03Acceleration and Motion Graphs
Explore acceleration, interpret position-time and velocity-time graphs, and apply the kinematics equations to solve problems involving constant acceleration.
Acceleration connects force to motion. Reading and drawing motion graphs is a core physics skill that appears on every major exam and in every unit that follows.
How can a graph tell you everything about an object's motion — where it is, how fast it's going, and whether it's speeding up or slowing down?
Acceleration is the rate at which velocity changes over time. When an object speeds up, slows down, or changes direction, it is accelerating. In this lesson we formalise that idea with the equation a = Δv / Δt and then build the skill of reading motion from graphs. Position-time (x-t), velocity-time (v-t), and acceleration-time (a-t) graphs each reveal different aspects of the same motion, and the relationships between them — slope and area — are the key tools you will use throughout the entire course.
Graph Interpretation Summary
Position-Time (x-t)
- Slope = velocity
- Straight line → constant velocity
- Curved (parabola) → accelerating
- Horizontal line → object at rest
- Steeper slope → faster speed
Velocity-Time (v-t)
- Slope = acceleration
- Area = displacement (Δx)
- Horizontal line → constant velocity (a = 0)
- Positive slope → speeding up
- Negative slope → decelerating
Acceleration-Time (a-t)
- Area = change in velocity (Δv)
- Horizontal line → constant acceleration
- Line at zero → constant velocity
- Uniform acceleration → rectangle shape
Key Equations
a = Δv / ΔtDefinition of average accelerationslope of x-t = vVelocity from position-time graphslope of v-t = aAcceleration from velocity-time grapharea under v-t = ΔxDisplacement from velocity-time grapharea under a-t = ΔvChange in velocity from a-t graphWorked Examples
A v-t graph shows an object's velocity increasing from 0 m/s to 20 m/s over 4 seconds. What is the object's acceleration?
Identify the formula: a = Δv / Δt
Find Δv: Δv = v_f − v_i = 20 m/s − 0 m/s = 20 m/s
Find Δt: Δt = 4 s
Calculate: a = 20 m/s ÷ 4 s = 5 m/s²
The slope of the v-t graph equals the acceleration.
A v-t graph shows an object moving at 10 m/s for the first 3 s, then its velocity increases linearly to 22 m/s over the next 3 s. Find the total displacement.
Displacement = area under the v-t graph.
Segment 1 (0–3 s): Rectangle. Area = base × height = 3 s × 10 m/s = 30 m.
Segment 2 (3–6 s): Trapezoid. Area = ½ × (v_i + v_f) × Δt = ½ × (10 + 22) × 3 = ½ × 32 × 3 = 48 m.
Total displacement = 30 m + 48 m = 78 m.
An x-t graph has two segments: from t = 0 to t = 3 s the line is horizontal at x = 5 m; from t = 3 s to t = 7 s the line rises steeply to x = 25 m. Describe the motion.
Segment 1 (0–3 s): Horizontal line → slope = 0 → velocity = 0 m/s. The object is at rest.
Segment 2 (3–7 s): Rising straight line → constant positive slope → constant velocity.
Slope of segment 2 = Δx / Δt = (25 − 5) m / (7 − 3) s = 20 / 4 = 5 m/s.
The object is stationary for 3 s, then moves forward at a constant 5 m/s.
An a-t graph shows a constant acceleration of 3 m/s² for 5 seconds. If the object starts with an initial velocity of 2 m/s, what is its final velocity?
Area under the a-t graph = Δv.
The graph is a rectangle: Δv = a × Δt = 3 m/s² × 5 s = 15 m/s.
Apply: v_f = v_i + Δv = 2 m/s + 15 m/s = 17 m/s.
A v-t graph shows an object decelerating from 30 m/s to 0 m/s in 6 seconds. Find (a) the acceleration and (b) the stopping distance.
(a) Acceleration: a = Δv / Δt = (0 − 30) / 6 = −30 / 6 = −5 m/s².
The negative sign confirms deceleration (acceleration opposite to motion).
(b) Stopping distance = area under v-t graph = triangle.
Area = ½ × base × height = ½ × 6 s × 30 m/s = 90 m.
Guided Problems
A car's velocity changes from 8 m/s to 20 m/s in 3 seconds. Calculate the average acceleration.
Hint: Use a = Δv / Δt. Remember Δv = v_f − v_i. Check your units — the answer should be in m/s².
A v-t graph shows a straight line from (0 s, 15 m/s) to (5 s, 0 m/s). What is the acceleration? Is the object speeding up or slowing down?
Hint: Slope = rise / run = (v_f − v_i) / (t_f − t_i). A negative slope means the velocity is decreasing — that's deceleration.
Using the same v-t graph as GP2, calculate the displacement of the object during the 5 seconds.
Hint: Displacement = area under the v-t graph. The shape is a triangle. Area = ½ × base × height.
An a-t graph shows a = 4 m/s² from t = 0 to t = 3 s, then a = 0 from t = 3 s to t = 6 s. If v_i = 0, find the velocity at t = 6 s.
Hint: Find the area under each segment separately. Area under a-t = Δv. Add each Δv to the running velocity.
