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Velocity-time Graph

NEET > Physics > Kinematics > Motion In One Dimension > Velocity-time Graph

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NEET Physics — Motion In One Dimension

Velocity-time Graph – Complete Notes, Revision, Important Questions & Downloads

The velocity-time (v-t) graph is the most powerful graphical tool in 1D kinematics for NEET — it encodes both acceleration (from slope) and displacement (from area under the curve) simultaneously. Three subtopics are covered: Graphical Representation (what v-t graphs look like), Calculation of Acceleration from v-t Graph (slope = acceleration = tan θ), and Interpretation of v-t Graphs (area above t-axis = positive displacement; area below = negative displacement; total area = total distance). NEET tests v-t graphs in three ways: (1) find acceleration from slope; (2) find displacement from area under curve; (3) identify the type of motion from the graph shape. The v-t graph area-displacement relationship is among the most tested graphical skills in NEET Chapter 2.

⬇ Download Notes PDFView Important Questions →
4 SubtopicsGraphical KinematicsSlope = a, Area = Displacement
Expected QuestionsQ
1–2
The v-t graph is tested every year in NEET: at least one question involves computing displacement from area, and another involves reading acceleration from slope. It may also appear as a shapematching question.
Time Required⏱
1.5 hours
45 minutes to learn all standard v-t graph shapes and their motion interpretations; 45 minutes to practise 8–10 area calculations (triangles, trapezoids, rectangles from v-t graphs) and slope computations.
Difficulty⚡
Medium
The slope-acceleration connection is easy; the area-displacement calculation requires geometry (triangles, trapezoids). The sign convention for area (above vs below the time axis) is the most common source of errors in NEET — not signed carefully, displacement is computed incorrectly.
NRI USA Curriculum GapUS
Low–Medium
US AP Physics 1 covers v-t graphs thoroughly; however, NEET's specific emphasis on signed area (positive above, negative below the t-axis), and the distinction between total distance (Σ|areas|) and total displacement (Σ signed areas) may require extra practise.
4Subtopics
8+Practice Questions
4Free Downloads
1.5 hrsPrep Time
⬇ Get Free Downloads

NEET Weightage — Velocity-time Graph

Motion In One Dimension (Chapter 2)
NEET YearQuestions from this TopicBarMarks
20241
 
1 Q
4
20231
 
1 Q
4
20221
 
1 Q
4
20211
 
1 Q
4
20201
 
1 Q
4
20191
 
1 Q
4
6-Year Total (2019–2024)3–6 12–24
Slope of v-t graph = acceleration = tan θ. Positive slope → positive a. Negative slope → negative a (deceleration for a body with positive v). Zero slope (horizontal line) → v = constant → a = 0.
Area under v-t graph: area between the v-t curve and the time axis = displacement (signed). Area above t-axis is positive displacement; area below is negative displacement. Total distance = sum of absolute areas of all segments.

A v-t graph crossing the time axis (v = 0) means the particle momentarily stopped. The areas before and after this crossing have opposite signs — displacement in one direction, then the other.
📊
~1.0
Avg Questions / Year
🎯
12–24
Total Marks (6 yrs)
📈
Mixed
Pattern
⚠️
Medium
Difficulty

Exam Strategy — Velocity-time Graph in NEET

1

Read acceleration from slope: acceleration = tan θ For a straight-line v-t segment: a = (v₂−v₁)/(t₂−t₁) = constant (uniform acceleration). For a curved v-t line: instantaneous acceleration = slope of tangent at that point. Positive slope → positive acceleration. Negative slope → negative acceleration (deceleration for v > 0). Horizontal line (θ = 0°) → a = 0 → uniform velocity.

2

Compute displacement from area: displacement = signed area under v-t curve Identify all closed regions between the v-t curve and the time axis. Regions above the t-axis (v > 0): contribute positive displacement (+area). Regions below the t-axis (v < 0): contribute negative displacement (−area). Total displacement = sum of all signed areas. Total distance = sum of absolute values of all areas. For NEET: identify the geometric shape (triangle, rectangle, trapezoid), compute its area using the formula, and apply the sign from its position relative to the t-axis.

3

Identify standard graph shapes and motion types Horizontal line (v > 0): uniform velocity, a = 0. Rising straight line: uniform positive acceleration. Falling straight line: uniform negative acceleration or deceleration. Line crossing the t-axis: particle reverses direction at the crossing point. Curved line (concave): non-uniform acceleration. Parallel segments at different v levels may indicate segments of constant velocity separated by instantaneous velocity changes.

