This Year 11 rates of change and areas under graphs revision notes is written for pupils aged 15-16 and follows the National Curriculum. In short, it shows you how to interpret distance-time and velocity-time graphs and estimate gradients and areas. Before you start, it helps to revise functions and graph transformations, because this lesson builds on it.
Year 11 rates of change and areas under graphs revision notes: what you will learn
Real-life graphs tell a story. The steepness shows how fast something is changing, and the area underneath can show a total amount. Gradients and areas for curved graphs are Higher tier only (marked Higher).
Starter activity
Imagine you walk to school. Sketch a graph of your distance from home against time. What does a steep line mean? What does a flat section show? Suppose you jog the second half. Where will the line change? Now sketch a graph of your speed against time for the same journey and decide what you think the area under it represents.
Distance-time graphs
Definition: distance-time graph
On a distance-time graph, time is on the horizontal axis and distance from the start is on the vertical axis. A straight sloping line means constant speed, a horizontal line means the object is stationary and a line sloping back down means it is returning to the start. The gradient is the speed.
Rule: speed and average speed
\(\text{speed}=\dfrac{\text{distance}}{\text{time}}\). Average speed for a whole journey is \(\dfrac{\text{total distance travelled}}{\text{total time taken}}\). Include stops in the total time.
Worked example 1: a runner
A runner travels 6 km in the first 30 minutes, rests for 10 minutes, then runs another 4 km in 20 minutes. Find the speed in the first part and the average speed.
Speed in the first part: \(6\div0.5=12\) km/h. Total distance \(=10\) km, total time \(=60\) minutes \(=1\) hour. Average speed \(=10\div1=10\) km/h.
Velocity-time graphs
Definition: acceleration
On a velocity-time (speed-time) graph the gradient is the acceleration: \(\dfrac{\text{change in velocity}}{\text{time}}\), in \(\text{m/s}^2\). A horizontal line means constant speed and a negative gradient means deceleration.
Rule: area under a velocity-time graph
The area under the graph is the distance travelled. Use these areas: rectangle \(=bh\), triangle \(=\tfrac12bh\), trapezium \(=\tfrac12(a+b)h\), where \(a\) and \(b\) are the parallel sides.
Worked example 2: a car journey
A car accelerates from 0 to 12 m/s in 4 s, travels at 12 m/s for 10 s, then slows to rest in 6 s. Find the acceleration, the total distance and the average speed.
Acceleration \(=\dfrac{12}{4}=3\ \text{m/s}^2\). Distances: \(\tfrac12\times4\times12=24\) m, then \(12\times10=120\) m, then \(\tfrac12\times6\times12=36\) m. Total \(=180\) m in 20 s, so the average speed is \(9\) m/s.
Other real-life graphs and units
Key fact: gradient is a rate
The gradient of any straight line graph is a rate. For a taxi, cost against distance has gradient equal to the cost per kilometre, and the \(y\)-intercept is the fixed charge. Write the units of a rate as "units of \(y\) per unit of \(x\)", for example pounds per kilometre.
Worked example 3: converting units
Convert 72 km/h into m/s, and 15 m/s into km/h.
To change km/h into m/s, multiply by 1000 and divide by 3600, which is the same as dividing by 3.6: \(72\ \text{km/h}=20\ \text{m/s}\). To go the other way, multiply by 3.6: \(15\ \text{m/s}=54\ \text{km/h}\).
(Higher) Curved graphs: gradient and area
Definition (Higher): rate of change at a point
For a curve, the gradient changes from point to point. The rate of change at a point is the gradient of the tangent to the curve at that point. On a distance-time graph it is the speed at that instant; on a velocity-time graph it is the acceleration at that instant.
Method (Higher): gradient of a curve
- Place a ruler so it touches the curve at the point and does not cut it.
- Draw the tangent line, making it long.
- Choose two points far apart on the tangent and calculate \(\dfrac{\text{change in }y}{\text{change in }x}\).
Rule (Higher): the trapezium estimate
To estimate the area under a curve, split it into strips and treat each strip as a trapezium. If the strips have width \(w\) and heights \(y_0,y_1,\dots,y_n\), the area is \(\tfrac12w\,(y_0+2y_1+2y_2+\dots+y_n)\). When the curve bends upwards (convex), trapezia overestimate; when it bends downwards (concave), they underestimate. More strips give a better estimate.
Worked example 4 (Higher): estimating distance
A velocity-time curve has \(v=0,5,8,9\) m/s at \(t=0,2,4,6\) s. Estimate the distance with trapezia.
Strip areas: \(\tfrac{0+5}{2}\times2=5\), \(\tfrac{5+8}{2}\times2=13\), \(\tfrac{8+9}{2}\times2=17\). Total \(=35\) m. The curve is bending downwards, so this is an underestimate.
Worked example 5 (Higher): different scales
1 cm across is 5 s and 1 cm up is 4 m/s. What does 3 cm\(^2\) of area represent?
The area represents (horizontal scale) \(\times\) (vertical scale) \(\times\) (area in squares): \(3\times5\times4=60\) m.
Watch out: common mistakes
- Reading a horizontal line on a distance-time graph as constant speed. It means stopped.
