Describing Motion Around Us · Lesson 7 of 10
Velocity-time graphs
“This graph gives velocity, acceleration and displacement without asking for overtime.”
• Read velocity from a velocity-time graph. • Identify zero, positive and negative acceleration from graph slope. • Calculate acceleration from graph coordinates. • Calculate displacement from the area under a graph. • Break a trapezium into rectangles and triangles. • Distinguish graph height, slope and area.
A velocity-time graph contains three layers of information. Its height tells the velocity at an instant. Its slope tells how quickly velocity changes. The area between the graph and the time axis tells the displacement. Keeping these three meanings separate is the key to reading the graph correctly.
A velocity-time graph shows how an object’s velocity changes with time.
| Graph form | Velocity | Acceleration |
|---|---|---|
| Horizontal line | Constant | Zero |
| Straight line rising with time | Increases equally in equal intervals | Constant and positive under the selected convention |
| Straight line falling with time | Decreases equally in equal intervals | Constant and negative under the selected convention |
The slope of a velocity-time graph is change in velocity divided by change in time. It therefore gives acceleration. A zero slope means velocity does not change. A positive slope means velocity increases in the positive direction. A negative slope means acceleration points in the negative direction.
Problem
Velocity increases from 5 m s⁻¹ at 10 s to 10 m s⁻¹ at 20 s. Find acceleration.
- 1.Change in velocity = (10-5=5 m s⁻¹).
- 2.Change in time = (20-10=10 s).
- 3.Acceleration = (5÷10=0.5 m s⁻²).
- 4.The positive value agrees with the upward slope.
Problem
Velocity decreases from 10 m s⁻¹ at 10 s to 5 m s⁻¹ at 20 s. Find acceleration.
- 1.Change in velocity = (5-10=-5 m s⁻¹).
- 2.Change in time = (20-10=10 s).
- 3.Acceleration = (-5÷10=-0.5 m s⁻²).
- 4.The negative value agrees with the downward slope.
The area under the graph has units ((m s^{-1})(s)=m), so it represents displacement. For constant velocity, the region is a rectangle. For a straight line showing constant acceleration, the region may be treated as a rectangle plus a triangle, or as a trapezium.
Problem
A car moves at 20 m s⁻¹ for 6 s. Find displacement from its velocity-time graph.
- 1.The graph is a horizontal line at 20 m s⁻¹.
- 2.The region under it is a rectangle.
- 3.Width = (6 s); height = (20 m s⁻¹).
- 4.Area = (6×20=120 m).
- 5.Therefore, displacement = 120 m.
Problem
Velocity changes uniformly from 5 m s⁻¹ to 10 m s⁻¹ between 10 s and 20 s. Find displacement.
- 1.Time interval = (20-10=10 s).
- 2.Rectangle area = initial velocity × time = (5×10=50 m).
- 3.Triangle height = change in velocity = (10-5=5 m s⁻¹).
- 4.Triangle area = (½×10×5=25 m).
- 5.Total displacement = (50+25=75 m).
On a velocity-time graph, height gives velocity, slope gives acceleration and area gives displacement.
Quiz
Which description best matches Velocity-Time Graph?
Which term matches this description: A velocity-time graph shows how an object’s velocity changes with time.
Which statement is a key takeaway from this lesson?
Which additional statement is also a key takeaway from this lesson?
Which further statement is also a key takeaway from this lesson?
Practice Problems
- Find acceleration when velocity rises from 2 to 14 m s⁻¹ in 6 s.
- Find displacement at constant velocity 8 m s⁻¹ for 15 s.
- Velocity rises uniformly from 4 to 10 m s⁻¹ in 3 s. Find displacement using areas.
- Explain why the area unit reduces to metre.
Key Takeaways
• Graph height gives velocity. • Graph slope gives acceleration. • A horizontal line shows zero acceleration. • The area under the graph gives displacement. • Complex areas can be divided into familiar shapes.