For motion in a straight line, the area between the velocity–time curve and the time axis, between two instants, equals the displacement of the object over that interval. This geometric interpretation is the basis for graphical derivations of the kinematic equations.
-- NCERT Class 11 Physics, Ch. 2, p. 17Velocity Time Graph
8 MCQs6 revision cards9-step worked example
Source: NCERT KinematicsOfficial key: NTA-verifiedLast updated: 25 Sep 2026
Velocity Time Graph, explained for NEET
The velocity-time (v-t) graph is one of the most information-dense representations in kinematics. NEET questions built on it test whether you can extract three quantities correctly: velocity at an instant (read the y-value), acceleration (compute the slope), and displacement (compute the area under the curve). The common trap is treating the graph like a position-time graph and misreading what slope and area represent.
What the v-t graph encodes. For straight-line motion under uniform acceleration, v = v₀ + at (NCERT Class 11 Physics Chapter 2, page 17). On the v-t graph this is a straight line whose slope equals the constant acceleration a, and whose y-intercept is the initial velocity v₀. A horizontal line means zero acceleration (uniform motion). A line sloping downward means deceleration (negative a if positive direction is upward).
Displacement as area. The displacement between times t₁ and t₂ equals the area under the v-t curve between those times. For constant acceleration, the shape is a trapezoid (or triangle if v₀ = 0). The second kinematic equation s = v₀t + ½at² is exactly this area. A high-frequency NEET trap: for free fall from rest, successive-second distances follow the odd-number ratio 1 : 3 : 5 : 7, not the linear ratio 1 : 2 : 3 : 4. The quadratic dependence of displacement on time (s = ½gt²) makes equal time intervals yield unequal distances.
When the graph is NOT a straight line. If the v-t curve is nonlinear, acceleration is not constant. You cannot use v = v₀ + at or v² = v₀² + 2as in that region — those three kinematic equations require constant acceleration. For a curved v-t graph, acceleration at any instant is the tangent's slope, and displacement is still the area (computed by integration or geometric estimation).
Watch-out for NEET. When a problem states an object is "thrown downward" with initial speed u, the v-t graph starts at v₀ = u (not zero). Using v² = 2gh instead of v² = u² + 2gh costs the full 4 marks plus the 1-mark negative penalty.
Can you answer these Velocity Time Graph 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
The slope of a velocity-time graph for a body moving in a straight line represents:
Show answer and why every option is right or wrong
Answer: A. The slope of the v-t graph is Δv/Δt, which by definition is acceleration (NCERT Class 11 Physics Chapter 2, page 17).
Why B is wrong: B is wrong because velocity is the y-coordinate of the v-t graph itself, not its slope.
Why C is wrong: C is wrong because displacement is the area under the v-t graph, not the slope. (trap: confusing slope with area)
Why D is wrong: D is wrong because distance covered is the absolute area under the v-t graph, not the slope.
MCQ 2Easy RecallPractice
The area under a velocity-time graph between two instants gives the:
Show answer and why every option is right or wrong
Answer: B. By definition, ∫v dt over a time interval equals the displacement during that interval. For a straight-line v-t graph, this area is a geometric shape (trapezoid or triangle) whose value equals s = v₀t + ½at² (NCERT Class 11 Physics Chapter 2, page 17).
Why A is wrong: A is wrong because acceleration is the slope of the v-t graph, not the area beneath it.
Why C is wrong: C is wrong because average velocity is the total displacement divided by total time — the area gives displacement, not average velocity directly. You would need to divide the area by the time interval to get average velocity.
Why D is wrong: D is wrong because the v-t graph's area has dimensions of length (m/s × s = m), not m/s², so it cannot represent a change in acceleration.
MCQ 3Easy RecallPractice
A velocity-time graph is a horizontal straight line at v = 5 m/s. Which statement is correct?
Show answer and why every option is right or wrong
Answer: D. A horizontal v-t line means v is constant at 5 m/s and slope = 0, so acceleration is zero — this is uniform velocity (NCERT Class 11 Physics Chapter 2, page 17).
Why A is wrong: A is wrong because a horizontal line has zero slope, meaning zero acceleration, not 5 m/s².