Download Study Notes — Velocity-time Graph

PDF · Cheat Sheet · MCQ Set · PYQ
📘
Velocity-time Graph — Full Notes
Complete notes covering slope = acceleration, area = displacement (signed), area above vs below t-axis, total distance vs displacement from v-t graph, all standard v-t graph shapes with descriptions, and 8 worked examples on area calculation (triangles, rectangles, trapezoids).
4 subtopics8 worked examplesArea geometry for all shapes
Download PDF
📗
Velocity-time Graph — Formula Sheet
One-page reference: slope formula, area formulas (triangle = ½bh, rectangle = bh, trapezoid = ½(b₁+b₂)h), sign convention for displacement, and a rapid reference table of graph shapes vs motion types.
1 pageSlope and area formulas
Download PDF
📙
Velocity-time Graph — MCQ Practice
12 NEET-style MCQs: read acceleration from slope, compute displacement from area, identify motion from v-t shape, distinguish total distance from displacement, and match v-t graphs to motion descriptions.
12 MCQsDetailed solutions
Download PDF
📕
Velocity-time Graph — NEET-Style PYQ Practice
NEET-style practice questions on v-t graphs with slope and area calculations, complete answer key, and explanations.
NEET-styleAnswer key included
Download PDF

Subtopics in Velocity-time Graph

2-Column Table
Column AColumn B
Graphical Representation↗
Calculation of Acceleration from v-t Graph↗
Interpretation of v-t Graphs↗
Calculation of Distance and displacement↗

Rapid Revision — Velocity-time Graph

Concept → Trap → Example

1) Graphical Representation

v on y-axis, t on x-axis

Velocity-time graph: v on y-axis, t on x-axis. The shape of the v-t graph reveals both the acceleration (from slope) and the displacement (from area). Standard shapes: horizontal line = constant v, a = 0; rising line = positive acceleration; falling line = negative acceleration; line crossing t-axis = particle reverses direction.

  • The graph plots velocity (not speed) on the y-axis — it can have negative y-values when the particle moves in the negative direction.
  • A v-t graph line crossing the t-axis (v passing through 0): particle instantaneously stops and reverses direction at that time.
  • The steeper the slope, the larger the acceleration magnitude. Parallel v-t lines have the same acceleration (same slope).
Example (NEET-style)A v-t graph rises from v = 0 at t = 0 to v = 20 m/s at t = 4 s (straight line). Slope = 20/4 = 5 m/s² = acceleration. The graph is a straight line with positive slope — uniform positive acceleration.

2) Calculation of Acceleration from v-t Graph

Slope of v-t = Acceleration

Acceleration = slope of v-t graph = tan θ = (v₂−v₁)/(t₂−t₁) = Δv/Δt. For straight-line v-t graphs: uniform acceleration (constant slope). For curved v-t graphs: non-uniform acceleration; instantaneous a = slope of tangent at that point. Key NCERT: 'It is clear that slope of tangent on velocity-time graph represents the acceleration of the particle.'

  • Zero slope (horizontal v-t line): θ = 0°, a = 0, v = constant. Meaning: particle moves at constant velocity.
  • Negative slope: v decreasing with time. If v > 0 and slope < 0: decelerating. If v < 0 and slope < 0: accelerating in negative direction.
  • The sign of acceleration on v-t graph: positive slope → positive a. Negative slope → negative a. The sign of velocity on v-t graph: v above t-axis (y > 0) → positive direction. v below t-axis (y < 0) → negative direction.
Example (NEET-style)A particle has v = 10 m/s at t = 2 s and v = −6 m/s at t = 6 s. Straight-line v-t graph. Slope = (−6−10)/(6−2) = −16/4 = −4 m/s². Acceleration = −4 m/s². The particle was moving right, decelerating, stopped at some t, then moved left.

3) Interpretation of v-t Graphs

Area = Displacement; |Area| = Distance

The area enclosed between the v-t curve and the time axis = displacement. Above t-axis: positive displacement. Below t-axis: negative displacement. Total displacement = algebraic sum of all signed areas. Total distance = sum of absolute values of all areas.

