These eight original questions follow the four lessons in the chapter. Sketch each graph, write the vertical axis, then open the answer.
Questions
Question 1. A distance-time graph passes through (0, 0) and (8, 48), with distance in metres and time in seconds. Find the speed.
Answer
Speed is the gradient: 48 ÷ 8 = 6 m/s.
Question 2. A distance-time graph has points (0, 0), (10, 40), (16, 40) and (24, 0). Describe each stage and find the speed in the last stage.
Answer
Stage 1 moves away at 40 ÷ 10 = 4 m/s. Stage 2 is flat, so the object is stopped for 6 seconds.
Stage 3 returns to the start. The distance falls by 40 m in 8 s, so the speed is 5 m/s back toward the start.
Question 3. A speed-time graph passes through (0, 3) and (5, 18), with speed in m/s. Find the acceleration.
Answer
Acceleration = (18 − 3) ÷ 5 = 15 ÷ 5 = 3 m/s².
Question 4. A speed-time graph has points (0, 0), (5, 20), (15, 20) and (20, 0). Find the deceleration in the last stage.
Answer
Change in speed = 0 − 20 = −20 over 5 s. The gradient is −4 m/s², so the deceleration is 4 m/s².
Question 5. Using the graph in Question 4, find the total distance.
Answer
First triangle: ½ × 5 × 20 = 50. Rectangle: 10 × 20 = 200. Last triangle: ½ × 5 × 20 = 50.
Total distance = 50 + 200 + 50 = 300 m.
Check with the trapezium: ½ × (10 + 20) × 20 = 300.
Question 6. A speed-time graph has points (0, 6), (4, 6) and (10, 24). Find the total distance travelled in the 10 seconds.
Answer
From 0 to 4 s: rectangle, 4 × 6 = 24 m.
From 4 to 10 s: trapezium with parallel sides 6 and 24 and width 6. Area = ½ × (6 + 24) × 6 = 90 m.
Total = 24 + 90 = 114 m.
Question 7. A student says, “The graph from (0, 0) to (10, 50) has gradient 5, so the acceleration is 5 m/s².” The graph is a distance-time graph. What is wrong?
Answer
On a distance-time graph the gradient is speed, not acceleration.
The correct statement is that the speed is 5 m/s, and it is constant because the line is straight.
Question 8. A car’s speed-time graph has points (0, 0), (8, 16), (20, 16) and (28, 0). Find the acceleration in the first stage, the distance in the first 8 seconds, and the total distance.
Answer
Acceleration = 16 ÷ 8 = 2 m/s² (gradient).
Distance in the first 8 s: ½ × 8 × 16 = 64 m (area).
Rectangle from 8 to 20 s: 12 × 16 = 192. Last triangle: ½ × 8 × 16 = 64.
Total distance = 64 + 192 + 64 = 320 m.
If you got these wrong
Slips in questions 1 and 2 point to reading distance-time graphs. Questions 3 and 4 need reading speed-time graphs.
Questions 5, 6 and 8 use calculating distance from a speed-time graph. Question 7 and the choice in question 8 come from distinguishing gradient from area in motion graphs.
Change the numbers of a journey in the motion-graph explorer and log your slips in the mistake log and paper-error review. Then build a timed set with the timed original practice session builder.
If you want a teacher to go through your graph reasoning, see online one-to-one Mathematics tuition.