Why B is wrong: B is wrong because the line is at v = 5 m/s, not at v = 0. The object is moving, not at rest.
Why C is wrong: C is wrong because with constant velocity, the displacement per unit time is the same every second (5 m each second), not increasing.
MCQ 4Direct ApplicationPractice
A body starts from rest and accelerates uniformly. The distances covered in the 1st, 2nd, and 3rd seconds of its motion are in the ratio:
Show answer and why every option is right or wrong
Answer: C. For motion from rest with constant acceleration, distance in the nth second is proportional to (2n − 1). So the ratio is 1 : 3 : 5. This follows from s = ½at²: cumulative distances at t = 1, 2, 3 are ½a, 2a, 9a/2, giving successive intervals ½a, 3a/2, 5a/2, i.e. ratio 1 : 3 : 5 (NCERT Class 11 Physics Chapter 2, page 17).
Why A is wrong: A is wrong because 1 : 2 : 3 assumes distance grows linearly with time. Distance grows as t² under uniform acceleration, so successive intervals follow the odd-number pattern, not a linear one. (trap: linear-time intuition from trap: free fall linear vs quadratic)
Why B is wrong: B is wrong because 1 : 4 : 9 represents cumulative distances at t = 1, 2, 3 (proportional to t²), not distances in individual successive seconds.
Why D is wrong: D is wrong because equal distances each second would mean uniform velocity (zero acceleration), contradicting the given uniform acceleration.
MCQ 5Direct ApplicationPractice
A ball is thrown vertically downward from a tower with an initial speed of 10 m/s. It strikes the ground with a speed of 30 m/s. Taking g = 10 m/s², the height of the tower is:
Show answer and why every option is right or wrong
Answer: D. Using v² = u² + 2gh: (30)² = (10)² + 2(10)h → 900 = 100 + 20h → h = 800/20 = 40 m. The initial velocity u = 10 m/s cannot be dropped (NCERT Class 11 Physics Chapter 2).
Why A is wrong: A is wrong because 20 m comes from (v − u)²/(2g) = 20²/20 = 20 m — squaring the difference of the speeds instead of taking the difference of their squares (trap: nonzero initial velocity dropped).
Why B is wrong: B is wrong because 25 m comes from subtracting (2u)² = 400 instead of u² = 100: (900 − 400)/20 = 25 m.
Why C is wrong: C is wrong because 45 m comes from v² / (2g) = 900/20 = 45, which ignores the subtraction of u² = 100 from the numerator.
MCQ 6Direct ApplicationPractice
The v-t graph of a body is a straight line passing through the origin with a slope of 3 m/s². The displacement of the body in the first 4 seconds is:
Show answer and why every option is right or wrong
Answer: C. Since the line passes through the origin, v₀ = 0 and a = 3 m/s². Displacement = area under v-t graph = ½ × base × height = ½ × 4 × (3 × 4) = ½ × 4 × 12 = 24 m. Equivalently, s = ½at² = ½(3)(16) = 24 m (NCERT Class 11 Physics Chapter 2, page 17).
Why A is wrong: A is wrong because 12 m comes from using s = at instead of s = ½at², omitting the factor of ½. The v-t graph area is a triangle, not a rectangle.
Why B is wrong: B is wrong because 48 m results from using s = at² = 3 × 16 = 48, missing the ½ factor in the area formula for the triangle.
Why D is wrong: D is wrong because 6 m likely comes from ½ × a × t = ½ × 3 × 4 = 6, which treats the area as ½ × slope × time instead of ½ × time × velocity-at-that-time.
MCQ 7Concept TrapPractice
The velocity-time graph of a particle is a curve (not a straight line) between t = 0 and t = 5 s. A student uses v = v₀ + at to find the velocity at t = 5 s. This approach is:
Show answer and why every option is right or wrong
Answer: A. The equation v = v₀ + at is derived under the assumption that acceleration a is constant. A curved v-t graph means acceleration varies with time, violating this condition. The equation cannot be applied here (NCERT Class 11 Physics Chapter 2, page 17).
Why B is wrong: B is wrong because v = v₀ + at applies only when acceleration is constant throughout the interval. A curved v-t graph explicitly indicates variable acceleration. (trap: mistake: kinematic nonuniform)
Why C is wrong: C is wrong because the acceleration at t = 0 holds only at that instant; on a curved v-t graph it changes, so v₀ + a(0)·t does not give the velocity at 5 s.
