Gradients & Areas Under Graphs

1 What Gradient Tells Us

The gradient of a graph tells you how quickly something is changing. It is the rate of change of the quantity on the y-axis with respect to the quantity on the x-axis.

Gradient Formula (straight line)

$$\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{\text{change in } y}{\text{change in } x} = \frac{y_2 - y_1}{x_2 - x_1}$$

The units of the gradient are always the y-unit divided by the x-unit. This is what gives gradient its real-world meaning:

Graph type	x-axis	y-axis	Gradient = ?
Distance–time	Time (s, min, h)	Distance (m, km)	Speed (m/s, km/h)
Velocity–time	Time (s)	Velocity (m/s)	Acceleration (m/s²)
Cost–units	Units used	Cost (£)	Cost per unit (£/unit)
Profit–items	Items sold	Profit (£)	Profit per item (£/item)

🔑 Key signs: A positive gradient means y is increasing as x increases. A negative gradient means y is decreasing. A zero gradient (horizontal line) means y is not changing at all.

How to calculate gradient:
1. Pick two clear points on the line (use gridline intersections where possible).
2. Draw a right-angled triangle — the horizontal leg is the "run", the vertical leg is the "rise".
3. Gradient = rise ÷ run. Include the sign (negative if the line slopes downward).

Line A: gradient = +2 (steep positive). Line B: gradient = 0 (horizontal, no change). Line C: gradient = −1 (negative slope).

2 Distance–Time Graphs

Key Rule

$$\text{Gradient of a distance–time graph} = \text{Speed}$$

Gradient = $\dfrac{\text{distance}}{\text{time}}$ — the same as the speed formula $v = \dfrac{d}{t}$.

What the graph looks like	What it means
Steep positive slope	Moving fast away from the start
Gentle positive slope	Moving slowly away from the start
Horizontal line (gradient = 0)	Stationary — not moving
Negative slope	Moving back towards the starting point
Curved line	Speed is changing (accelerating or decelerating)

A distance–time graph for a car journey: 60km in 3h (blue), rest for 1h (green), 30km more in 2h (orange).

Worked Example 1 — Reading a Distance–Time Graph

Use the graph above. Find the speed in each moving section.

①Section 1 (blue, 0→3h): gradient $= \dfrac{60 - 0}{3 - 0} = \dfrac{60}{3} = \mathbf{20}$ km/h

②Section 2 (green, 3→4h): horizontal line → gradient $= 0$ → stationary

③Section 3 (orange, 4→6h): gradient $= \dfrac{90 - 60}{6 - 4} = \dfrac{30}{2} = \mathbf{15}$ km/h

✓Section 1 is steeper than section 3, confirming the car was faster in section 1 (20 km/h vs 15 km/h).

💡 Tip: To compare speeds on a distance-time graph without calculating: the steeper the line, the faster the object. A negative gradient means the object is returning to its starting point.

Worked Example 2 — Finding Speed When Returning

A cyclist travels 24km from home in 2 hours, stops for 30 minutes, then returns home in 1.5 hours. Find the speed on the return journey.

①On the return journey, distance goes from 24km back to 0km. Change in distance = $0 - 24 = -24$ km.

②Time for return = 1.5 hours. Gradient $= \dfrac{-24}{1.5} = -16$ km/h.

③The speed is the magnitude: $\mathbf{16}$ km/h. The negative sign tells us the cyclist is heading back towards home.

3 Velocity–Time Graphs

Gradient

$$\text{Gradient} = \text{Acceleration}$$

$a = \dfrac{\Delta v}{\Delta t}$ in m/s²

Area Under Graph

$$\text{Area} = \text{Distance travelled}$$

Count squares or use shape formulas

What the v–t graph looks like	What it means
Positive slope (rising)	Accelerating (speeding up)
Horizontal line (gradient=0)	Constant velocity (not accelerating)
Negative slope (falling)	Decelerating (slowing down)
Line touching x-axis (v=0)	Object momentarily at rest

Velocity–time graph. Shaded area = total distance = 24 + 48 + 12 = 84 m. Orange triangle shows how to find acceleration.

Worked Example 3 — Velocity–Time: Acceleration & Distance

Using the graph above: (a) Find the acceleration in the first 4 seconds. (b) Find the deceleration in the last section. (c) Find the total distance travelled.

