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Cambridge International AS & A Level · 2 · structured

Use gradient and signed area on a velocity graph

Use gradient and signed area on a velocity graph · Original GioPhysics question with a detailed, mark-by-mark answer guide.

Demand
demanding
Marks
7
Topics
2
Answer
Complete
structured7 marks

A velocity–time graph consists of straight segments joining (t/s,v/m s⁻¹)=(0,4),(2,0),(4,−2),(6,−2). The particle starts at displacement zero. Use the specified straight segments, not a smooth interpolated curve.

  1. (a)

    Find acceleration on 0–2s,2–4s and4–6s. Explain the meaning of the horizontal final segment.

    3 marks
  2. (b)

    Use areas to determine total displacement and total distance travelled over 6s. Show how you treat areas below the time axis.

    4 marks
Ready to self-mark?Reveal the detailed answer guide
Answer overview(a) −2.0,−1.0,0 m s⁻² respectively. Final segment is constant negative velocity, not rest. (b) Signed areas:+4,−2,−4m; displacement−2m; distance 4+2+4=10m.
01

(a)

3 marks

Find acceleration on 0–2s,2–4s and4–6s. Explain the meaning of the horizontal final segment.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  1. 1

    Gradients:(0−4)/2=−2.0.

  2. 2

    (−2−0)/2=−1.0.

  3. 3

    (−2−(−2))/2=0; v remains−2m s⁻¹.

02

(b)

4 marks

Use areas to determine total displacement and total distance travelled over 6s. Show how you treat areas below the time axis.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  1. 1

    0–2s area is½×2×4=4m.

  2. 2

    2–4s area is−½×2×2=−2m.

  3. 3

    4–6s area is−2×2=−4m.

  4. 4

    Displacement adds signed areas; distance adds their magnitudes.

Complete marking guidance

Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.

  1. (a) Gradients:(0−4)/2=−2.0.
  2. (a) (−2−0)/2=−1.0.
  3. (a) (−2−(−2))/2=0; v remains−2m s⁻¹.
  4. (b) 0–2s area is½×2×4=4m.
  5. (b) 2–4s area is−½×2×2=−2m.
  6. (b) 4–6s area is−2×2=−4m.
  7. (b) Displacement adds signed areas; distance adds their magnitudes.

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Your self-assessment

Compare your work with the guide. This is your own rating, not an awarded exam mark.