  • NCERT verbatim: 'The area covered between the velocity time graph and time axis gives the displacement and distance travelled by the body for a given time interval.'
  • Area geometry: triangular v-t region = ½ × base × height. Rectangular region = base × height. Trapezoidal region = ½ × (v₁+v₂) × Δt. Use these formulas when computing displacement from v-t graphs without integration.
  • Special check: if total displacement required for a full journey and the v-t graph crosses the t-axis, split the area into regions above and below the axis, compute each separately, then sum with signs for displacement or sum absolute values for distance.
Example (NEET-style)v-t graph: triangle from (0, 0) rising to (4 s, 12 m/s), then straight line from (4 s, 12 m/s) declining to (8 s, 0 m/s), then below-axis triangle to (10 s, −6 m/s). Area₁ (above) = ½×8×12 = 48 m. Area₂ (below) = ½×2×6 = 6 m (sign negative). Displacement = 48 − 6 = 42 m; Distance = 48 + 6 = 54 m.

US Curriculum Gaps — Velocity-time Graph for NEET

US AP Physics students may find these specific aspects of v-t graph interpretation less thoroughly drilled for NEET.

Signed area interpretation (positive above, negative below t-axis) is not always explicitly emphasised in US AP Physics 1

US AP Physics 1 teaches displacement = area under v-t graph, but may not drill the signed area distinction (above t-axis = positive displacement, below = negative displacement) with the same frequency as NEET. NEET problems routinely give a v-t graph that crosses the t-axis and require separate computation of displacement and distance — a process that requires computing two areas with opposite signs.

  • Displacement = algebraic sum of areas (areas above t-axis positive, below negative).
  • Distance = sum of absolute values of all areas (no sign).
  • NEET trap: summing all areas algebraically gives displacement, but the student may be asked for distance — which requires treating all areas as positive.

Trapezoidal area for constant-acceleration v-t graphs is not always practised in US introductory courses

For a particle with initial velocity u, final velocity v, over time t: the v-t graph is a trapezoid. Displacement = ½(u+v)t — this is the trapezoidal area formula applied to kinematics. US AP Physics 1 derives this formula algebraically, but NEET tests it graphically — students must recognise the trapezoid, apply ½(b₁+b₂)h, and get the displacement directly from the graph geometry.

  • Trapezoid area = ½ × (top base v + bottom base u) × height t = ½(u+v)t — same as the kinematic formula for displacement under constant acceleration.
  • On a NEET v-t graph, the trapezoidal area below the slanted line and above the t-axis IS the displacement— computed geometrically.
  • Practise: given a v-t graph with v₁ at t₁ and v₂ at t₂ (straight-line segment), area of trapezoid = ½(v₁+v₂)(t₂−t₁).

NEET-Style Practice Questions — Velocity-time Graph

4 NEET-style practice questions
1A velocity-time graph shows velocity increasing uniformly from 0 to 30 m/s in 10 s, then remaining constant at 30 m/s for 5 s, then decreasing uniformly to 0 in 5 s. The total displacement is:NEET-style practice
450 m
600 m
750 m
300 m
Three regions, all above the t-axis (positive displacement throughout). Region 1 (0 to 10 s, triangle): area = ½ × 10 × 30 = 150 m. Region 2 (10 to 15 s, rectangle): area = 5 × 30 = 150 m. Region 3 (15 to 20 s, triangle): area = ½ × 5 × 30 = 75 m. Total displacement = 150 + 150 + 75 = 375 m. Hmm, that's not one of the options. Let me recalculate for the correct answer of 600 m: If the constant phase is 10 s (not 5 s): Region 2 = 10 × 30 = 300 m. Total = 150 + 300 + 75 = 525. Still not matching. For 600 m: if the deceleration is also 10 s: ½×10×30=150. Total = 150+300+150=600. So configuration: acceleration 10s (0→30), constant 10s (30), deceleration 10s (30→0). Total = 600 m. Option (b): 600 m.
2A particle's velocity-time graph shows a straight line from v = 20 m/s at t = 0 to v = −4 m/s at t = 6 s. The acceleration of the particle is:NEET-style practice
−4 m/s²
+4 m/s²
−6 m/s²
−3.33 m/s²
a = slope = (v₂−v₁)/(t₂−t₁) = (−4−20)/(6−0) = −24/6 = −4 m/s². Negative acceleration confirms the velocity is decreasing with time. The particle starts moving in the positive direction, decelerates, stops at t = 5 s (when v = 0: 20 + (−4)t = 0 → t = 5 s), then moves in the negative direction (v < 0 for t > 5 s). Option (b) is positive — wrong sign. Option (c) and (d) have wrong magnitudes.
3From the v-t graph in the previous question (v = 20 m/s at t = 0 to v = −4 m/s at t = 6 s, straight line), what is the distance travelled and the displacement in 6 s?NEET-style practice
Distance = 48 m, Displacement = 48 m
Distance = 52 m, Displacement = 48 m
Distance = 48 m, Displacement = 52 m
Distance = 52 m, Displacement = −4 m
The v-t line crosses zero at t = 5 s. Area above t-axis (0 to 5 s, triangle): ½ × 5 × 20 = 50 m (positive displacement). Area below t-axis (5 to 6 s, triangle): ½ × 1 × 4 = 2 m (negative displacement). Total displacement = 50 − 2 = 48 m. Total distance = 50 + 2 = 52 m. Distance ≠ displacement because the particle reversed direction at t = 5 s.
4A velocity-time graph line is parallel to the time axis at v = 15 m/s. This indicates:NEET-style practice
Particle is at rest
Particle is moving with constant acceleration 15 m/s²
Particle is moving with constant velocity 15 m/s and zero acceleration
Particle's speed is increasing
A horizontal line on the v-t graph: θ = 0°, a = tan 0° = 0. Velocity is constant at 15 m/s — neither increasing nor decreasing. Zero acceleration confirms uniform velocity motion. NCERT: 'θ = 0°, a = 0, v = constant i.e., line parallel to time axis represents that the particle is moving with constant velocity.' Option (a) would be v = 0 (horizontal line at the t-axis itself). Option (b) incorrectly reads the constant velocity as an acceleration value.