①(a) Acceleration (0→4s): gradient $= \dfrac{12 - 0}{4 - 0} = \dfrac{12}{4} = \mathbf{3 \text{ m/s}^2}$

②(b) Deceleration (8→10s): gradient $= \dfrac{0 - 12}{10 - 8} = \dfrac{-12}{2} = -6 \text{ m/s}^2$ → deceleration of $\mathbf{6 \text{ m/s}^2}$

③(c) Total distance = total area under the graph:
Triangle (0–4s): $\frac{1}{2} \times 4 \times 12 = 24$ m
Rectangle (4–8s): $4 \times 12 = 48$ m
Triangle (8–10s): $\frac{1}{2} \times 2 \times 12 = 12$ m

✓Total distance $= 24 + 48 + 12 = \mathbf{84 \text{ m}}$

🔑 Remember: On a velocity-time graph, the area of each shape gives a distance. Always identify whether each region is a triangle ($\frac{1}{2} \times \text{base} \times \text{height}$), a rectangle ($\text{length} \times \text{height}$), or a trapezium ($\frac{1}{2}(a+b) \times h$).

4 Gradient of a Curve — Drawing a Tangent

A curved distance-time or velocity-time graph means the rate of change is not constant (e.g., an object is accelerating). The gradient changes at every point. To find the gradient at one specific point, we draw a tangent to the curve at that point and then find the gradient of that straight tangent line.

Method

Draw a straight line that just touches the curve at the point in question — this is the tangent. Then calculate gradient = rise ÷ run on that straight line.

The curve $d = t^2$ with a tangent drawn at $t = 3$ s. The gradient of the tangent ≈ 6 m/s = speed at that instant.

Mark the point on the curve where you want the gradient.
Use a ruler to draw a straight line that touches the curve only at that point (tangent). The line should not cut through the curve.
Choose two points that are far apart on the tangent line (not necessarily on the curve) and read off their coordinates from the graph.
Calculate gradient = $\dfrac{y_2 - y_1}{x_2 - x_1}$ using those two points. Include the units!

⚠️ Common mistake: Students sometimes calculate the gradient between the point of tangency and another point on the curve. Always use two points on the tangent line, not on the curve.

Worked Example 4 — Estimating Speed from a Curved Distance-Time Graph

A tangent is drawn to a distance-time curve at $t = 3$ s. The tangent passes through the points $(1, 0)$ and $(5, 24)$. Find the speed at $t = 3$ s.

①Two points on the tangent: $(1, 0)$ and $(5, 24)$.

②Gradient $= \dfrac{24 - 0}{5 - 1} = \dfrac{24}{4} = \mathbf{6}$

✓Speed at $t = 3$ s $\approx \mathbf{6}$ m/s. (This is an estimate — drawing a tangent by hand introduces a small error.)

5 Area Under a Graph — The Trapezium Rule

For a velocity-time graph, the area between the line and the x-axis tells you the distance travelled. When the graph is a curve (not made up of straight lines), we can estimate the area using the trapezium rule.

Trapezium Rule (area estimation)

$$\text{Area} \approx \frac{h}{2}\bigl(y_0 + 2y_1 + 2y_2 + \cdots + 2y_{n-1} + y_n\bigr)$$

where $h$ = width of each strip, $y_0$ = first value, $y_n$ = last value, and all middle values are doubled.

How to Use the Trapezium Rule

Divide the x-axis into equal strips of width $h$.
Read off the y-values at each strip boundary ($y_0, y_1, y_2, \ldots$).
Use the formula: add the first and last values once, and all middle values twice.
Multiply by $\frac{h}{2}$. Remember to include units (y-unit × x-unit).

Worked Example 5 — Trapezium Rule for Distance

A car's velocity is recorded every 2 seconds:

$t$ (s)	0	2	4	6	8
$v$ (m/s)	0	6	10	12	8

Estimate the total distance using the trapezium rule.

①Strip width: $h = 2$ s. There are 4 strips.

②Identify values: $y_0 = 0$, $y_1 = 6$, $y_2 = 10$, $y_3 = 12$, $y_4 = 8$

③Apply the rule: $$\text{Area} \approx \frac{2}{2}(0 + 2(6) + 2(10) + 2(12) + 8) = 1 \times (0 + 12 + 20 + 24 + 8) = 64$$

✓Estimated distance $\approx \mathbf{64 \text{ m}}$. (This is an approximation — the more strips used, the more accurate.)

💡 Counting squares: Another method is to count the squares under the graph. Whole squares count as 1; partial squares are estimated. Multiply by the value each square represents (x-unit × y-unit). This is less precise but useful for a quick check.

