The fastest 100m sprint ever was 9.58 seconds. But the runner was not running at 10.44 m/s for the entire race. His speed varied: slow at the start, explosive in the middle, then a slight fade at the end. In this lesson, you will learn how to measure average speed over an interval — and glimpse how calculus will let us find his exact speed at any single instant.
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A car travels 120 km in 2 hours, so its average speed is 60 km/h. Does this mean the speedometer showed exactly 60 km/h at every moment? What do you think happens to the accuracy of the average speed as we measure it over shorter and shorter time intervals — say, over 1 minute, then 10 seconds, then 1 second?
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Wrong: Average rate of change over an interval equals instantaneous rate of change at every point in that interval.
Right: Average rate of change is the overall slope between two points. Instantaneous rate of change is the slope at a single point. They are only equal for linear functions.
📚 Core Content
When a quantity changes, we often want to know how fast it changes. If a car travels from $x_1$ to $x_2$ and its position changes from $f(x_1)$ to $f(x_2)$, the average rate of change is:
$$\text{Average rate of change} = \frac{\Delta y}{\Delta x} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$$
This is exactly the same calculation as finding the gradient of a straight line through the two points $(x_1, f(x_1))$ and $(x_2, f(x_2))$. On a curve, this line is called a secant — it cuts through the curve at two points.
The same idea applies to any changing quantity:
Average speed over a whole journey tells us very little about what the speedometer reads at any single moment. To get a better estimate of the speed at a particular instant, we calculate the average speed over a shorter interval around that instant.
Imagine watching a runner at the 50-metre mark. If we time them from 45 m to 55 m, we get a good estimate of their speed at 50 m. If we time them from 49 m to 51 m, the estimate gets even better. As the interval shrinks toward zero, the average speed approaches the instantaneous speed — the exact reading on the speedometer.
Geometrically, as the second point on the curve moves closer to the first, the secant line pivots and approaches a limiting position: the tangent line at that point. The gradient of this tangent is the instantaneous rate of change.
🧮 Worked Examples
🧪 Activities
1 $f(x) = 3x + 5$ over $[2, 6]$
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2 $g(x) = x^2 - 2x$ over $[1, 5]$
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3 $h(t) = t^3$ over $[1, 3]$
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4 The height of a ball is $s(t) = 5t^2$ metres after $t$ seconds. Find the average speed over $[2, 4]$.
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1 A company's profit $P$ (in thousands of dollars) after selling $n$ items is $P(n) = 2n - 100$. The average rate of change of profit over $[50, 150]$ is 2. Explain what this means in practical terms.
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2 A population of bacteria is modelled by $P(t) = 100 \times 2^t$ where $t$ is in hours. The average rate of change over $[0, 2]$ is 150 bacteria per hour. Why might the instantaneous growth rate at $t = 0$ be different from 150?
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Earlier you were asked: Does an average speed of 60 km/h mean the speedometer showed exactly 60 km/h at every moment?
No. Average speed is calculated over a whole interval and smooths out all the variation. The speedometer shows instantaneous speed, which can be faster or slower than the average at any given moment. As we measure over shorter and shorter intervals, the average speed gets closer and closer to the instantaneous speed — this is the fundamental idea that will lead us to the derivative.
Now revisit your initial response. What did you get right? What has changed in your thinking?
Look back at your initial response in your book. Annotate it with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
6. The position of a cyclist along a straight track is given by $s(t) = t^2 + 4t$ metres, where $t$ is in seconds. Find the average speed over $[1, 5]$, and estimate the instantaneous speed at $t = 3$ by using the interval $[2.9, 3.1]$. Show all working. 4 MARKS
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7. The graph of $y = f(x)$ passes through $(1, 2)$ and $(5, 10)$. Explain why the average rate of change over $[1, 5]$ does not tell us the exact value of $f'(3)$, and describe what would happen to the estimate of $f'(3)$ if we used the interval $[2.9, 3.1]$ instead. 3 MARKS
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8. Usain Bolt's 100m world record of 9.58 seconds gives an average speed of approximately 10.44 m/s. However, video analysis shows his peak speed was closer to 12.3 m/s around the 60–80 metre mark. By considering the difference between average and instantaneous rates of change, explain why his average speed is lower than his peak speed. 3 MARKS
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1. $\frac{f(6) - f(2)}{6 - 2} = \frac{23 - 11}{4} = 3$
2. $\frac{g(5) - g(1)}{5 - 1} = \frac{15 - (-1)}{4} = 4$
3. $\frac{h(3) - h(1)}{3 - 1} = \frac{27 - 1}{2} = 13$
4. $\frac{s(4) - s(2)}{4 - 2} = \frac{80 - 20}{2} = 30$ m/s
1. For each additional item sold between 50 and 150, the company makes an average profit of $2,000 (since $P$ is in thousands). This is a constant rate because the function is linear.
2. The average rate of 150 bacteria/hour is an overall measure across 2 hours. At $t = 0$, the population is just starting to grow, so the instantaneous growth rate is lower than the average over $[0, 2]$.
1. A — $\frac{f(3) - f(1)}{3 - 1} = \frac{9 - 1}{2} = 4$
2. A — $\frac{s(4) - s(2)}{4 - 2} = \frac{80 - 20}{2} = 30$ m/s
3. A — Average rate of change = $\frac{\\Delta y}{\\Delta x}$, which is the gradient of the secant line.
4. A — As the interval shrinks, the average rate approaches the instantaneous rate.
5. A — $\frac{f(4) - f(0)}{4 - 0} = \frac{32 - 0}{4} = 8$
6. A — For a straight line, the gradient is constant, so the average rate of change always equals $m$.
7. A — If the function is increasing, $f(b) > f(a)$, so $\frac{f(b) - f(a)}{b - a} > 0$.
Q6 (4 marks): $s(1) = 5$ and $s(5) = 45$, so average speed $= \frac{45 - 5}{5 - 1} = 10$ m/s [2]. For instantaneous speed at $t = 3$: $s(2.9) = 20.01$ [1], $s(3.1) = 22.01$ [1]. Average speed over $[2.9, 3.1] = \frac{22.01 - 20.01}{0.2} = 10$ m/s [1]. This estimates instantaneous speed because the interval is very narrow [1].
Q7 (3 marks): The average rate of change over $[1, 5]$ equals the gradient of the secant through $(1, 2)$ and $(5, 10)$, which is $\frac{10 - 2}{5 - 1} = 2$ [1]. This measures the overall steepness across the whole interval and does not account for how the curve may bend between the points [1]. Using $[2.9, 3.1]$ shrinks the interval so the secant approximates the tangent at $x = 3$ much more closely, giving a better estimate of $f'(3)$ [1].
Q8 (3 marks): Average speed is calculated over the entire 100 m race and smooths out all changes in speed [1]. Bolt started from rest and accelerated, so his speed was below the average at the start and end [1]. His peak instantaneous speed of about 12.3 m/s around the 60–80 m mark is the speed at a single instant and is not dragged down by the slower sections, which is why it is higher than the overall average of 10.44 m/s [1].
Average and Instantaneous Rates of Change
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