You already know how to find a gradient. The new idea here is that on a real-life graph the gradient always means something: it is the rate at which the vertical quantity changes for each unit of the horizontal quantity.
If $y$ is in metres and $x$ is in seconds, the gradient is in metres per second — a speed.
If $y$ is in pounds and $x$ is in hours, the gradient is in pounds per hour — a rate of pay.
Always state the units when you interpret a gradient.
- Choose two points on the line that sit on clear grid intersections — the further apart the better.
- Draw a right-angled triangle beneath the line.
- Read the rise (vertical change) and the run (horizontal change) using the axis scales, not the number of squares.
- Divide: gradient $=$ rise $\div$ run.
- Attach the units: (unit of $y$) per (unit of $x$).
A distance–time graph passes through $(1,\ 40)$ and $(4,\ 160)$, where time is in hours and distance in km. Find the speed.
Speed $= 40$ km/h
| Feature of the graph | What it means |
|---|---|
| Steep upward line | Moving away quickly |
| Shallow upward line | Moving away slowly |
| Horizontal line | Stationary (gradient $= 0$) |
| Downward line | Returning towards the start |
| Curve getting steeper | Speeding up (accelerating) |
| Curve getting shallower | Slowing down |
Using the graph above, describe the journey and find the speed on each moving stage.
Total distance travelled $= 60$ km, over $6$ hours, so the average speed for the whole trip is $10$ km/h.
On a speed–time graph the gradient means something different again.
A car accelerates uniformly from $0$ to $24$ m/s in $8$ seconds, then travels at $24$ m/s for $10$ seconds.
(a) Find the acceleration. (b) Find the total distance travelled.
| Graph | Gradient means | Units |
|---|---|---|
| Distance against time | Speed | m/s, km/h |
| Speed against time | Acceleration | m/s² |
| Volume against time | Flow rate | litres/min |
| Cost against quantity | Price per item / per unit | £ per kg |
| Pay against hours | Hourly rate | £/hour |
| Temperature against time | Rate of heating or cooling | °C per minute |
| Money in a currency conversion | Exchange rate | € per £ |
A phone bill graph is a straight line passing through $(0,\ 12)$ and $(200,\ 32)$, where $x$ is the number of minutes used and $y$ is the bill in pounds.
(a) Interpret the intercept. (b) Find and interpret the gradient. (c) Write the equation.
Because of the £$12$ intercept, the bill is not proportional to the minutes used.
Inverse proportion ($y = \dfrac{k}{x}$): a smooth curve falling from top-left to bottom-right, approaching but never touching either axis. The rate of change is not constant — it becomes gentler as $x$ increases.
Sketch the graph of "the time taken for a journey of $120$ km against the average speed".
Gradient
$\dfrac{\text{change in }y}{\text{change in }x}$ — always give the units.
Distance–time
Gradient $=$ speed. Horizontal $=$ stopped.
Speed–time
Gradient $=$ acceleration. Area $=$ distance.
Read the scale
Use the axis values, not the number of squares.
Negative gradient
The quantity is decreasing — returning, cooling, emptying.
Intercept
A non-zero $y$-intercept means a fixed charge or starting value.
Direct proportion
Straight line through the origin; constant rate.
Inverse proportion
Hyperbola; rate of change gets gentler.
A distance–time graph passes through $(2,\ 90)$ and $(5,\ 300)$, with time in hours and distance in km. Find the speed.
▶ Show solution
Rise $= 300 - 90 = 210$ km
Run $= 5 - 2 = 3$ hours
Gradient $= 210 \div 3 = 70$
Speed $= 70$ km/h
On a speed–time graph, a train's speed rises steadily from $5$ m/s to $35$ m/s over $12$ seconds. Find the acceleration.
▶ Show solution
Gradient $= \dfrac{35 - 5}{12} = \dfrac{30}{12} = 2.5$
Acceleration $= 2.5\text{ m/s}^2$
A tank empties steadily. A volume–time graph passes through $(0,\ 900)$ and $(30,\ 300)$, with volume in litres and time in minutes. Find and interpret the gradient, and say when the tank will be empty.
▶ Show solution
Gradient $= \dfrac{300 - 900}{30 - 0} = \dfrac{-600}{30} = -20$
The tank is emptying at $20$ litres per minute (the minus sign shows a decrease).
Starting from $900$ litres: $900 \div 20 = 45$ minutes.
The tank is empty after $45$ minutes.