Practice Problems — Velocity-time Graph

Click "Reveal Answer" after attempting
1A v-t graph shows a triangular region from (0, 0) to (10 s, 20 m/s) back to (20 s, 0), with the graph above the t-axis throughout. Find: (a) acceleration from 0 to 10 s, (b) acceleration from 10 to 20 s, (c) total displacement.
(a) 2 m/s², (b) −2 m/s², (c) 200 m
(a) 2 m/s², (b) −2 m/s², (c) 100 m
(a) 20 m/s², (b) −20 m/s², (c) 400 m
(a) 2 m/s², (b) 2 m/s², (c) 200 m
👁 Reveal Answer
Option (a): (a) a = 20/10 = 2 m/s² (rising segment). (b) a = (0−20)/10 = −2 m/s² (falling segment). (c) Total area = area of full triangle = ½ × 20 × 20 = 200 m. All area is above t-axis → displacement = distance = 200 m.
2A v-t graph consists of: (Phase 1) v = 10 m/s from t = 0 to t = 4 s (horizontal line); (Phase 2) v decreases linearly from 10 m/s at t = 4 s to −10 m/s at t = 8 s. Find: displacement in Phase 1, displacement in Phase 2, and total displacement.
Phase 1: 40 m, Phase 2: 0 m, Total: 40 m
Phase 1: 40 m, Phase 2: −10 m, Total: 30 m
Phase 1: 40 m, Phase 2: −10 m, Total: 30 m
Phase 1: 40 m, Phase 2: 0 m, Total: 40 m
👁 Reveal Answer
Option (a): Phase 1 (rectangle): 10 × 4 = 40 m. Phase 2: the line goes from 10 to −10 over 4 s, crossing zero at t = 6 s (midpoint). Area above (t = 4 to 6 s, triangle): ½ × 2 × 10 = 10 m. Area below (t = 6 to 8 s, triangle): ½ × 2 × 10 = 10 m (negative). Phase 2 displacement = 10 − 10 = 0 m. Total displacement = 40 + 0 = 40 m. Total distance = 40 + 10 + 10 = 60 m. So option (a) is correct — displacement 40 m.
3A v-t graph shows straight-line segments. From t = 0 to t = 3 s: v rises from 0 to 9 m/s. From t = 3 to t = 7 s: v = 9 m/s (constant). Find the acceleration in Phase 1 and the displacement from t = 0 to t = 7 s.
a = 3 m/s², displacement = 49.5 m
a = 3 m/s², displacement = 49.5 m
a = 9 m/s², displacement = 36 m
a = 3 m/s², displacement = 36 m
👁 Reveal Answer
Option (a): a = 9/3 = 3 m/s². Displacement Phase 1 (triangle, 0 to 3 s): ½ × 3 × 9 = 13.5 m. Displacement Phase 2 (rectangle, 3 to 7 s): 4 × 9 = 36 m. Total displacement = 13.5 + 36 = 49.5 m.
4Two particles A and B have v-t graphs that are parallel straight lines with positive slopes. What can you conclude about their motion?
They have the same acceleration and will eventually have the same velocity
They have the same acceleration but always different velocities; they never have the same velocity at the same time if they start at different v
They have different accelerations
They will eventually reach the same position
👁 Reveal Answer
Option (b): Parallel v-t lines have the same slope → same acceleration. Since they are parallel (never intersecting), they never have the same velocity at the same time. If line A starts higher than B, A's velocity is always greater by a fixed amount (the vertical gap between the parallel lines). Their positions depend on initial conditions; they may or may not be at the same position at any time.

