Motion Graphs
3. Area Under a Graph
Learning outcomes
- I can explain the significance of the area under a velocity-time graph.
- I can calculate displacement from the area under a velocity-time graph.
- I can calculate areas using rectangles and triangles.
- I can distinguish between slope and area when interpreting graphs.
- I can solve motion problems using graphical methods.
Area Under a Velocity-Time Graph
A velocity-time graph tells us how an object's velocity changes over time.
There are two especially important features to interpret:
- the slope tells us the object's acceleration
- the area between the graph and the time axis tells us the object's displacement
This means a single velocity-time graph can provide information about both how the velocity changes and how far the object moves from its starting position.
Why Does Area Represent Displacement?
Start with the equation for constant velocity: \( v = \frac{ \Delta x }{ \Delta t } \)
Rearranging: Δx = vΔt
where:
- Δx = displacement
- v = velocity
- Δt = time interval
On a velocity-time graph:
- velocity is the height
- time is the width
Therefore: Area = (velocity)(time)
So: Area under a velocity-time graph = displacement
The units confirm this: \( (\frac{m}{s})(s) = m \)
The result is measured in metres, which is the unit of displacement.
Constant Velocity: Rectangle Area
Suppose a car travels at a constant velocity of 8 m/s for 5s
The velocity-time graph forms a rectangle.
The area of a rectangle is: A = (base)(height)
Therefore: A = (5)(8) = 40m
The car's displacement is 40m.
Accelerating Motion: Triangle Area
Suppose an object starts from rest and accelerates uniformly to 12 m/s over 4s.
The graph forms a triangle.
The area of a triangle is: \( A = \frac{1}{2}(base)(height) \)
Therefore: \( A = \frac{1}{2}(4)(12) = 24 m \)
So the displacement is 24 m.
Why a Triangle Appears During Constant Acceleration
If an object starts from rest and accelerates at a constant rate, its velocity increases steadily.
For example:
| Time | Velocity |
|---|---|
| 0 s | 0 m/s |
| 1 s | 3 m/s |
| 2 s | 6 m/s |
| 3 s | 9 m/s |
| 4 s | 12 m/s |
Because the velocity changes at a constant rate, the velocity-time graph is a straight sloping line.
The area beneath this line forms a triangle.
Combining Shapes
More complicated velocity-time graphs can often be divided into simple shapes.
Common shapes include:
- rectangles
- triangles
- trapezoids
Calculate each area separately and then add them together.
Worked Example: Accelerate, Then Constant Velocity
A cyclist:
- accelerates from 0 to 10 m/s during the first 4 seconds
- then travels at 10 m/s for another 6 seconds
The graph can be divided into:
- one triangle
- one rectangle
Triangle
\( A = \frac{1}{2}(4)(10) = 20m \)
Rectangle
A = 6(10) = 60m
Total Displacement
20 + 60 = 80
Displacement = 80 m
A Complete Journey
Suppose a car has the following motion:
- accelerates from 0 to 12 m/s in 4 s
- travels at 12 m/s for 5 s
- slows uniformly to rest over 3 s
The graph contains:
- Triangle A
- Rectangle B
- Triangle C
Triangle A
\( A = \frac{1}{2}(4)(12) = 24 m\)
Rectangle B
A = 5(12) = 60 m
Triangle C
\( A = \frac{1}{2}(3)(12) = 18m \)
Total
24 + 60 + 18 = 102
Total displacement = 102 m
Using a Trapezoid
Sometimes an object begins with a non-zero velocity and then accelerates.
For example:
Initial velocity: 4 m/s
Final velocity: 10 m/s
Time: 3 s
The area under the graph forms a trapezoid.
The trapezoid area can be calculated using: \( A = \frac{1}{2}(a + b)h \)
For a velocity-time graph: \( A = \frac{1}{2}(v_i + v_f)t \)
Therefore: \( A = \frac{1}{2}(4 + 10)(3) = \frac{1}{2}(14)(3) = 21 m \)
The displacement is 21 m.
Another Way to Handle a Trapezoid
Instead of using the trapezoid formula, we can divide it into:
- a rectangle
- a triangle
Rectangle
A = 4(3) = 12 m
Triangle
Difference in velocity: 10 - 4 = 6 m/s
Area: \( A = \frac{1}{2}(3)(6) = 9 m \)
Total: 12 + 9 = 21 m
The answer is the same.
Positive Area
If the graph is above the time axis, the velocity is positive.
Therefore, the area represents positive displacement.
For example: +6 m/s for 5 s
gives: Δx = 6(5) = + 30 m
The object moves 30 m in the positive direction.
Negative Area
If the graph is below the time axis, the velocity is negative.
The area therefore represents negative displacement.
For example: -4 m/s for 5 s
gives: Δx = (-4)(5) = -20 m
The negative sign means the displacement is in the negative direction.
When the Graph Crosses the Axis
A velocity-time graph can contain both positive and negative areas.
Suppose an object:
- moves forward
- stops
- reverses direction
- moves backward
The area above the time axis represents displacement in the positive direction.
The area below the axis represents displacement in the negative direction.
To calculate the final displacement:
Net displacement = positive area + negative area
Remember that the area below the axis has a negative sign.
