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This page sets out the method for real-life & distance–time graphs in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Real-life & distance–time graphs topic page.
When you are asked for the speed of one part of a journey, use only that straight section.
When a graph shows cost against amount used, find what the gradient and intercept mean.
Distance is usually up the side. Time is along the bottom.
A point on the graph is a distance at a time.
Speed = distance ÷ time, which is the gradient of that section.
A steeper section is a higher speed.
A horizontal line has gradient 0.
The object stays at the same distance while time passes.
A line sloping down means the distance from the start is decreasing.
The speed is still positive. The direction has changed.
If distance is in metres and time in seconds, the gradient is in m/s.
Convert units before you divide if the question asks for km/h.
Describe each section: set off, stop, return.
Use the numbers on the axes, not a guess from the picture alone.
Before any calculation, write down what one square is worth on each axis. A time axis labelled every hour, with five squares between labels, is 12 minutes per square. Counting squares as if they were hours makes every later answer five times too big. Draw the reading lines on the graph so the marker can see which square you used.
Read to the accuracy the scale allows. If each square is 2 km, a point halfway up a square is 1 km, not a guess of 0.4. If the value you need is past the end of a conversion graph, read a value that is on the graph and scale it. A graph that converts 10 miles might be used for 25 miles by multiplying the reading for 5 miles by 5. Show that scaling as its own line.
Method: 1. Write down the value of one square on each axis before you calculate. 2. If the time labels are one hour apart and five squares lie between them, one square is 12 minutes. 3. Counting those squares as whole hours makes every later answer five times too big.
On a distance-time graph the vertical axis is how far the object is from the start, not how fast it is going. A straight section has a steady speed, equal to distance divided by time for that section only. A horizontal section has zero speed: the distance is not changing, so the object is stopped. It does not mean "steady speed". A steeper section is faster. You can often name the fastest part by looking, before you divide anything.
Time on the axis may be in minutes while the speed is asked for in kilometres per hour. Change the minutes first. 30 minutes is 0.5 hours, because 30÷ 60=0.5. 20 minutes is 1/3 hour. 45 minutes is 0.75 hours. Dividing 18 km by 20 and calling the result 0.9 km/h has divided by minutes. Dividing 18 by 1/3 gives 54 km/h, which is the speed the story deserves.
Method: 1. Read the vertical axis as distance from the start, not as speed. 2. On a straight section, divide that section’s distance by that section’s time. 3. A horizontal section means the distance is not changing, so the object is stopped.
A straight tariff such as a hire charge often has the form C=mx+f. The gradient m is the cost per unit, and the intercept f is the standing charge you pay even if you buy nothing. Read two points on the line, divide the change in cost by the change in quantity, and read the intercept where the quantity is zero. Say what each number means in the story: "£2.50 per mile, plus a £3 call-out".
Average speed for a whole journey uses total distance and total time, including any stops. A stop adds time and adds no distance, so it lowers the average. Do not average the speeds of the moving sections. Two sections at 54 km/h and 48 km/h are not averaged to 51 km/h unless they lasted the same time, and even then a stop would change it. Total distance over total time is the definition that survives.
Method: 1. Write a straight hire charge in the form C=mx+f. 2. Here m is the cost per unit and f is the charge you pay when the quantity is zero. 3. Find m from the change between two points, and read f on the vertical axis.
A distance-time graph tells a story in segments. A steep straight line means a fast, steady speed, a flat line means the object is stationary and a line sloping back down towards zero means the return journey. In an exam you are often given the points where the line bends, such as (0, 0), (1, 15), (1.5, 15) and (3, 45) with time in hours and distance in kilometres. Work section by section.
Speed in each section is the change in distance divided by the change in time. The flat section here lasts 1.5-1=0.5 hours, which is 30 minutes. Be careful with units: 80 minutes is 11/3 hours, not 1.20 hours. Average speed for a whole journey is total distance divided by total time, and the time includes any stops. Two travellers meet where their lines cross on the graph.
Method: 1. List the points where the line changes direction or steepness. 2. For each section work out change in distance divided by change in time. 3. Convert minutes to hours by dividing by 60 before you divide distance by time. 4. For average speed use the total distance and the total time, including any rest.
Tip: A longer flat section does not change the speed of the moving parts, but it does lower the average speed for the whole trip.
A conversion graph links two units, such as miles and kilometres or pounds and dollars. If the line is straight and passes through the origin, the two quantities are in proportion. A line through (0, 0) and (5, 8) tells you that 5 miles is 8 kilometres. To convert 25 miles, notice that 25 is 5 times 5, so multiply 8 by 5 to get 40 kilometres.
Use scaling when the value you need is off the grid. To change 72 kilometres to miles, divide by 8 to find how many lots of 8 kilometres there are, which is 9, then multiply by 5 to get 45 miles. Some conversion graphs do not pass through the origin. The Celsius to Fahrenheit graph starts at (0, 32), so doubling the Celsius value does not double the Fahrenheit value. Check the starting point before you scale.
Method: 1. Read two exact points from the line, such as (5, 8). 2. Decide whether the line goes through the origin, because only then can you scale directly. 3. Multiply or divide both numbers of a known pair by the same amount. 4. Say which unit your answer is in.
On a cost graph the gradient is the cost for each extra unit, and the point where the line meets the vertical axis is the fixed charge. A straight line through (0, 3) and (10, 23) has gradient (23-3)/(10-0)=2, so the cost is £2 per mile with a fixed charge of £3. The equation is C=3+2m. A journey of 14 miles costs 3+2×14=31 pounds.
To compare two plans, write a formula for each and decide when the costs are equal. If company A charges £25 plus £8 per day and company B charges £14 per day, then 25+8d=14d gives d=25/6, about 4.17 days. Days are whole numbers, so test 4 and 5 days: B is cheaper for 4 days and A is cheaper from 5 days. Always finish with a sentence that answers the question.
Method: 1. Find the gradient from two points on the line, using change in cost divided by change in quantity. 2. Write the formula as fixed charge plus rate times quantity. 3. For two plans, set the formulae equal to find where the costs match. 4. Test whole numbers either side of the crossing point and state which is cheaper.
On a speed-time graph the gradient is acceleration and the area under the graph is the distance travelled. A car that speeds up from 0 to 20 m/s in 10 seconds has acceleration 20/10=2 m/s^2. The distance in that section is the area of a triangle, 1/2×10×20=100 m. A flat section at 20 m/s for 30 seconds is a rectangle with area 20×30=600 m.
Split awkward shapes into triangles, rectangles and trapezia and add the areas. Tank questions use the same idea. If the depth is 18 cm after 4 minutes and 36 cm after 10 minutes, the rate is (36-18)/(10-4)=3 cm per minute, and the depth at time zero is 18-3×4=6 cm. Check that any rate you find is a sensible size for the context.
Method: 1. Gradient of a speed-time graph gives acceleration, so divide change in speed by change in time. 2. Find the area under the graph by splitting it into simple shapes. 3. For filling or emptying, find the rate from two readings, then the starting amount from the formula. 4. State the unit, such as m, m/s^2 or cm per minute.
Tip: The unit of the area gives it away: seconds times metres per second gives metres.
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