6 Real-Life & Financial Graphs

Gradients and areas appear in many real-life contexts beyond physics. In financial graphs, the gradient tells you a rate of change of cost, profit, or value. The y-intercept often represents a fixed charge or starting value.

Electricity Bill: $C = 10 + 0.30n$

Reading this graph:
y-intercept = £10 — this is the standing charge (the fixed amount you pay even if you use no electricity).

Gradient = £0.30 per kWh — this is the unit price (cost per kWh of electricity used).

For any number of kWh used ($n$): $$C = 10 + 0.30n$$

💡 In financial graphs:
• gradient = rate of change of cost/profit
• y-intercept = starting value or fixed cost
• steeper line = higher rate

Worked Example 6 — Financial Graph: Gradient & Interpretation

A mobile phone plan charges a £15 monthly fee plus £0.08 per minute of calls. (a) Write the equation for the monthly cost $C$. (b) What is the gradient of the cost graph and what does it represent? (c) What is the y-intercept and what does it mean?

①(a) Let $m$ = minutes used. Cost $C = 15 + 0.08m$ (£).

②(b) Gradient = 0.08. It means the cost increases by 8p for every minute of calls.

③(c) y-intercept = £15. This is the fixed monthly charge — the amount you pay even if you make no calls.

Worked Example 7 — Comparing Two Tariffs

Plan A: £5/month + £0.15 per minute. Plan B: £20/month + £0.05 per minute. Which is cheaper for 100 minutes? For 200 minutes?

①Plan A at 100 min: $5 + 0.15 \times 100 = 5 + 15 = \mathbf{£20}$

②Plan B at 100 min: $20 + 0.05 \times 100 = 20 + 5 = \mathbf{£25}$

③Plan A at 200 min: $5 + 0.15 \times 200 = 5 + 30 = \mathbf{£35}$

④Plan B at 200 min: $20 + 0.05 \times 200 = 20 + 10 = \mathbf{£30}$

✓Plan A is cheaper for 100 minutes; Plan B is cheaper for 200 minutes. The plans are equal when $5 + 0.15m = 20 + 0.05m \Rightarrow 0.10m = 15 \Rightarrow m = 150$ minutes.

7 Quick Reference

📐 Gradient of any straight-line graph

$\text{gradient} = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{\text{rise}}{\text{run}}$
Units = y-unit ÷ x-unit

🚗 Distance–time graph

Gradient = speed
Positive → moving away
Zero → stationary
Negative → returning
Steeper = faster

🚀 Velocity–time graph

Gradient = acceleration
Area = distance
Positive slope → speeding up
Negative slope → slowing down

📈 Curve: drawing a tangent

Touch curve at one point only. Use two widely-spaced points on the tangent to calculate gradient = instantaneous rate of change.

🔲 Trapezium rule

$\text{Area} \approx \dfrac{h}{2}(y_0 + 2y_1 + \cdots + 2y_{n-1} + y_n)$
More strips → more accurate

💷 Financial graphs

Gradient = rate of change (cost per unit, profit per item)
y-intercept = fixed cost or starting value

Shape in v–t graph	Area formula	Example
Triangle	$\frac{1}{2} \times \text{base} \times \text{height}$	Accelerating from rest to $v$ in time $t$: area $= \frac{1}{2}tv$
Rectangle	$\text{width} \times \text{height}$	Constant speed $v$ for time $t$: area $= tv$
Trapezium	$\frac{1}{2}(a + b) \times h$	Speed goes from $u$ to $v$ in time $t$: area $= \frac{1}{2}(u+v)t$

8 Practice Questions

Question 1 — Distance–Time: Multiple Sections

A distance-time graph for a train journey has the following key points: $(0\text{ h},\ 0\text{ km})$, $(2.5\text{ h},\ 100\text{ km})$, $(3\text{ h},\ 100\text{ km})$, $(4.5\text{ h},\ 55\text{ km})$.

(a) Find the speed during the first section (0 to 2.5 h).

(b) Describe what is happening between 3 h and 4.5 h.

▶ Show Solution

(a) Speed $= \dfrac{\text{change in distance}}{\text{change in time}} = \dfrac{100 - 0}{2.5 - 0} = \dfrac{100}{2.5} = \mathbf{40}$ km/h

(b) Between 3 h and 4.5 h the distance decreases from 100 km to 55 km. The train is travelling back towards the start.

