm would be wrong too — the correct remaining fraction is v₁²/(u²) of the original KE consumed, leaving d₂ = (v₁²/u²) × (u²/(2a)) ... the safest approach is to compute a from stage 1 and use it directly.
- 9
Similar NEET-style question
A car braking uniformly from 60 m/s has its speed reduced to 20 m/s after covering 80 m. What additional distance will it cover before coming to rest?
*(Answer: Using the same method, a = (3600 − 400)/(2 × 80) = 20 m/s². Then d₂ = 400/(2 × 20) = 10 m.)*
---
What to remember before solving Relations Uniform Acceleration questions
Formula
Second kinematic equation
x − x₀ = v₀ t + ½ a t². Displacement equals initial-velocity-times-time plus half acceleration-times-time-squared, for uniform acceleration.
-- NCERT Class 11 Physics, Ch. 2, p. 18Formula
Third kinematic equation
v² = v₀² + 2 a (x − x₀). Velocity-squared relation, useful when time is not known.
-- NCERT Class 11 Physics, Ch. 2, p. 18Worked Example
Example 2.6 — Stopping distance
When brakes apply a constant deceleration -a to a vehicle with initial velocity v₀, the stopping distance is d = v₀² / (2a). Derived from v² = v₀² + 2(-a) d with v = 0.
-- NCERT Class 11 Physics, Ch. 2, p. 20Which Relations Uniform Acceleration formulas do you need for NEET?
3 formulas — click to collapse
First kinematic equation (uniform acceleration)
Final velocity equals initial velocity plus acceleration times the time elapsed, for motion under constant acceleration.
| Symbol | Quantity | SI Unit |
|---|---|---|
| v | Final velocity | m/s |
| v0 | Initial velocity | m/s |
| a | Constant (uniform) acceleration | m/s^2 |
| t | Elapsed time | s |
Valid when
- Acceleration a is CONSTANT (uniform) in both magnitude and direction
- All quantities measured in the same inertial reference frame
- Motion is along a straight line; signs encode direction along chosen axis
Do NOT use when
- Acceleration changes in magnitude or direction (use a(t) integration)
- Motion is uniformly circular at constant speed (a is centripetal, not tangential)
Second kinematic equation (displacement under uniform acceleration)
Displacement equals initial-velocity-times-time plus half of acceleration-times-time-squared. The (1/2) factor is the area of the triangle on the v-t graph.
| Symbol | Quantity | SI Unit |
|---|---|---|
| x | Final position | m |
| x0 | Initial position | m |
| v0 | Initial velocity | m/s |
| a | Constant acceleration | m/s^2 |
| t | Time elapsed | s |
Valid when
- Acceleration constant (magnitude and direction)
- Sign convention consistent across x, v, a (one chosen positive direction)
Third kinematic equation (velocity-squared)
Relates final velocity to initial velocity, displacement, and acceleration without using time. Most useful when t is unknown or unwanted.
| Symbol | Quantity | SI Unit |
|---|---|---|
| v | Final velocity | m/s |
| v0 | Initial velocity | m/s |
| a | Constant acceleration | m/s^2 |
| x - x0 | Displacement | m |
Valid when
- Constant acceleration
- Use signed values for v, v0, a, and (x - x0) consistently
Do NOT use when
- Time-dependent acceleration
- Curvilinear motion where acceleration is not parallel to displacement
Where do students lose marks on Relations Uniform Acceleration?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
5 items — click to collapse
Category: Similar Terms
Student answers 1:2:3:4 for distances in successive 1-second intervals (linear) instead of 1:3:5:7 (Galileo's odd numbers).
When it triggers
Question asks about ratios of distances traversed in successive 1-s intervals during free fall from rest.
How to avoid
Distance grows quadratically (y = ½ g t²); successive interval distances are y_n - y_{n-1} = ½ g (t_n² - t_{n-1}²) = ½ g (2n-1) seconds. The factor (2n-1) gives 1, 3, 5, 7, ...
Category: Overthinking
Student attempts to invert t(x) algebraically before differentiating, getting tangled in messy algebra; misses chain rule.
