You’ll need graph paper, a ruler, a sharp pencil and a calculator. Plot each graph from the data, then use it to calculate speeds, accelerations and distances. Work each answer out first, then tap to check. Everything here is for Combined and Separate; questions marked HT are Higher tier only. ← Forces topic
📝 Copy into your book
DISTANCE–TIME graphs:
• Gradient = SPEED
• Horizontal line = stationary (not moving)
• Straight sloping line = constant speed (steeper = faster)
• Curving upwards = accelerating (HT: find the speed at one moment from the gradient of a tangent)
VELOCITY–TIME graphs:
• Gradient = ACCELERATION
• Horizontal line = constant velocity
• Sloping up = accelerating; sloping down = decelerating (negative acceleration)
• AREA under the graph = DISTANCE travelled
gradient = change in y ÷ change in x
acceleration = change in velocity ÷ time a = Δv ÷ t (Δv = final − initial)
📝 How to plot a graph that gets full marks
- Time always goes on the x-axis (across).
- Label both axes with the quantity and unit, e.g. “Distance (m)” and “Time (s)”.
- Choose an even scale (e.g. 1 big square = 5 s) so your graph fills more than half the paper.
- Plot each point with a small neat cross (×), accurate to half a small square.
- Join the points with a ruler for these journeys (each stage is a straight line).
- To find a gradient, draw a large triangle on the line – big triangles give more accurate answers.
Activity 1: What is it doing? (5 minutes)
For each graph description, say what the object is doing.
- A distance–time graph is a horizontal line.
- A velocity–time graph is a horizontal line.
- A distance–time graph is a steep straight line going up.
- A velocity–time graph is a straight line sloping down to zero.
- A distance–time graph curves upwards, getting steeper.
- A velocity–time graph is a straight line going up from zero.
- A distance–time graph curves but gets less steep until it is horizontal.
- A velocity–time graph goes up steeply, then less steeply, then becomes horizontal.
Show answers
1 Stationary (not moving) · 2 Moving at a constant velocity · 3 Moving at a high constant speed · 4 Decelerating (slowing down) until it stops · 5 Accelerating (speeding up) · 6 Accelerating steadily (uniform acceleration) from rest · 7 Slowing down until it stops · 8 Accelerating, but the acceleration is decreasing until it reaches a steady speed – like a skydiver reaching terminal velocity
Watch out for the classic trap: a horizontal line means STOPPED on a distance–time graph, but CONSTANT VELOCITY on a velocity–time graph! Always check the y-axis label first.
Activity 2: The walk to school – distance–time graph (12 minutes)
Sam walks to school. On the way, she stops to wait for a friend, then they both hurry because they’re late.
| Time (s) | Distance (m) |
|---|---|
| 0 | 0 |
| 10 | 15 |
| 20 | 30 |
| 30 | 30 |
| 40 | 30 |
| 50 | 60 |
| 60 | 90 |
- Plot the distance–time graph. Join the points with straight lines.
- Calculate Sam’s speed for the first 20 s.
- What is Sam doing between 20 s and 40 s? How can you tell from the graph?
- Calculate the speed between 40 s and 60 s.
- Which part of the graph is steepest? What does that tell you?
- Calculate Sam’s average speed for the whole 60 s.
- Challenge: Sam’s friend Jo walked the whole 90 m at a steady speed, arriving at the same time. Draw Jo’s line on the same axes. At what time does Jo overtake Sam?
Show answers
(b) gradient = 30 ÷ 20 = 1.5 m/s
(c) She is stationary (waiting for her friend) – the line is horizontal, so the distance isn’t changing.
(d) gradient = (90 − 30) ÷ (60 − 40) = 60 ÷ 20 = 3.0 m/s
(e) 40–60 s is steepest, so that’s when she is moving fastest.
(f) average speed = total distance ÷ total time = 90 ÷ 60 = 1.5 m/s
(g) Jo’s line is a straight line from (0, 0) to (60, 90) – speed 1.5 m/s. It lies exactly on Sam’s line for the first 20 s. Sam then stops, so Jo is ahead from 20 s. The lines meet again at 60 s. So Jo overtakes Sam at 20 s, and Sam only catches up right at the end.
Activity 3: The car journey – velocity–time graph (15 minutes)
A car pulls away from traffic lights, drives along a straight road, then stops at the next set of lights.
| Time (s) | Velocity (m/s) |
|---|---|
| 0 | 0 |
| 5 | 10 |
| 10 | 20 |
| 15 | 20 |
| 20 | 20 |
| 25 | 10 |
| 30 | 0 |
- Plot the velocity–time graph. Join the points with straight lines.
- Calculate the acceleration for the first 10 s.
- Describe the motion between 10 s and 20 s.
- Calculate the acceleration between 20 s and 30 s. What does the sign tell you?
- Use the area under the graph to find the distance travelled in each of the three stages.
- Calculate the total distance and the average speed for the whole journey.
- Check it another way: use v² − u² = 2as to calculate the braking distance for the last 10 s. Does it match your answer from the area?
Hint: split the area into shapes. Area of a triangle = ½ × base × height; area of a rectangle = base × height.
Show answers
(b) a = Δv ÷ t = (20 − 0) ÷ 10 = 2.0 m/s²
(c) Constant velocity of 20 m/s (acceleration = 0).
(d) a = (0 − 20) ÷ 10 = −2.0 m/s². The minus sign means it is decelerating (slowing down) – a deceleration of 2.0 m/s².
