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Science lesson plan · 10th grade (Year 11 in England) · 30 minutes

Distance from the area under a velocity–time graph

Forces and motion

CCSS NGSS.HS-PS2-1CCSS NGSS.SEP 5England: GCSE Combined Science: Forces (Higher: area under velocity–time graph)

Objective and success criteria

Students will: Distance from the area under a velocity–time graph.

  • I can answer questions on today's skill on my own, without help.
  • I can explain each step and say why it works.
  • I can check my answer and spot a mistake.

Before and after

Needs first: Read velocity–time graphs: gradient and motion

Leads to: Uniform acceleration: v² − u² = 2 a s

Materials: Force meters, Toy cars and ramps, Mini whiteboards, A diagram or short video clip of the idea, Printed student sheet (below).

Warm-up · 4 min

  • Quick retrieval from earlier learning (Read velocity–time graphs: gradient and motion). Students answer on mini whiteboards.
  • Show answers together and note who needs support in the main task.

Hook · 3 min

  • Drop a sheet of paper and a scrunched-up sheet together and ask why one falls faster.
  • Share the objective: Distance from the area under a velocity–time graph. Read the success criteria together.

I do: model it · 7 min

  • Work through the example out loud, saying why you do each step, not just what you do.
  • Example: A velocity–time graph rises in a straight line from (0 s, 0 m/s) to (4 s, 4 m/s). How far does the object travel in this time, in m?
  • Step 1: ½ × 4 × 4 = 8
  • Step 2: So the answer is: 8 m
  • Answer: 8 m
  • Pause after each step and ask a student to say what comes next.

We do: guided practice · 6 min

  • Do these together. Students try each step on whiteboards before you reveal it.
  • Use the hints as prompts rather than giving the answer.
  1. A velocity–time graph rises in a straight line from (0 s, 0 m/s) to (5 s, 10 m/s). How far does the object travel in this time, in m? → 25 mPrompt: Distance travelled = area under a velocity–time graph.
  2. A train travels at a constant 20 m/s for 60 s. On a velocity–time graph, what is the area under the line, and so the distance travelled, in m? → 1200 mPrompt: A constant velocity gives a horizontal line, so the area is a rectangle.

You do: independent practice · 7 min

  • Students work on the student sheet. Circulate and check the first two answers of every student early.
  • Students who finish: write a question of their own for a partner, with the answer.

The 6 questions are on the student sheet below, with answers in the key.

Exit ticket · 3 min

  • Two questions on their own. Sort answers into got it / nearly / not yet to plan the next lesson.
  1. A train travels at a constant 25 m/s for 40 s. On a velocity–time graph, what is the area under the line, and so the distance travelled, in m? → 1000 mPrompt: A constant velocity gives a horizontal line, so the area is a rectangle.
  2. A velocity–time graph rises in a straight line from (0 s, 0 m/s) to (5 s, 15 m/s). How far does the object travel in this time, in m? → 37.5 mPrompt: Distance travelled = area under a velocity–time graph.

Watch out for

  • Everyday meanings of words getting in the way of the science meaning: define key words explicitly.
  • Remembering facts without the reason: ask "why?" and "what is the evidence?"

Support and stretch

Support

  • Review first: Read velocity–time graphs: gradient and motion.
  • Use concrete materials: force meters.
  • Give the first step of each question, or a partly worked example to finish.
  • Pair with a confident partner for the guided practice.

Stretch

  • Move on to: Uniform acceleration: v² − u² = 2 a s.
  • Explain a common mistake in writing and how to fix it.
  • Write a word problem that uses today's skill, then solve a partner's.

Homework and more practice

Free online practice with hints and printable worksheets for this skill: eurekami.com/science/grade-10.

Free lesson plan from eurekami.com/lesson-plan-builder

eurekami

Distance from the area under a velocity–time graph

Name

Date

Score/ 6

  1. Question 1

    A velocity–time graph rises in a straight line from (0 s, 0 m/s) to (4 s, 12 m/s). How far does the object travel in this time, in m?

    Answerm

  2. Question 2

    A train travels at a constant 10 m/s for 20 s. On a velocity–time graph, what is the area under the line, and so the distance travelled, in m?

    Answerm

  3. Question 3

    Velocity–time graph: (0 s, 0 m/s) → (5 s, 15 m/s) → (10 s, 15 m/s) → (14 s, 0 m/s). What is the total distance travelled, in m?

    xy24681012141624681012141618200(0, 0)ABC

    Answerm

  4. Question 4

    A velocity–time graph rises in a straight line from (0 s, 0 m/s) to (4 s, 4 m/s). How far does the object travel in this time, in m?

    Answerm

  5. Question 5

    A train travels at a constant 12 m/s for 40 s. On a velocity–time graph, what is the area under the line, and so the distance travelled, in m?

    Answerm

  6. Question 6

    Velocity–time graph: (0 s, 0 m/s) → (10 s, 10 m/s) → (15 s, 10 m/s) → (20 s, 0 m/s). What is the total distance travelled, in m?

    xy612186120(0, 0)ABC

    Answerm

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Answer key: Distance from the area under a velocity–time graph

  1. 1. 24 m

  2. 2. 200 m

  3. 3. 142.5 m

  4. 4. 8 m

  5. 5. 480 m

  6. 6. 125 m