A taxi's charge graph is a straight line through $(0,\ 3.50)$ and $(8,\ 17.50)$, where $x$ is miles and $y$ is cost in pounds.
(a) Interpret the $y$-intercept. (b) Find the cost per mile. (c) Write the equation. (d) Find the cost of a $15$-mile trip.
▶ Show solution
(a) A fixed hire charge of £$3.50$ before any distance is travelled.
(b) Gradient $= \dfrac{17.50 - 3.50}{8} = \dfrac{14}{8} = 1.75$, so £$1.75$ per mile.
(c) $y = 1.75x + 3.50$
(d) $y = 1.75 \times 15 + 3.50 = 26.25 + 3.50 = £29.75$
A car accelerates uniformly from rest to $18$ m/s in $6$ seconds, holds that speed for $14$ seconds, then brakes uniformly to a stop in $4$ seconds. Find the total distance travelled.
▶ Show solution
Stage 1 (triangle): $\tfrac{1}{2} \times 6 \times 18 = 54$ m
Stage 2 (rectangle): $14 \times 18 = 252$ m
Stage 3 (triangle): $\tfrac{1}{2} \times 4 \times 18 = 36$ m
Total $= 54 + 252 + 36 = \mathbf{342}$ m
Two straight-line graphs of cost against weight are drawn for two shops. Shop A's line passes through the origin and $(4,\ 10)$. Shop B's line passes through $(0,\ 2)$ and $(4,\ 9)$. Weights are in kg and costs in pounds.
(a) Which shop's pricing is directly proportional to weight? (b) Which shop is cheaper for $10$ kg?
▶ Show solution
(a) Shop A — its line passes through the origin, so cost $\propto$ weight. Shop B has a £$2$ fixed charge.
(b) Shop A: gradient $= 10 \div 4 = 2.5$, so $y = 2.5x$; at $10$ kg, cost $= £25$.
Shop B: gradient $= \dfrac{9-2}{4} = 1.75$, so $y = 1.75x + 2$; at $10$ kg, cost $= 17.50 + 2 = £19.50$.
Shop B is cheaper for $10$ kg.
A conversion graph for pounds and euros is a straight line through the origin and $(50,\ 58)$. Find the exchange rate and use it to convert €$203$ into pounds.
▶ Show solution
Gradient $= \dfrac{58}{50} = 1.16$, so the rate is €$1.16$ per £$1$.
$203 \div 1.16 = £175$
Sketch the shape of a graph showing the number of days a fixed amount of food lasts against the number of animals eating it. What type of proportion is it?
▶ Show solution
If the total food is $F$, then days $= \dfrac{F}{\text{animals}}$, so days $\times$ animals is constant.
This is inverse proportion.
The graph is a hyperbola: steeply falling at first (adding a second animal halves the time) and then levelling off, never reaching zero days and never touching either axis.
A cyclist's distance–time graph shows: $0$ to $30$ minutes, $12$ km; $30$ to $45$ minutes, no change; $45$ to $90$ minutes, a further $18$ km.
(a) Find the speed on each moving stage in km/h. (b) Find the average speed for the whole $90$ minutes.
▶ Show solution
(a) Stage 1: $30$ min $= 0.5$ h, so speed $= 12 \div 0.5 = 24$ km/h.
Stage 3: $45$ min $= 0.75$ h, so speed $= 18 \div 0.75 = 24$ km/h.
(b) Total distance $= 12 + 18 = 30$ km; total time $= 90$ min $= 1.5$ h.
Average speed $= 30 \div 1.5 = 20$ km/h.
(The average is lower than $24$ km/h because of the $15$-minute stop.)
Two water tanks are filled at constant rates. Tank P's volume–time graph passes through $(0,\ 0)$ and $(8,\ 240)$. Tank Q's passes through $(0,\ 60)$ and $(8,\ 220)$, with time in minutes and volume in litres.
(a) Find both filling rates. (b) After how many minutes do the two tanks hold the same volume? (c) What volume is that?
▶ Show solution
(a) P: $\dfrac{240}{8} = 30$ litres/min. Q: $\dfrac{220 - 60}{8} = \dfrac{160}{8} = 20$ litres/min.
(b) $P: V = 30t$; $Q: V = 20t + 60$.
Set equal: $30t = 20t + 60 \Rightarrow 10t = 60 \Rightarrow t = 6$ minutes.
(c) $V = 30 \times 6 = 180$ litres.
Check with Q: $20 \times 6 + 60 = 180$ ✓