2dVelocity–Time Graphs
Master v–t graphs: slope gives acceleration, area under the curve gives displacement — two powerful tools in one graph.
Velocity–time graphs are used in crash analysis, sports science, and aerospace engineering. Reading them fluently lets you extract acceleration and displacement from any motion data — bridging graphs and equations of motion.
Lesson Overview
A velocity–time (v–t) graph reveals two key pieces of information simultaneously: the slope tells you the acceleration, and the area under the curve tells you the displacement. In this lesson you will master reading v–t graphs, calculating acceleration from slope, and finding displacement from area — skills that connect kinematics graphs to equations of motion.
Key Concepts
Slope of v–t graph
Equals acceleration: a = Δv / Δt
Positive slope
Positive acceleration — object is speeding up (if v > 0) or slowing down (if v < 0)
Negative slope
Negative acceleration — object is slowing down (if v > 0) or speeding up in negative direction
Zero slope
Constant velocity — zero acceleration
Area under v–t graph
Equals displacement: Δx = area between the curve and the time axis
Area below time axis
Represents negative displacement (motion in the negative direction)
A v–t graph shows a straight line from (0 s, 0 m/s) to (5 s, 20 m/s). Find (a) the acceleration and (b) the displacement.
A v–t graph shows a horizontal line at v = 15 m/s from t = 0 to t = 6 s. Find (a) the acceleration and (b) the displacement.
A v–t graph shows a line from (0 s, 30 m/s) to (6 s, 0 m/s). Find (a) the acceleration and (b) the displacement.
A v–t graph has two segments: (0–4 s) v increases from 0 to 12 m/s; (4–10 s) v stays constant at 12 m/s. Find the total displacement.
A v–t graph shows velocity going from +10 m/s at t = 0 to −10 m/s at t = 4 s in a straight line. Find (a) the acceleration and (b) the net displacement.
A v–t graph shows a line from (0 s, 5 m/s) to (4 s, 13 m/s). Calculate the acceleration.
Hint: Acceleration = slope = Δv / Δt.
A v–t graph shows a rectangle from t = 2 s to t = 8 s at v = 6 m/s. What is the displacement during this interval?
Hint: Area of a rectangle = length × width.
On a v–t graph, the line crosses the time axis (v = 0). What does this crossing represent physically?
Hint: Think about what v = 0 means for the object's motion.
An object has a v–t graph with a negative slope and positive velocity. Is the object speeding up or slowing down?
Hint: Negative slope = negative acceleration. If velocity is positive and acceleration is negative, what happens to speed?
How does the area under a v–t graph relate to the slope of the corresponding x–t graph?
Hint: Think about what both quantities represent in terms of motion.
Key Vocabulary
Velocity–Time Graph (v–t graph)
A graph with time on the horizontal axis and velocity on the vertical axis. Slope = acceleration; area = displacement.
Example: A straight line with positive slope on a v–t graph shows constant positive acceleration.
Acceleration
The rate of change of velocity: a = Δv / Δt. A vector quantity with SI unit m/s².
Example: A car going from 0 to 20 m/s in 4 s has acceleration = 5 m/s².
Area Under the Curve
On a v–t graph, the area between the velocity curve and the time axis equals the displacement.
Example: A rectangle of width 5 s and height 10 m/s has area = 50 m of displacement.
Deceleration
A common term for negative acceleration — when an object slows down. Technically, deceleration means acceleration opposite to the direction of motion.
Example: A braking car decelerates: velocity decreases, so Δv is negative if motion is in the positive direction.
Interactive Practice — 5 Questions
On a velocity–time graph, what does the slope represent?
A horizontal line on a v–t graph means the object has:
A v–t graph shows a line from (0 s, 0 m/s) to (8 s, 24 m/s). What is the acceleration?
The area under a v–t graph represents:
A v–t graph shows velocity decreasing from +20 m/s to 0 m/s in 5 s. The displacement is:
Independent Practice
A v–t graph shows three segments: (0–3 s) v increases from 0 to 9 m/s; (3–7 s) v = 9 m/s constant; (7–10 s) v decreases from 9 m/s to 0. Find (a) acceleration in each segment and (b) total displacement.
Explain why the area below the time axis on a v–t graph represents negative displacement.
A car accelerates from rest at 2 m/s² for 6 s, then brakes at −4 m/s² until it stops. Find the total displacement.
Sketch the v–t graph for an object that starts at rest, accelerates to 10 m/s in 4 s, maintains that speed for 3 s, then decelerates to rest in 2 s. Label all slopes and areas.
★ A v–t graph shows the line v = −3t + 12 (m/s, seconds). Find (a) the initial velocity, (b) the acceleration, (c) the time when the object stops, and (d) the total displacement from t = 0 to t = 6 s, accounting for the sign change in velocity.
ChallengeCommon Mistakes
Reading the height of the v–t graph as displacement.
The HEIGHT of the v–t graph gives velocity. The AREA gives displacement. Always calculate area, not height.
Assuming negative slope always means the object is slowing down.
Negative slope means negative acceleration. If the object is already moving in the negative direction, negative acceleration means it is speeding up.
Forgetting that area below the time axis is negative displacement.
When velocity is negative (below the axis), the object moves in the negative direction. That area subtracts from total displacement.
Math Tips
Two rules for v–t graphs: (1) Slope = acceleration = Δv / Δt. (2) Area = displacement. For triangles: ½bh. For rectangles: lw. For trapezoids: ½(b₁ + b₂)h.
To find total distance (not displacement) from a v–t graph, add the absolute values of all areas — do not subtract the area below the axis.