(e) 0–10 s: triangle = ½ × 10 × 20 = 100 m; 10–20 s: rectangle = 10 × 20 = 200 m; 20–30 s: triangle = ½ × 10 × 20 = 100 m
(f) Total = 100 + 200 + 100 = 400 m; average speed = 400 ÷ 30 = 13.3 m/s
(g) v² − u² = 2as → 0² − 20² = 2 × (−2.0) × s → −400 = −4.0s → s = 100 m ✓ – it matches the area.
Activity 4: Write the story (8 minutes)
Plot this velocity–time graph for a cyclist, then write a short story describing the journey. Include the acceleration for each stage and the total distance.
| Time (s) | Velocity (m/s) |
|---|---|
| 0 | 2 |
| 4 | 6 |
| 8 | 6 |
| 12 | 6 |
| 16 | 8 |
| 20 | 0 |
Hint: the first and fourth stages don’t start at zero, so their areas are trapeziums: ½ × (start velocity + end velocity) × time.
Show a model answer
The cyclist is already moving at 2 m/s and speeds up to 6 m/s in 4 s (a = 4 ÷ 4 = 1.0 m/s²). She rides at a steady 6 m/s for 8 s. She speeds up again to 8 m/s in 4 s (a = 2 ÷ 4 = 0.5 m/s²), then brakes hard and stops in 4 s (a = −8 ÷ 4 = −2.0 m/s², a deceleration of 2.0 m/s²).
Distance: 0–4 s: ½ × (2 + 6) × 4 = 16 m · 4–12 s: 6 × 8 = 48 m · 12–16 s: ½ × (6 + 8) × 4 = 28 m · 16–20 s: ½ × 8 × 4 = 16 m · Total = 108 m
Activity 5: Link the graphs – challenge (8 minutes)
Sketch the distance–time graph for the car journey in Activity 3. Use your distances from part (e) to mark the distance at 10 s, 20 s and 30 s. Think carefully about the shape of each section.
Show answer
0–10 s: a curve getting steeper (accelerating), reaching 100 m at 10 s.
10–20 s: a straight line with gradient 20 m/s (constant speed), reaching 300 m at 20 s.
20–30 s: a curve getting less steep (decelerating), levelling off to horizontal at 400 m at 30 s, when the car stops.
Key idea: the gradient of the distance–time graph at any moment equals the value on the velocity–time graph at that moment.
Activity 6: Exam practice (18 minutes)
Q1 (2 marks) A distance–time graph is a straight line from the origin to (8 s, 40 m). Calculate the speed.
Show answer
gradient = 40 ÷ 8 (1) = 5.0 m/s (1)
Q2 (4 marks) A velocity–time graph is a straight line from 4 m/s at 0 s to 16 m/s at 6 s. Calculate (a) the acceleration and (b) the distance travelled.
Show answer
(a) a = (16 − 4) ÷ 6 (1) = 2.0 m/s² (1)
(b) area = ½ × (4 + 16) × 6 (1) = 60 m (1) (or rectangle 4 × 6 = 24 m + triangle ½ × 6 × 12 = 36 m)
Q3 (4 marks) A bus travelling at 15 m/s brakes steadily and stops in 5.0 s. Sketch the velocity–time graph, then calculate (a) the acceleration and (b) the braking distance.
Show answer
Sketch: a straight line from 15 m/s down to 0 at 5.0 s.
(a) a = (0 − 15) ÷ 5.0 (1) = −3.0 m/s² (1) (a deceleration of 3.0 m/s²)
(b) area = ½ × 5.0 × 15 (1) = 37.5 m (1)
Q4 (3 marks) A sprinter accelerates uniformly from rest to 10 m/s in 2.0 s, then runs at 10 m/s for 9.0 s. Sketch the velocity–time graph and show that the race is 100 m long.
Show answer
Triangle: ½ × 2.0 × 10 = 10 m (1)
Rectangle: 9.0 × 10 = 90 m (1)
Total = 10 + 90 = 100 m (1)
Q5 (3 marks) A distance–time graph for a train is a straight line showing it travels 3.6 km in 2.0 minutes. Calculate its speed in m/s.
Show answer
Convert: 3.6 km = 3600 m and 2.0 minutes = 120 s (1)
speed = gradient = 3600 ÷ 120 (1)
= 30 m/s (1)
Q6 (3 marks – HT) A distance–time graph for an accelerating car is a curve. A tangent drawn to the curve at 4 s passes through the points (2 s, 0 m) and (6 s, 24 m). Calculate the speed of the car at 4 s.
Show answer
The speed at one moment = gradient of the tangent (1)
= (24 − 0) ÷ (6 − 2) (1)
= 6.0 m/s (1)
Q7 (2 marks) On a velocity–time graph, line A is steeper than line B. Both start from rest. Which object has the greater acceleration? Explain.
Show answer
A (1). The gradient of a velocity–time graph is the acceleration, and A has the steeper gradient (1).
✅ Add up your exam practice marks out of 21 (or out of 18 if you skipped the HT question).
⚠️ Where students lose marks on motion graphs
- Mixing up the two graphs. Read the y-axis label before you do anything else.
- Using distance = speed × time when the speed is changing. On a velocity–time graph, always use the area.
- Forgetting the ½ for triangles when finding the area.
- Doing initial − final instead of final − initial. Slowing down must give a negative acceleration.
- Wrong units: acceleration is in m/s², not m/s. Convert km → m and minutes → s first.
- Tiny gradient triangles give inaccurate answers – draw them as large as possible.
Want more? Try the Motion graphs questions, the Forces fill the gaps page or the Skydive Challenge lesson.
