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ScienceAges 16-18

Speed-time graphs: area & distance

Calculate total distance travelled from the area under a two-stage speed-time graph (acceleration triangle plus constant-speed rectangle). Review suitability for your learners before teaching.

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Worksheet StudioPractice pack · Ages 16-18

Speed-time graphs: area & distance

Take your time. Show your thinking.

  1. 1. On a speed-time graph, a train accelerates uniformly from rest to 36 m/s in 12 s and then continues at a constant speed of 36 m/s for a further 18 s. Calculate the total distance travelled.

    v_max = 36 m/s (t₁ = 12 s, t₂ = 18 s) · Distance = Area = ½vt₁ + vt₂
  2. 2. On a speed-time graph, a train accelerates uniformly from rest to 34 m/s in 10 s and then continues at a constant speed of 34 m/s for a further 19 s. Calculate the total distance travelled.

    v_max = 34 m/s (t₁ = 10 s, t₂ = 19 s) · Distance = Area = ½vt₁ + vt₂
  3. 3. On a speed-time graph, a train accelerates uniformly from rest to 16 m/s in 3 s and then continues at a constant speed of 16 m/s for a further 9 s. Calculate the total distance travelled.

    v_max = 16 m/s (t₁ = 3 s, t₂ = 9 s) · Distance = Area = ½vt₁ + vt₂
  4. 4. On a speed-time graph, a train accelerates uniformly from rest to 22 m/s in 12 s and then continues at a constant speed of 22 m/s for a further 13 s. Calculate the total distance travelled.

    v_max = 22 m/s (t₁ = 12 s, t₂ = 13 s) · Distance = Area = ½vt₁ + vt₂
  5. 5. On a speed-time graph, a train accelerates uniformly from rest to 36 m/s in 4 s and then continues at a constant speed of 36 m/s for a further 5 s. Calculate the total distance travelled.

    v_max = 36 m/s (t₁ = 4 s, t₂ = 5 s) · Distance = Area = ½vt₁ + vt₂
  6. 6. On a speed-time graph, a train accelerates uniformly from rest to 36 m/s in 10 s and then continues at a constant speed of 36 m/s for a further 8 s. Calculate the total distance travelled.

    v_max = 36 m/s (t₁ = 10 s, t₂ = 8 s) · Distance = Area = ½vt₁ + vt₂
  7. 7. On a speed-time graph, a train accelerates uniformly from rest to 28 m/s in 3 s and then continues at a constant speed of 28 m/s for a further 4 s. Calculate the total distance travelled.

    v_max = 28 m/s (t₁ = 3 s, t₂ = 4 s) · Distance = Area = ½vt₁ + vt₂
  8. 8. On a speed-time graph, a train accelerates uniformly from rest to 48 m/s in 5 s and then continues at a constant speed of 48 m/s for a further 20 s. Calculate the total distance travelled.

    v_max = 48 m/s (t₁ = 5 s, t₂ = 20 s) · Distance = Area = ½vt₁ + vt₂
  9. 9. On a speed-time graph, a train accelerates uniformly from rest to 36 m/s in 12 s and then continues at a constant speed of 36 m/s for a further 17 s. Calculate the total distance travelled.

    v_max = 36 m/s (t₁ = 12 s, t₂ = 17 s) · Distance = Area = ½vt₁ + vt₂
  10. 10. On a speed-time graph, a train accelerates uniformly from rest to 10 m/s in 8 s and then continues at a constant speed of 10 m/s for a further 4 s. Calculate the total distance travelled.

    v_max = 10 m/s (t₁ = 8 s, t₂ = 4 s) · Distance = Area = ½vt₁ + vt₂
  11. 11. On a speed-time graph, a train accelerates uniformly from rest to 22 m/s in 6 s and then continues at a constant speed of 22 m/s for a further 14 s. Calculate the total distance travelled.

    v_max = 22 m/s (t₁ = 6 s, t₂ = 14 s) · Distance = Area = ½vt₁ + vt₂
  12. 12. On a speed-time graph, a train accelerates uniformly from rest to 8 m/s in 6 s and then continues at a constant speed of 8 m/s for a further 4 s. Calculate the total distance travelled.

    v_max = 8 m/s (t₁ = 6 s, t₂ = 4 s) · Distance = Area = ½vt₁ + vt₂
12 questions · Seed 53 · Age guidance, not curriculum certification
View the answer key
Worksheet StudioAnswer key · Ages 16-18

Speed-time graphs: area & distance

Answers and reasoning. Review before teaching.

  1. 1. On a speed-time graph, a train accelerates uniformly from rest to 36 m/s in 12 s and then continues at a constant speed of 36 m/s for a further 18 s. Calculate the total distance travelled.