Why D is wrong: D is wrong because v-t graphs can absolutely represent non-uniform acceleration — the curve's slope at any point gives the instantaneous acceleration. The issue is with applying constant-acceleration formulas, not with the graph itself.
MCQ 8CalculationPractice
A particle starts from rest and moves with constant acceleration. If it covers 10 m in the first 2 seconds, what distance does it cover in the next 2 seconds (from t = 2 s to t = 4 s)?
Show answer and why every option is right or wrong
Answer: B. From s = ½at²: 10 = ½a(4), so a = 5 m/s². Distance at t = 4 s: s₄ = ½(5)(16) = 40 m. Distance in next 2 s = 40 − 10 = 30 m. Alternatively, using the odd-number ratio: distances in successive 2-s intervals from rest are in ratio 1 : 3 : 5 : ..., so the second interval covers 3 × 10 = 30 m.
Why A is wrong: A is wrong because 10 m assumes equal distance in equal time intervals, which holds only for uniform velocity (zero acceleration). Under constant acceleration, later intervals cover more distance. (trap: linear-time intuition from trap: free fall linear vs quadratic)
Why C is wrong: C is wrong because 20 m assumes distance doubles each interval (geometric scaling), but under uniform acceleration distance grows quadratically — the ratio of successive-interval distances follows the odd-number rule, not doubling.
Why D is wrong: D is wrong because 40 m is the total distance from t = 0 to t = 4 s, not the distance covered in the interval from t = 2 s to t = 4 s alone.
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Velocity Time Graph: quick recall before you leave
How do you solve a Velocity Time Graph question? A worked example
- 1
Given
• Initial velocity: v₀ = 0 (released from rest)• Acceleration: a = g = 10 m/s² (constant, downward)• Required: ratio of distances in the 1st, 2nd, 3rd, and 4th individual seconds
- 2
Required
Distance covered in each successive 1-second interval: d₁, d₂, d₃, d₄ and their ratio.
- 3
Concept
Under constant acceleration from rest, cumulative distance grows as s = ½gt². The distance in the nth second equals sₙ − sₙ₋₁ = ½g(n² − (n−1)²) = ½g(2n − 1). The factor (2n − 1) produces the odd-number sequence 1, 3, 5, 7, ...
- 4
Formula
Distance in nth second: dₙ = ½g(2n − 1)
This is derived from the second kinematic equation s = ½at² by subtraction of consecutive cumulative distances. - 5
Substitution
• d₁ = ½(10)(2×1 − 1) = 5 × 1 = 5 m• d₂ = ½(10)(2×2 − 1) = 5 × 3 = 15 m• d₃ = ½(10)(2×3 − 1) = 5 × 5 = 25 m• d₄ = ½(10)(2×4 − 1) = 5 × 7 = 35 m
- 6
Calculation
Ratio d₁ : d₂ : d₃ : d₄ = 5 : 15 : 25 : 35 = 1 : 3 : 5 : 7
Note on exact constants: g = 10 m/s² is a problem-defined exact value, and the integers 1, 2, 3, 4 are counting numbers. These do not limit significant figures in the ratio. - 7
Final answer
The distances in the 1st, 2nd, 3rd, and 4th seconds are in the ratio 1 : 3 : 5 : 7.
- 8
Common trap
Students often answer 1 : 2 : 3 : 4, assuming distance grows linearly with time. This is wrong because distance grows as t² (quadratic), making successive-interval distances follow the odd-number rule, not a linear sequence. On the v-t graph, the area of each successive strip (between t = n−1 and t = n) is a trapezoid whose area grows as (2n − 1), not as n.
- 9
Similar NEET-style question
A stone is dropped from a tall building. If it covers 20 m in the first 2 seconds, what distance does it cover in the 3rd and 4th seconds combined? (Use the odd-number ratio for 2-second intervals, or compute directly from s = ½gt².)
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What to remember before solving Velocity Time Graph questions
More in Kinematics: 14 exam traps and mistakes · 6 formulas · 10 question patterns from its other lessons.
Velocity Time Graph 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.17
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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