Worked Example: Changing Direction
Suppose the positive area is: +50 m
and the negative area is: -20 m
The net displacement is: 50 + (-20) = 30 m
The object finishes 30 m in the positive direction from its starting point.
Displacement vs Distance
This is an important distinction.
Displacement includes direction.
Distance is the total amount of ground travelled.
Suppose an object moves: +50m then -20 m
Displacement
50 - 20 = 30 m
Distance
50 + 20 = 70 m
Therefore:
Distance = 30 m
but:
Distance = 70 m
On a velocity-time graph:
Displacement = signed area
For total distance, add the magnitudes of all areas.
Slope and Area Are Different
One of the most important skills is distinguishing between the slope and area of a velocity-time graph.
Slope
The slope tells us: Acceleration because:
\( a = \frac{ \Delta v }{ \Delta t } \)
Area
The area tells us: Displacement
because: Δx = vΔt
A useful memory rule is:
Slope → Acceleration
Area → Displacement
Don't Confuse the Two
Consider a horizontal line on a velocity-time graph.
Its slope is: 0
Therefore: a = 0
But there may still be a large area beneath the graph.
So the object can have:
Zero acceleration
while still having:
Large displacement
For example, travelling at a constant 20 m/s for 10 seconds:
Acceleration: 0 m/s2
Displacement: 20(10) = 200 m
This is why slope and area must be interpreted separately.
Solving Motion Problems Graphically
Consider this journey:
A train:
- accelerates from 0 to 20 m/s in 10 s
- travels at 20 m/s for 30 s
- slows to rest in 10 s
We can solve the displacement entirely from the graph.
Acceleration Section
Triangle:
\( A = \frac{1}{2}(10)(20) = 100 m \)
Constant Velocity Section
Rectangle:
A = (30)(20) = 600 m
Deceleration Section
Triangle:
\( A = \frac{1}{2}(10)(20) = 100 m \)
Total Displacement
100 + 600 + 100 = 800 m
No displacement equation was needed beyond calculating the areas of geometric shapes.
Graphical Methods Can Give Multiple Answers
A velocity-time graph can often tell us several things about the same motion.
For example:
Slope
Provides:
Area
Provides:
Displacement
Height
Provides:
Velocity
Position above or below axis
Provides:
Direction of motion
Crossing the axis
May indicate:
A change in direction
Velocity-time graphs therefore contain a large amount of information.
Constructing and Analysing a Graph
Suppose a cyclist:
- starts from rest
- accelerates uniformly to 6 m/s in 3 s
- travels at 6 m/s for 4 s
- slows uniformly to rest in 2 s
First identify the points:
| Time | Velocity |
|---|---|
| 0 s | 0 m/s |
| 3 s | 6 m/s |
| 7 s | 6 m/s |
| 9 s | 0 m/s |
After constructing the graph, calculate the areas.
First Triangle
\( A = \frac{1}{2}(3)(6) = 9m \)
Rectangle
A = (4)(6) = 24 m
Final Triangle
\( A = \frac{1}{2}(2)(6) = 6 m \)
Total:
9 + 24 + 6 = 39 m
The cyclist's displacement is 39 m.
A Useful Problem-Solving Method
When asked to find displacement from a velocity-time graph:
Step 1: Identify the time interval.
Step 2: Divide the area into simple shapes.
Step 3: Calculate each area.
Step 4: Give areas below the time axis a negative sign.
Step 5: Add the areas.
Step 6: Give the answer in metres.
For distance rather than displacement, add the absolute values of the areas.
Common Misconceptions
The slope does not give displacement.
Slope gives acceleration.
The area does not give acceleration.
Area gives displacement.
Areas below the axis are not automatically ignored.
They represent negative displacement.
Distance and displacement are not always equal.
If an object changes direction, distance is usually greater than the magnitude of displacement.
Did You Know?
The area-under-the-graph idea is an early example of a much more powerful mathematical concept called integration.
In calculus, displacement can be found from a velocity function by integrating velocity over time.
For straight-line sections, however, there is no need for calculus. We can simply calculate the areas of familiar geometric shapes such as rectangles, triangles, and trapezoids.
Key Terms
Area under a graph – The region between a graph line and the horizontal axis.
Displacement – Change in position, including direction.
Distance – Total length of the path travelled.
Velocity-time graph – A graph showing velocity against time.
Slope – The steepness of a graph; on a velocity-time graph it represents acceleration.
Rectangle – A shape with area A = bh.
Triangle – A shape with area A = \( \frac{1}{2}bh \).
Trapezoid – A four-sided shape with one pair of parallel sides.
Net displacement – The overall change in position after positive and negative displacements are combined.
Key Takeaways
- The area under a velocity-time graph represents displacement.
- This works because vt gives units of metres.
- A rectangle has area: A = bh
- A triangle has area: A = \( \frac{1}{2}bh \)
- More complicated graphs can be divided into rectangles, triangles, and trapezoids.
- Area above the time axis represents positive displacement.
- Area below the time axis represents negative displacement.
- Net displacement is found by adding positive and negative areas.
- Total distance is found by adding the magnitudes of all areas.
- The slope of a velocity-time graph gives acceleration.
- The area under a velocity-time graph gives displacement.
- Velocity-time graphs can be used to solve motion problems graphically, often without needing more advanced equations.