An x-t graph is a parabola curving upward. What does this tell you about the object's velocity and acceleration?
Hint: Think about what the slope of the x-t graph represents. As the parabola gets steeper, what is happening to the slope? Is the slope constant or changing?
Key Vocabulary
Acceleration
The rate of change of velocity with respect to time. Acceleration occurs whenever an object speeds up, slows down, or changes direction.
Example: A car going from 0 to 60 km/h in 5 s has an average acceleration of 12 km/h/s (or about 3.3 m/s²).
Average Acceleration
The total change in velocity divided by the total time interval: a_avg = Δv / Δt. It describes the overall rate of velocity change across a time period.
Example: If a ball's velocity changes from 2 m/s to 10 m/s in 4 s, average acceleration = (10−2)/4 = 2 m/s².
Instantaneous Acceleration
The acceleration at a single instant in time. On a v-t graph, it equals the slope of the tangent line drawn at that point.
Example: At t = 3 s, the tangent to a curved v-t graph may have a slope of 6 m/s², so instantaneous acceleration = 6 m/s².
Position-Time Graph (x-t)
A graph with position on the y-axis and time on the x-axis. The slope at any point equals the instantaneous velocity. A straight line means constant velocity; a curve means changing velocity (acceleration).
Example: A horizontal line on an x-t graph means the object is not moving (v = 0).
Velocity-Time Graph (v-t)
A graph with velocity on the y-axis and time on the x-axis. The slope equals acceleration, and the area between the line and the time axis equals displacement.
Example: A straight line with positive slope on a v-t graph means constant positive acceleration (uniform acceleration).
Acceleration-Time Graph (a-t)
A graph with acceleration on the y-axis and time on the x-axis. The area under the curve equals the change in velocity (Δv) over that time interval.
Example: A horizontal line at a = 9.8 m/s² on an a-t graph represents free-fall acceleration near Earth's surface.
Slope
The steepness of a line on a graph, calculated as rise ÷ run (Δy / Δx). On motion graphs, slope carries physical meaning: slope of x-t = velocity; slope of v-t = acceleration.
Example: A line rising 10 m/s over 2 s has a slope of 5 m/s², which is the acceleration.
Area Under the Curve
The region between a graph line and the horizontal axis. On a v-t graph, this area equals displacement. On an a-t graph, this area equals the change in velocity.
Example: A triangle on a v-t graph with base 4 s and height 8 m/s has area = ½ × 4 × 8 = 16 m of displacement.
Workbook Check
Interactive Practice — 5 Questions
What does the slope of a velocity-time graph represent?
An object has a horizontal line on its x-t graph. What is the object doing?
A v-t graph shows a straight line from (0 s, 0 m/s) to (10 s, 50 m/s). What is the acceleration?
What does the area under a velocity-time graph equal?
A curved (parabolic) line on an x-t graph indicates the object is:
Independent Practice
A velocity-time graph shows v = 0 at t = 0 and v = 24 m/s at t = 6.0 s (straight line). Find (a) the acceleration and (b) the displacement during this interval.
An object has a constant velocity of −5.0 m/s for 4.0 s. Sketch the v-t graph and calculate the displacement.
A ball is dropped from rest. Using g = 9.8 m/s², calculate its speed and distance fallen after (a) 1.0 s, (b) 2.0 s, and (c) 3.0 s.
A v-t graph shows a straight line from (0 s, 10 m/s) to (5 s, −10 m/s). Find the acceleration and the total displacement.
★ A car accelerates from rest at 2.0 m/s² for 5.0 s, then travels at constant velocity for 10 s, then decelerates at 4.0 m/s² until it stops. (a) Sketch the complete v-t graph with labeled axes. (b) Calculate the total distance traveled. (c) Find the average speed for the entire trip.
ChallengeCommon Mistakes
Reading the slope of an x-t graph as acceleration instead of velocity
Slope of x-t graph = velocity; slope of v-t graph = acceleration; area under v-t graph = displacement
Assuming a curved x-t graph means the object is accelerating in a curve (2D)
A curved x-t graph simply means velocity is changing — the object is accelerating in 1D
Treating a horizontal line on a v-t graph as 'stopped'
A horizontal v-t line means constant velocity (which could be non-zero); v = 0 only when the line touches the time axis
Using g = 9.8 m/s² as positive when taking downward as positive
If downward is your positive direction, then a = +9.8 m/s²; if upward is positive, then a = −9.8 m/s²
Math Tips
Area under a v-t graph = displacement. Split the area into rectangles and triangles: A_rectangle = base × height, A_triangle = ½ × base × height
To find acceleration from a v-t graph, pick two clearly readable points and compute slope: a = (v₂ − v₁)/(t₂ − t₁)
For free fall from rest: v = gt and Δy = ½gt². After 1 s: v ≈ 10 m/s, Δy ≈ 5 m. After 2 s: v ≈ 20 m/s, Δy ≈ 20 m
Sketch a motion diagram (dots spaced by equal time intervals) before solving. Closer dots = slower; farther dots = faster; evenly spaced = constant velocity