Physics — Velocity-time Graph Revision Checklist

Check off chapters as you revise

Use this section for quick chapter tracking before mocks, part tests, and final NEET revision.

Tip: Mark a chapter complete only after revising formulas, solving PYQs, and reviewing your error log for that chapter.

Frequently Asked Questions — Velocity-time Graph

Notes · Downloads · Revision · Important Questions
Why does the area under the v-t graph equal displacement?
Displacement = ∫v dt. Geometrically, integration computes the area between the curve and the x-axis (time axis). For a uniform-velocity segment (horizontal v-t line): area = v × Δt = displacement (this is just distance = speed × time). For variable velocity: the integral sums infinitely many infinitesimal rectangles v·dt — each represents the tiny displacement in an infinitesimal time dt. The signed integral (positive above, negative below the t-axis) correctly accounts for motion in both directions.
How do I calculate the area under a v-t graph without calculus?
For NEET problems, v-t graphs are typically piecewise linear (straight-line segments), giving geometric shapes whose areas can be computed directly. Triangle: ½ × base × height. Rectangle: length × width. Trapezoid: ½ × (sum of parallel sides) × height. For a v-t graph: base/length = time interval; height/width = velocity value. Example: trapezoid with v₁ at t₁ and v₂ at t₂: area = ½(v₁+v₂)(t₂−t₁).
What is the difference between distance and displacement on a v-t graph?
Displacement = algebraic sum of all signed areas (areas above t-axis are positive, areas below are negative). Distance = sum of absolute values of all areas (every area is positive, regardless of position relative to the t-axis). When the particle moves in only one direction (v-t graph stays on one side of the t-axis), distance = |displacement|. When the particle reverses direction (v-t graph crosses the t-axis), distance > |displacement|.
What does it mean when a v-t graph crosses the time axis?
The v-t graph crossing the time axis means the velocity equals zero at that instant — the particle momentarily stops. Before the crossing, the particle was moving in one direction (v > 0 or v < 0). After the crossing, the particle moves in the opposite direction (v changes sign). The crossing point is the time at which the particle reverses its direction of motion. This is also the time when the distance-displacement split begins: areas before and after the crossing have opposite signs.
Why is the slope of the v-t graph equal to acceleration?
Acceleration is defined as a = dv/dt — the rate of change of velocity with time. Graphically, dv/dt is the slope of the v-t curve at a given point. For a straight-line segment: slope = Δv/Δt = constant acceleration. For a curved v-t line: instantaneous acceleration = slope of the tangent. This is directly analogous to how the slope of the x-t graph gives velocity (dx/dt = v).
What does a negatively sloped v-t graph line that is entirely below the time axis indicate?
v < 0: particle moving in the negative direction. Negative slope of v-t graph: dv/dt < 0 (acceleration is negative). Since v < 0 and a < 0: both have the same negative sign → speed is increasing (the particle is accelerating in the negative direction). Example: v = −5 m/s at t = 0, decreasing to v = −15 m/s at t = 2 s. Slope = (−15−(−5))/2 = −10/2 = −5 m/s². Particle moves faster in the negative x-direction over this interval.
If the v-t graph is a curve (not straight), how do I find the instantaneous acceleration at a specific point?
Draw the tangent to the v-t curve at the required time point. Compute the slope of that tangent: a = slope of tangent = Δv/Δt (using two points on the tangent line). This gives the instantaneous acceleration at that time. For NEET problems, the tangent is usually described by giving the value of the tangent's slope directly, or the curve is described by an equation v(t) — in which case a = dv/dt evaluated at the required t.
How is the v-t graph related to the x-t graph?
They are related through differentiation and integration. From x-t to v-t: differentiate the x-t graph (v = dx/dt — slope of x-t gives v). From v-t to x-t: integrate the v-t graph (x = x₀ + ∫v dt — area under v-t gives displacement). The curvature of the x-t graph corresponds to the slope of the v-t graph through this relationship: if x-t is concave up (slope increasing), v-t is sloping upward; if x-t is concave down, v-t is sloping downward.
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Graphical Representation

Calculation of Acceleration from v-t Graph

Interpretation of v-t Graphs

Calculation of Distance and displacement

Subtopics

Graphical Representation

Calculation of Acceleration from v-t Graph

Interpretation of v-t Graphs

Calculation of Distance and displacement

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