Instantaneous vs. Average Speed Revision Notes for HSC SSCE Mathematics Advanced

Instantaneous vs. Average Speed

What is average speed?

Average speed tells us how fast something travels over a period of time. It gives us an overall measure of motion by dividing the total distance travelled by the total time taken.

The formula for average speed over an interval $[t_1, t_2]$ is:

$\text{Average speed} = \frac{\Delta d}{\Delta t} = \frac{d(t_2) - d(t_1)}{t_2 - t_1}$

where $d$ represents distance and $t$ represents time.

This calculation gives us a single value that represents the average rate of motion across the entire time interval.

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Think of average speed as the overall pace of a journey. For example, if you drive 120 km in 2 hours, your average speed is 60 km/h, even though you might have driven faster or slower at different points during the trip.

What is instantaneous speed?

Instantaneous speed is quite different from average speed. It tells us the speed at a specific moment in time, rather than over a period of time. Think of it like looking at a speedometer in a car - it shows your speed at that exact instant.

Mathematically, instantaneous speed is represented by the gradient of the tangent line to the distance-time graph at a particular point. This represents the rate at which distance is changing at that exact moment.

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The key distinction: Average speed measures motion over an interval, while instantaneous speed captures the exact speed at a single moment. This difference is crucial for understanding variable motion and acceleration.

Comparing average and instantaneous speed

The key difference between these two concepts lies in what they measure:

Average speed provides an overall measure of how fast something travels during a time interval
Instantaneous speed captures the exact speed at a given moment, which is useful for analysing variable motion such as acceleration

On a distance-time graph, these two types of speed are represented differently:

Average speed is shown by the gradient of a secant line connecting two points on the curve
Instantaneous speed is shown by the gradient of a tangent line touching the curve at a single point

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Understanding the Graph

In the graph above, the secant line from $t = 1$ to $t = 3$ represents average speed over that interval, while the gradient at the point where $t = 2$ represents the instantaneous speed at that moment.

Notice how the tangent line only touches the curve at one point, while the secant line cuts through it at two points. This visual difference mirrors the conceptual difference between the two types of speed.

Worked example: Comparing speeds

Let's look at a practical example to understand how these concepts work together.

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Worked Example: Cyclist Motion Analysis

Question: A cyclist's distance is modelled by $d(t) = t^3$ , where $d$ is in metres and $t$ is in seconds. Compare the average speed from $t = 1$ to $t = 3$ with an approximation of the instantaneous speed at $t = 2$ using the interval $[2, 2.5]$ .

Part 1: Calculating average speed

First, we'll find the average speed from $t = 1$ to $t = 3$ .

Write the equation: $d(t) = t^3$

Substitute $t = 1$ : $d(1) = 1^3 = 1$

Write the equation again: $d(t) = t^3$

Substitute $t = 3$ : $d(3) = 3^3 = 27$

Write the formula for average speed: $\text{Average speed} = \frac{d(t_2) - d(t_1)}{t_2 - t_1}$

Substitute $t_2 = 3$ and $t_1 = 1$ : $= \frac{d(3) - d(1)}{3 - 1}$

Substitute the values $d(3) = 27$ and $d(1) = 1$ : $= \frac{27 - 1}{3 - 1}$

Evaluate the subtraction: $= \frac{26}{2}$

Evaluate: $= 13$

The average speed is 13 m/s.

Part 2: Approximating instantaneous speed

Now we'll approximate the instantaneous speed at $t = 2$ using the interval $[2, 2.5]$ .

Write the equation: $d(t) = t^3$

Substitute $t = 2$ : $d(2) = 2^3 = 8$

Write the equation again: $d(t) = t^3$

Substitute $t = 2.5$ : $d(2.5) = 2.5^3 = 15.625$

Write the formula for approximate instantaneous speed: $\text{Approximate instantaneous speed} = \frac{d(t_2) - d(t_1)}{t_2 - t_1}$

Substitute $t_2 = 2.5$ and $t_1 = 2$ : $= \frac{d(2.5) - d(2)}{2.5 - 2}$

Substitute $d(2.5) = 15.625$ and $d(2) = 8$ : $= \frac{15.625 - 8}{2.5 - 2}$

Evaluate the subtraction: $= \frac{7.625}{0.5}$

Evaluate: $= 15.25$

The approximate instantaneous speed at t = 2 is 15.25 m/s.