- Mixing minutes and hours in a speed calculation.
- Using the area under a distance-time graph. It has no meaning here.
- Drawing a tangent that crosses the curve, or using points too close together.
Worked example 6: Average speed for a two-part journey
A runner goes 6 km at 12 km/h and then 6 km at 8 km/h. Find the average speed.
Times: \(\dfrac6{12}=0.5\) h and \(\dfrac68=0.75\) h. Total time \(1.25\) h for 12 km. Average speed \(=\dfrac{12}{1.25}=9.6\) km/h. This is not the mean of 12 and 8 (which is 10) because more time is spent at the slower speed. Always use total distance divided by total time.
Worked example 7: Reading acceleration from a velocity-time graph
A velocity-time graph goes from \((0,5)\) to \((8,29)\) in a straight line. Find the acceleration and the distance in the 8 seconds.
Gradient \(=\dfrac{29-5}8=3\) m/s\(^2\). The area under the line is a trapezium: \(\tfrac12(5+29)\times8=136\) m. The units of the area are metres because velocity (m/s) times time (s) gives m.
Worked example 8: Constant speed from a distance-time graph
On a distance-time graph a straight line goes through \((1,20)\) and \((4,50)\). What is the speed?
Gradient \(=\dfrac{50-20}{4-1}=10\). The units are those of distance divided by those of time: if the distance is in km and the time in hours the speed is 10 km/h. A steeper line means a higher speed, and a line sloping down means returning.
Worked example 9: A tangent to a curve
A distance-time curve has a tangent at \(t=3\) passing through \((1,2)\) and \((5,14)\). Estimate the speed at \(t=3\).
The speed at an instant is the gradient of the tangent: \(\dfrac{14-2}{5-1}=3\). Draw the tangent so that it just touches the curve at the point and use a large triangle to calculate the gradient accurately.
Worked example 10: Using the area under a graph for other quantities
A graph shows the flow rate of water in litres per minute against time in minutes. Between 0 and 10 minutes the flow is steady at 20 L/min. How much water has flowed?
Area \(=20\times10=200\) litres. The area under a rate graph gives the total amount: litres per minute multiplied by minutes. Likewise, the area under a speed-time graph gives distance and the gradient of a distance-time graph gives speed.
Worked example 11: Converting between m/s and km/h
Show that 1 m/s is 3.6 km/h and use this to convert 30 m/s.
1 m/s is 3 600 m in an hour, which is 3.6 km/h. So \(30\times3.6=108\) km/h. To convert the other way, divide by 3.6: 54 km/h is \(54\div3.6=15\) m/s.
Worked example 12: An estimate for unequal strips
The speed of a bike at 0, 1, 3 seconds is 0, 4, 8 m/s. Estimate the distance.
Two trapezia with different widths: \(\tfrac12\times1\times(0+4)=2\) and \(\tfrac12\times2\times(4+8)=12\). The total is 14 m. Each trapezium uses its own width, which is the time gap for that strip.
Note: the three key links between graphs
Remember three links. On a distance-time graph the gradient is speed. On a velocity-time graph the gradient is acceleration and the area under the graph is distance. For a curved graph, the gradient at a point is found with a tangent and the area is estimated with trapezia. Write the units of every answer: m/s for speed, m/s² for acceleration and m for distance. A flat line means constant value, which on a distance-time graph means stopped, and on a velocity-time graph means constant speed.
Method: estimating the gradient of a curve
- Place a ruler so that it just touches the curve at the point and appears to follow its direction.
- Draw a long straight line through the point.
- Choose two points far apart on the tangent and draw a right-angled triangle.
- Work out the vertical change divided by the horizontal change, and include the units.
Watch out: common mistakes with rate graphs
- Mixing units: if time is in minutes and distance in km, the speed is km per minute, not km/h.
- Confusing a horizontal line on a distance-time graph (stopped) with one on a velocity-time graph (steady speed).
- Using the mean of speeds as the average speed instead of total distance divided by total time.
- Forgetting that area under a velocity-time graph is distance only when velocity and time units match.
Did you know?
Galileo Galilei, an Italian scientist who lived about 400 years ago, rolled balls down slopes to study acceleration. He found that the distance travelled grows with the square of the time, so a graph of distance against time is a curve. Modern speed cameras and fitness watches use the same idea: they measure position at regular times and calculate speed and acceleration from the gradient of the graph.
Otto’s tip
Write the units next to every number. If your answer for acceleration comes out in metres, something has gone wrong. Gradient of distance gives speed, gradient of speed gives acceleration, and area under speed gives distance.
How to use this Year 11 rates of change and areas under graphs revision notes
First, read the key points so that you know where the Year 11 rates of change and areas under graphs revision notes is going. Then work through each definition and worked example, and cover the solution before you read it. After that, practise with the rates of change and areas under graphs worksheet and check every question with the worksheet answers. When you feel ready, test yourself with the multiple choice questions or the quick quiz. Finally, sit the rates of change and areas under graphs test and mark it with the test answers.
Also, the starters are a good warm-up, so you can keep the key facts fresh. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, this Year 11 rates of change and areas under graphs revision notes gives you everything you need for the chapter.
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