Constant Rate Of Change
Master the constant rate of change for Queensland Year 11 Mathematical Methods (QCAA). A constant rate means a quantity changes by the same amount each unit of time, drawing a straight-line graph with a fixed gradient.
You will learn to calculate a rate with its units, build a linear model from a starting value, convert between units, and reverse a rate to find a time — the groundwork for gradient.
Every question with a fully worked solution.
- Constant Rate Of Change - Video - Constant and non constant rates of change Watch
Theory
In Year 11 Mathematical Methods (QCAA, Unit 2), a constant rate of change is one that is the same over the whole interval, so the change in a quantity divided by the change in time (or per unit) is fixed. It equals the gradient of a straight-line graph and leads to a linear model \(y=\text{initial value}+\text{rate}\times t\). This page shows how to compute a constant rate with units, read it from a graph, build the linear equation, and convert or reverse a rate.
A constant rate of change measures how much one quantity changes for each unit change in another, when that change is the same everywhere. You compute it as the total change in the quantity divided by the total change in time (or in the other variable), and you must state its units — for example metres per second, litres per minute or dollars per kilogram.
Because the rate never varies, the graph of the relationship is a straight line, and the rate is exactly the line's gradient. A positive rate means the quantity is increasing (filling, rising); a negative rate means it is decreasing (draining, falling).
A constant rate gives a linear model \(y=c+kt\), where \(c\) is the initial value and \(k\) is the rate. You can read off a later value, or reverse the calculation to find a time.
A constant rate over a change in the other variable is
The linear relationship built from an initial value \(c\) and a constant rate \(k\) is
How to work with a constant rate
- Identify: find the total change in the quantity and the total change in time (or the other variable).
- Divide: compute \(\text{rate}=\dfrac{\text{change in quantity}}{\text{change in time}}\) and attach the correct units.
- Apply: use the rate as a gradient — build \(y=c+kt\), predict a later value, or reverse it to find a time.
Speed is distance divided by time:
| \(\text{speed}\) | \(=\) | \(\dfrac{\text{distance}}{\text{time}}\) |
| \(=\) | \(\dfrac{150\text{ m}}{6\text{ s}}\) | |
| \(=\) | \(25\text{ m/s}\) |
The constant speed is \(25\) m/s.
The speed is the gradient — rise over run between the two points:
| \(\text{speed}\) | \(=\) | \(\dfrac{100-20}{5-1}\) |
| \(=\) | \(\dfrac{80}{4}\) | |
| \(=\) | \(20\text{ m/s}\) |
The line is straight, so this rate is the same at every instant.
The constant speed is \(20\) m/s.
Rule — start at \(24\) and lose \(3\) cm each hour (a negative rate):
| \(h\) | \(=\) | \(24+(-3)\,t\) |
| \(h\) | \(=\) | \(24-3t\) |
Height after \(5\) hours — substitute \(t=5\):
| \(h\) | \(=\) | \(24-3(5)\) |
| \(=\) | \(24-15\) | |
| \(=\) | \(9\text{ cm}\) |
Rule \(h=24-3t\); after \(5\) hours the candle is \(9\) cm tall.
Convert kilometres to metres and hours to seconds — multiply by \(\dfrac{1000}{3600}\):
| \(108\text{ km/h}\) | \(=\) | \(108\times\dfrac{1000}{3600}\) |
| \(=\) | \(\dfrac{108000}{3600}\) | |
| \(=\) | \(30\text{ m/s}\) |
Since the speed is constant, the distance-time graph is the straight line \(d=30t\).
\(108\) km/h \(=30\) m/s.
Common pitfalls
Frequently asked questions
How do you calculate a constant rate of change?
Divide the total change in the quantity by the total change in time (or the other variable), and give the answer its units, for example \(\dfrac{150\text{ m}}{6\text{ s}}=25\) m/s.
Why does a constant rate give a straight-line graph?
Because the quantity changes by the same amount each unit, the gradient is the same everywhere, and a graph with a constant gradient is a straight line.
What does a negative rate of change mean?
The quantity is decreasing — draining, cooling or falling — and its straight-line graph slopes downwards.
How do you write a linear model from a constant rate?
Use \(y=c+kt\), where \(c\) is the initial value and \(k\) is the constant rate. For a candle \(24\) cm tall burning \(3\) cm/h, \(h=24-3t\).
How do you convert km/h to m/s?
Multiply by \(\dfrac{1000}{3600}\), which is the same as dividing by \(3.6\). So \(108\) km/h \(=30\) m/s.
How do you reverse a rate to find a time?
Rearrange \(\text{change}=\text{rate}\times\text{time}\) to \(\text{time}=\dfrac{\text{change}}{\text{rate}}\).