Interpretation:

The average speed (13 m/s) is lower than the instantaneous speed (15.25 m/s), indicating the cyclist is accelerating. This makes sense because $d(t) = t^3$ is a non-linear function showing increasing speed over time.

Approximating instantaneous speed

Since calculating the exact instantaneous speed requires calculus techniques you'll learn later, we can approximate it using a clever method.

The instantaneous speed at time $t$ can be approximated by calculating the average speed over a very small interval. Rather than using an interval centred on $t$ , we choose one that starts at $t$ and extends a short distance to the right, such as $[t, t + h]$ , where $h$ is a small value.

This approach simplifies calculations and gives us a good approximation.

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Key Principle: The smaller the interval, the closer the approximation to the true instantaneous speed.

This is because as the interval gets smaller, the secant line approaches the tangent line. When the secant line nearly coincides with the tangent line, the gradient of the secant (average speed) closely matches the gradient of the tangent (instantaneous speed).

Improving the approximation

Let's see how reducing the interval size improves our approximation.

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Worked Example: Improved Approximation

Question: For the same cyclist's motion, $d(t) = t^3$ (metres, seconds), improve the accuracy of the approximation of instantaneous speed at $t = 2$ using the interval $[2, 2.1]$ . Explain how this improves from $[2, 2.5]$ .

Calculating with the smaller interval:

Write the equation: $d(t) = t^3$

Substitute $t = 2$ : $d(2) = 2^3 = 8$

Write the equation again: $d(t) = t^3$

Substitute $t = 2.1$ : $d(2.1) = 2.1^3 = 9.261$

Write the formula: $\text{Approximate instantaneous speed} = \frac{d(t_2) - d(t_1)}{t_2 - t_1}$

Substitute $t_2 = 2.1$ and $t_1 = 2$ : $= \frac{d(2.1) - d(2)}{2.1 - 2}$

Substitute the values: $= \frac{9.261 - 8}{2.1 - 2}$

Evaluate the subtraction: $= \frac{1.261}{0.1}$

Evaluate: $= 12.6$

The approximate instantaneous speed at t = 2 is 12.6 m/s.

Why is this better?

Using $h = 0.1$ (interval $[2, 2.1]$ ) gives a closer approximation than $h = 0.5$ (interval $[2, 2.5]$ ) because the secant line is nearer to the tangent. This reduces the error in estimating the gradient at $t = 2$ .

Smaller intervals yield better approximations because the secant line more closely aligns with the tangent at $t = 2$ .

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Visualizing the Improvement

Imagine drawing secant lines with progressively smaller intervals. Each time you reduce the interval size, the secant line tilts closer to matching the tangent line at $t = 2$ . The gradient gets more accurate with each reduction, approaching the true instantaneous speed as the interval approaches zero.

Exam tips

When working with speed problems in exams, keep these strategies in mind:

Always show your working when calculating average or instantaneous speed
Remember that instantaneous speed requires very small intervals for accurate approximation
On a distance-time graph, look for the steepness of the curve at a point to understand instantaneous speed
If average speed is different from instantaneous speed, this indicates acceleration or deceleration
Label your units clearly (e.g., m/s) in your final answer

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Common Mistake to Avoid

Don't confuse the formulas! While both average and instantaneous speed use the same formula structure $\frac{d(t_2) - d(t_1)}{t_2 - t_1}$ , the key difference is the interval size. Average speed uses any interval, while approximating instantaneous speed requires a very small interval starting at the point of interest.

Remember!

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Key Points to Remember:

Average speed is calculated by dividing total distance by total time: $\frac{\Delta d}{\Delta t} = \frac{d(t_2) - d(t_1)}{t_2 - t_1}$
Instantaneous speed is the speed at a specific moment, represented by the gradient of the tangent to the distance-time graph
To approximate instantaneous speed at time $t$ , calculate average speed over a small interval $[t, t + h]$ where $h$ is small
Smaller intervals produce more accurate approximations because the secant line approaches the tangent line
The difference between average and instantaneous speed helps us analyse variable motion and acceleration
On graphs: Secant line represents average speed, tangent line represents instantaneous speed

Instantaneous vs. Average Speed (HSC SSCE Mathematics Advanced): Revision Notes