When it triggers
Question gives t as function of x (instead of x as function of t), e.g. t = x² + x.
How to avoid
Differentiate the given relation directly: dt/dx = (function of x). Then v = dx/dt = 1/(dt/dx). For acceleration use chain rule: a = dv/dt = (dv/dx)(dx/dt) = v dv/dx.
Category: Sign Convention
Student treats a 'thrown vertically downward' problem as if the object were dropped (u = 0). The result is wrong by an additive u² term in v² = u² + 2gh. When the question explicitly states a launch speed, that speed is non-zero and CANNOT be ignored.
When it triggers
Question phrases: 'thrown vertically downward', 'projected with initial velocity', 'launched with speed u'. If u is given numerically, it MUST appear in the equation.
How to avoid
Always parse the launch verbal cue and write down u with its sign before reaching for v² = 2gh. Use the full v² = u² + 2gh (or u² - 2gh for upward motion).
Category: Overthinking
Student assumes proportionality of speed to remaining distance under uniform deceleration. In fact, KE drops linearly with distance (v² is the linear quantity, not v): v² = u² - 2as. Speed-vs-distance is a sqrt-curve, not a line.
When it triggers
Question describes a body decelerating through stages with given speed at one stage; asks for distance to stop or speed at another stage.
How to avoid
Always work with v², not v, when uniform deceleration is in play. The work-energy theorem gives the same answer faster: ½ m v² = work done against constant force over distance.
Root cause: formula misuse
Correction
The three kinematic equations require CONSTANT acceleration. For variable acceleration, use a = dv/dt and integrate, or use v dv = a dx for position-dependent acceleration. Verify constant-a before applying these formulas.
Wrong option pattern
Distractor uses constant-acceleration kinematic equations on a problem where the question explicitly says acceleration changes with time or position.
More in Kinematics: 9 exam traps and mistakes · 3 formulas · 6 question patterns from its other lessons.
Relations Uniform Acceleration questions from past NEET papers
2 questions from NEET 2023, 2025. Answers verified against NTA official keys. — click to collapse
NEET 2025
1+2/(2x + 1)
2−2/(x + 2)³
3−2/(2x + 1)³
4+2/(x + 1)³
NTA Answer: Option 3(final)
NEET 2023
124 cm
228 cm
330 cm
427 cm
NTA Answer: Option 4(final)
How does NEET ask about Relations Uniform Acceleration?
4 recurring patterns from past papers — click to collapse
For an object dropped from rest, distances traversed in successive equal time intervals are in the ratio 1:3:5:7:9:... (odd numbers). Derivable from y = ½ g t². Common shape: 'find ratio of distance covered in 1st, 2nd, 3rd, 4th seconds of free fall'.
Direct ApplicationEasy
Common distractors
uses arithmetic progression 1 2 3 4
Linear-time intuition
Time given as implicit function of position, e.g. t = x² + x. To find acceleration, differentiate twice: dt/dx = 2x + 1, so v = dx/dt = 1/(2x+1); a = dv/dt = -2v²·v = -2v³ (chain rule). Multi-step calculus.
Multi StepMedium
Common distractors
treats t x as explicit
Default to t-as-input thinking
A projectile (typically a bullet) penetrates a uniform medium with constant decelerating force; given initial speed and speed after a known distance, find the total stopping distance. Apply v² = u² + 2as to each segment, noting that 'a' is the same throughout. Common shape: bullet hits block at u, slows to u/k after distance d₁; how much further to stop?
Multi StepMedium
Common distractors
treats speed ratio as distance ratio
Linear thinking: 'speed went from u to u/3, so distance traversed should be 3× the original'
Object given a non-zero downward initial velocity from elevation; asked for the height fallen, time of impact, or final speed. Apply v² = u² + 2gh (or analogous) with proper signs. Common shape: a ball thrown vertically downward from a tower with initial speed u, hitting the ground at speed v; find the tower height. Distractors test (i) sign-of-u confusion, (ii) using v² = 2gh forgetting u², (iii) wrong g unit.
Multi StepEasy
Common distractors
drops initial velocity term
Student conflates 'thrown' with 'dropped' and uses v² = 2gh
Sources
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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