(c) Gradient $= \dfrac{55 - 100}{4.5 - 3} = \dfrac{-45}{1.5} = -30$. Speed $= \mathbf{30}$ km/h in the direction of the starting point (negative = returning).

Question 2 — Velocity–Time: Acceleration & Distance

A velocity-time graph has vertices at $(0\text{ s},\ 0\text{ m/s})$, $(6\text{ s},\ 30\text{ m/s})$, $(10\text{ s},\ 30\text{ m/s})$, and $(14\text{ s},\ 0\text{ m/s})$.

(a) Find the acceleration in the first 6 seconds.

(b) Find the deceleration in the last section (10 s to 14 s).

▶ Show Solution

(a) Acceleration $= \dfrac{30 - 0}{6 - 0} = \dfrac{30}{6} = \mathbf{5 \text{ m/s}^2}$

(b) Gradient $= \dfrac{0 - 30}{14 - 10} = \dfrac{-30}{4} = -7.5$ m/s² → deceleration of $\mathbf{7.5 \text{ m/s}^2}$

(c) Total area = Triangle + Rectangle + Triangle:

$\frac{1}{2} \times 6 \times 30 + 4 \times 30 + \frac{1}{2} \times 4 \times 30 = 90 + 120 + 60 = \mathbf{270 \text{ m}}$

Question 3 — Gradient of a Curve: Tangent Method

A distance-time graph is curved (the object is accelerating). A tangent is drawn to the curve at $t = 5$ s. The tangent passes through the coordinates $(2\text{ s},\ 3\text{ m})$ and $(8\text{ s},\ 39\text{ m})$.

(a) Calculate the gradient of the tangent line.

(b) What does this gradient represent?

▶ Show Solution

(a) Gradient $= \dfrac{39 - 3}{8 - 2} = \dfrac{36}{6} = \mathbf{6}$ m/s

(b) The gradient of the tangent to a distance-time graph gives the instantaneous speed at that moment — in this case, the speed of the object at $t = 5$ s is approximately 6 m/s.

(c) Drawing a tangent by hand is not perfectly accurate. The tangent is an approximation to the true slope of the curve at that point, so the calculated gradient is only an estimate.

Question 4 — Trapezium Rule

A car's velocity is measured every 3 seconds during braking:

$t$ (s)	0	3	6	9	12
$v$ (m/s)	24	18	12	6	0

(a) Use the trapezium rule with 4 strips to estimate the braking distance.

(b) Comment on whether your answer is an overestimate or underestimate if the velocity decreases at an ever-increasing rate.

▶ Show Solution

(a) $h = 3$ s, $y_0 = 24,\ y_1 = 18,\ y_2 = 12,\ y_3 = 6,\ y_4 = 0$

$$\text{Area} \approx \frac{3}{2}(24 + 2(18) + 2(12) + 2(6) + 0) = \frac{3}{2}(24 + 36 + 24 + 12 + 0) = \frac{3}{2} \times 96 = \mathbf{144 \text{ m}}$$

(b) If the velocity decreases at an ever-increasing rate (concave-down curve), each trapezium lies above the true curve, so the trapezium rule gives an overestimate.

Question 5 — Financial Graph: Interpreting Gradient

A car hire company charges customers. A graph of total cost ($C$, in £) against miles driven ($m$) is a straight line passing through $(0,\ 40)$ and $(200,\ 140)$.

(a) Find the gradient of the line and state its units.

(b) What does the gradient represent in context?

(d) Write the equation for $C$ in terms of $m$.

▶ Show Solution

(a) Gradient $= \dfrac{140 - 40}{200 - 0} = \dfrac{100}{200} = \mathbf{0.5}$ £/mile (or 50p per mile)

(b) The gradient means the hire cost increases by 50p for every mile driven. This is the cost per mile.

(c) The y-intercept is £40 — this is the base/fixed hire charge that applies regardless of how far you drive.

(d) $C = 40 + 0.5m$

Question 6 — Velocity–Time: Trapezoid with Non-Zero Start

A velocity-time graph is a straight line from $(0\text{ s},\ 5\text{ m/s})$ to $(8\text{ s},\ 21\text{ m/s})$.

(a) Find the acceleration.

(b) Calculate the distance travelled in the 8 seconds.