    864 m

    Distance is the area under the speed-time graph: triangle area = ½ × 12 × 36 = 216 m, plus rectangle area = 18 × 36 = 648 m. Total distance = 216 + 648 = 864 m.

  2. 2. On a speed-time graph, a train accelerates uniformly from rest to 34 m/s in 10 s and then continues at a constant speed of 34 m/s for a further 19 s. Calculate the total distance travelled.

    816 m

    Distance is the area under the speed-time graph: triangle area = ½ × 10 × 34 = 170 m, plus rectangle area = 19 × 34 = 646 m. Total distance = 170 + 646 = 816 m.

  3. 3. On a speed-time graph, a train accelerates uniformly from rest to 16 m/s in 3 s and then continues at a constant speed of 16 m/s for a further 9 s. Calculate the total distance travelled.

    168 m

    Distance is the area under the speed-time graph: triangle area = ½ × 3 × 16 = 24 m, plus rectangle area = 9 × 16 = 144 m. Total distance = 24 + 144 = 168 m.

  4. 4. On a speed-time graph, a train accelerates uniformly from rest to 22 m/s in 12 s and then continues at a constant speed of 22 m/s for a further 13 s. Calculate the total distance travelled.

    418 m

    Distance is the area under the speed-time graph: triangle area = ½ × 12 × 22 = 132 m, plus rectangle area = 13 × 22 = 286 m. Total distance = 132 + 286 = 418 m.

  5. 5. On a speed-time graph, a train accelerates uniformly from rest to 36 m/s in 4 s and then continues at a constant speed of 36 m/s for a further 5 s. Calculate the total distance travelled.

    252 m

    Distance is the area under the speed-time graph: triangle area = ½ × 4 × 36 = 72 m, plus rectangle area = 5 × 36 = 180 m. Total distance = 72 + 180 = 252 m.

  6. 6. On a speed-time graph, a train accelerates uniformly from rest to 36 m/s in 10 s and then continues at a constant speed of 36 m/s for a further 8 s. Calculate the total distance travelled.

    468 m

    Distance is the area under the speed-time graph: triangle area = ½ × 10 × 36 = 180 m, plus rectangle area = 8 × 36 = 288 m. Total distance = 180 + 288 = 468 m.

  7. 7. On a speed-time graph, a train accelerates uniformly from rest to 28 m/s in 3 s and then continues at a constant speed of 28 m/s for a further 4 s. Calculate the total distance travelled.

    154 m

    Distance is the area under the speed-time graph: triangle area = ½ × 3 × 28 = 42 m, plus rectangle area = 4 × 28 = 112 m. Total distance = 42 + 112 = 154 m.

  8. 8. On a speed-time graph, a train accelerates uniformly from rest to 48 m/s in 5 s and then continues at a constant speed of 48 m/s for a further 20 s. Calculate the total distance travelled.

    1080 m

    Distance is the area under the speed-time graph: triangle area = ½ × 5 × 48 = 120 m, plus rectangle area = 20 × 48 = 960 m. Total distance = 120 + 960 = 1080 m.

  9. 9. On a speed-time graph, a train accelerates uniformly from rest to 36 m/s in 12 s and then continues at a constant speed of 36 m/s for a further 17 s. Calculate the total distance travelled.

    828 m

    Distance is the area under the speed-time graph: triangle area = ½ × 12 × 36 = 216 m, plus rectangle area = 17 × 36 = 612 m. Total distance = 216 + 612 = 828 m.

  10. 10. On a speed-time graph, a train accelerates uniformly from rest to 10 m/s in 8 s and then continues at a constant speed of 10 m/s for a further 4 s. Calculate the total distance travelled.

    80 m

    Distance is the area under the speed-time graph: triangle area = ½ × 8 × 10 = 40 m, plus rectangle area = 4 × 10 = 40 m. Total distance = 40 + 40 = 80 m.

  11. 11. On a speed-time graph, a train accelerates uniformly from rest to 22 m/s in 6 s and then continues at a constant speed of 22 m/s for a further 14 s. Calculate the total distance travelled.

    374 m

    Distance is the area under the speed-time graph: triangle area = ½ × 6 × 22 = 66 m, plus rectangle area = 14 × 22 = 308 m. Total distance = 66 + 308 = 374 m.

  12. 12. On a speed-time graph, a train accelerates uniformly from rest to 8 m/s in 6 s and then continues at a constant speed of 8 m/s for a further 4 s. Calculate the total distance travelled.

    56 m

    Distance is the area under the speed-time graph: triangle area = ½ × 6 × 8 = 24 m, plus rectangle area = 4 × 8 = 32 m. Total distance = 24 + 32 = 56 m.

12 questions · Seed 53 · Age guidance, not curriculum certification