Module 1: Measurements and Motion in 1D
PHYS-2325 M1L6 Graphical Integration in Motion Analysis
"A picture is worth a thousand words, but a graph is worth a thousand equations."
— Physics Insight
Graphs tell the complete story of motion! While equations give you precise answers, graphs reveal the whole journey—showing you when objects speed up, slow down, or change direction. In this lesson, you'll learn to extract position and velocity information directly from motion graphs using the powerful technique of graphical integration. You'll discover that the area under a velocity-time graph equals the displacement, and the area under an acceleration-time graph equals the change in velocity. Master these visual tools, and you'll unlock the ability to analyze any motion scenario with confidence!
Course Competencies and Learning Objectives
A ★ indicates that this page contains content related to that LO.
CC1.1 Solve problems of motion in one dimension
LO1.1.1 Translate from scientific notation to regular numbers
LO1.1.2 Translate from different measurement systems
★ LO1.1.3 Investigate the quantities that define motion in one dimension
★ LO1.1.4 Analyze a problem in one dimension
Required Reading
Click the blue buttons to go to the Open Stax reading assignments.
Optional Reading
Media
Watch these videos to master the art of reading motion from graphs. Learn to see velocity in slopes and displacement in areas!
Practice and Apply - Conceptual
Steps for Graphical Integration
Order the Steps for Graphical Integration
Arrange these steps in the correct order for finding displacement from a velocity-time graph:
- Calculate the area of each geometric shape
- Identify the time interval of interest
- Add up all the areas (considering positive and negative)
- Break the area under the curve into simple shapes (rectangles, triangles, trapezoids)
- Determine which areas are positive and which are negative
- Interpret the final result as total displacement
Graph Features
Sort Graph Features
Match each graph characteristic with what it tells us about motion:
Graph Features
- Horizontal line on position-time graph
- Straight line with positive slope on position-time graph
- Curved line (concave up) on position-time graph
- Area under velocity-time graph
- Slope of velocity-time graph
- Negative area under velocity-time graph
- Peak or valley on velocity-time graph
- Zero crossing on velocity-time graph
No Motion
Constant Velocity
Acceleration Present
Displacement Information
Direction Changes
Graph Analysis
Master Graph Analysis
Click each card to test your understanding of motion graphs and their relationships:
Why does area under a velocity-time graph equal displacement?
Answer
Because displacement = velocity × time, and area = base × height. The area calculation naturally gives you the total displacement!
What does a negative area under a velocity-time graph mean?
Answer
It means displacement in the negative direction. The object moved backward during that time interval.
How do you find instantaneous velocity from a position-time graph?
Answer
Find the slope of the tangent line to the curve at that specific point. Steeper slope = greater velocity.
If velocity is positive but decreasing, what does the position graph look like?
Answer
The position graph has a positive slope that's getting less steep (curved, concave down). The object is slowing down but still moving forward.
Practice and Apply - Computational
Need a quick reference for area calculations? Click here for formulas!
Quick Reference: Area Formulas
Rectangle:
A = base × height
A = Δt × v
Triangle:
A = ½ × base × height
A = ½ × Δt × Δv
Trapezoid:
A = ½(b₁ + b₂) × h = ½(v₁ + v₂) × Δt
Remember: Positive area = motion in positive direction, Negative area = motion in negative direction
Important: Graph Area Calculations
When finding areas under velocity-time graphs:
- Rectangle: Area = base × height = Δt × v
- Triangle: Area = ½ × base × height = ½ × Δt × Δv
- Trapezoid: Area = ½(b₁ + b₂) × h = ½(v₁ + v₂) × Δt
- Positive area: Motion in positive direction
- Negative area: Motion in negative direction
Question 1
A velocity-time graph shows constant velocity of +8 m/s for 5 seconds. What is the displacement?
See Answer
A velocity-time graph shows constant velocity of +8 m/s for 5 seconds. What is the displacement?
The graph shows a horizontal line at v = +8 m/s from t = 0 to t = 5 s.
This forms a rectangle with:
- Base = Δt = 5 s
- Height = v = +8 m/s
Area = base × height = 5 s × 8 m/s = 40 m
The displacement is +40 m in the positive direction.
Question 2
A velocity-time graph shows velocity increasing linearly from 0 to 12 m/s over 4 seconds. Find the displacement.
See Answer
A velocity-time graph shows velocity increasing linearly from 0 to 12 m/s over 4 seconds. Find the displacement.
The graph shows a straight line from (0 s, 0 m/s) to (4 s, 12 m/s).
This forms a triangle with:
- Base = Δt = 4 s
- Height = v_final = 12 m/s
Area = ½ × base × height = ½ × 4 s × 12 m/s = 24 m
The displacement is +24 m in the positive direction.
Question 3
A velocity-time graph shows velocity changing from +6 m/s to +10 m/s over 3 seconds. What is the displacement?
See Answer
A velocity-time graph shows velocity changing from +6 m/s to +10 m/s over 3 seconds. What is the displacement?
The graph shows a line from (0 s, +6 m/s) to (3 s, +10 m/s).
This forms a trapezoid with:
- Parallel sides: v₁ = 6 m/s and v₂ = 10 m/s
- Height = Δt = 3 s
Area = ½(v₁ + v₂) × Δt = ½(6 + 10) × 3 = ½(16) × 3 = 24 m
Alternative: Rectangle (6 × 3 = 18) + Triangle (½ × 3 × 4 = 6) = 24 m
The displacement is +24 m in the positive direction.
- 0-2 s: constant velocity of +8 m/s
- 2-4 s: velocity decreases linearly from +8 m/s to -4 m/s
- 4-6 s: constant velocity of -4 m/s
Find the total displacement after 6 seconds.
Question 4
A complex velocity-time graph shows:
See Answer
A complex velocity-time graph shows:
- 0-2 s: constant velocity of +8 m/s
- 2-4 s: velocity decreases linearly from +8 m/s to -4 m/s
- 4-6 s: constant velocity of -4 m/s
Find the total displacement after 6 seconds.
Solution:
Segment 1 (0-2 s): Rectangle with base = 2 s, height = +8 m/s
Area₁ = 2 × 8 = +16 m
Segment 2 (2-4 s): Trapezoid with heights +8 m/s and -4 m/s, base = 2 s
Area₂ = ½(8 + (-4)) × 2 = ½(4) × 2 = +4 m
Segment 3 (4-6 s): Rectangle with base = 2 s, height = -4 m/s
Area₃ = 2 × (-4) = -8 m
Total Displacement = Area₁ + Area₂ + Area₃ = 16 + 4 + (-8) = +12 m
The object ends up 12 meters in the positive direction from its starting point, despite changing directions during the motion!
Ready to Move On?
Lesson Checklist
Test your understanding by sketching these scenarios:
- Draw a velocity-time graph for an object that travels 30 m forward in 5 seconds at constant speed, then returns to its starting point in the next 3 seconds.
- If an acceleration-time graph shows constant acceleration of 2 m/s² for 4 seconds, what's the change in velocity?
- How would you find the position at t = 6s if you know the position at t = 2s and have a velocity-time graph?
Think through these problems, then discuss with classmates or your instructor to verify your understanding!