▶ Show Solution

(a) Acceleration $= \dfrac{21 - 5}{8 - 0} = \dfrac{16}{8} = \mathbf{2 \text{ m/s}^2}$

(b) The area under the graph is a trapezium with parallel sides 5 and 21, and height (width) = 8 s:

$$\text{Distance} = \frac{1}{2}(5 + 21) \times 8 = \frac{1}{2} \times 26 \times 8 = \mathbf{104 \text{ m}}$$

Question 7 — Describing a Real-Life Graph

A graph shows the depth of water ($d$ cm) in a bathtub against time ($t$ minutes). The graph has three sections: (i) a steep positive slope for 4 minutes; (ii) a horizontal line for 10 minutes; (iii) a steep negative slope for 3 minutes.

(a) Describe what is happening in each section in the context of the scenario.

(b) The depth rises from 0 to 30 cm in 4 minutes. Find the rate of filling.

▶ Show Solution

(a) (i) The bath is filling — depth increases rapidly.
(ii) The bath is full and the water isn't moving (someone is bathing, tap off, plug in).
(iii) The bath is emptying — depth decreases rapidly.

(b) Rate of filling = gradient $= \dfrac{30 - 0}{4 - 0} = \mathbf{7.5}$ cm/min

(c) Rate of emptying = $\dfrac{0 - 30}{3} = -10$ cm/min → 10 cm/min (magnitude of gradient)

Question 8 — Velocity–Time: Complex Journey

A motorbike's journey is described by the following velocity-time graph points: $(0,\ 0)$, $(5,\ 20)$, $(15,\ 20)$, $(20,\ 0)$, $(25,\ 15)$.

(a) Find the acceleration between 0 and 5 seconds.

(b) Find the distance covered in the first 20 seconds.

▶ Show Solution

(a) Acceleration $= \dfrac{20 - 0}{5 - 0} = \mathbf{4 \text{ m/s}^2}$

(b) Distance in 0–20 s = area under graph:
Triangle (0–5 s): $\frac{1}{2} \times 5 \times 20 = 50$ m
Rectangle (5–15 s): $10 \times 20 = 200$ m
Triangle (15–20 s): $\frac{1}{2} \times 5 \times 20 = 50$ m
Total $= 50 + 200 + 50 = \mathbf{300 \text{ m}}$

(c) The motorbike starts from rest (v=0 at 20 s) and accelerates again to 15 m/s in 5 s. Acceleration $= \dfrac{15 - 0}{5} = \mathbf{3 \text{ m/s}^2}$

Question 9 — Profit Graph & Break-Even

A small business sells handmade candles. Each candle sells for £8. The cost to produce $n$ candles is given by $C = 120 + 3n$ (£). The revenue from selling $n$ candles is $R = 8n$ (£).

(a) Find the gradient of the cost graph and the revenue graph. What do they represent?

(b) How many candles must be sold to break even (profit = 0)?

▶ Show Solution

(a) Gradient of cost graph = £3/candle (cost to make each candle). Gradient of revenue graph = £8/candle (selling price per candle). Revenue graph is steeper → eventually profit is made.

(b) Break-even: Revenue = Cost
$8n = 120 + 3n \Rightarrow 5n = 120 \Rightarrow n = \mathbf{24}$ candles

(c) Profit $P = R - C = 8n - (120 + 3n) = 5n - 120$
At $n = 50$: $P = 5(50) - 120 = 250 - 120 = \mathbf{£130}$

Question 10 — Mixed: Distance–Time Curve & Tangent

A ball is rolled along a surface. Its distance from a sensor (in cm) at various times is recorded:

$t$ (s)	0	1	2	3	4	5
$d$ (cm)	0	8	28	54	80	100

(a) Without calculating, describe how the speed of the ball changes over time (look at the differences between consecutive d values).

(b) A tangent is drawn to the d-t curve at $t = 2$ s. It passes through $(0,\ -4)$ and $(4,\ 52)$. Find the speed at $t = 2$ s.

▶ Show Solution

(a) The differences are: 8, 20, 26, 26, 20. The ball speeds up between $t = 0$ and $t = 3$ s (differences increasing), then slows down from $t = 3$ to $t = 5$ s (differences decreasing). The speed increases then decreases.

(b) Two points on tangent: $(0, -4)$ and $(4, 52)$
Gradient $= \dfrac{52 - (-4)}{4 - 0} = \dfrac{56}{4} = \mathbf{14}$ cm/s

Speed at $t = 2$ s $\approx \mathbf{14}$ cm/s

(c) Average speed $= \dfrac{\text{total distance}}{\text{total time}} = \dfrac{100 - 0}{5 - 0} = \mathbf{20}$ cm/s

Note: average speed uses total distance ÷ total time, NOT the gradient of the tangent.