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Year 11 · 2017 syllabus · Module 1 pilot

Graphs that reveal and preserve patterns — Apply worksheet

Original practice. Write explanations that distinguish observations, evidence and claims.

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Name: __________________________ Date: ______________

Read a falling-object distance–time graph

Read a falling-object distance–time graphDownward distance (m)0.00.00.20.10.40.20.60.30.80.4Time since release (s)
Synthetic idealised falling-object data, generated using distance = 5 × time². Assumptions: release from rest, downward distance from release, constant acceleration 10 m/s² and negligible air resistance. Not measured data.
Text description
  • Time in seconds is on the horizontal axis and downward distance in metres on the vertical axis. Points are (0, 0), (0.1, 0.05), (0.2, 0.20), (0.3, 0.45) and (0.4, 0.80). The plotted points rise increasingly steeply. Straight segments join the sampled points as a visual guide, not exact motion between samples.

Time in seconds is on the horizontal axis and downward distance in metres on the vertical axis. Points are (0, 0), (0.1, 0.05), (0.2, 0.20), (0.3, 0.45) and (0.4, 0.80). The plotted points rise increasingly steeply. Straight segments join the sampled points as a visual guide, not exact motion between samples.

Synthetic sampled downward distance
Time since release (s)Downward distance (m)
00
0.10.05
0.20.2
0.30.45
0.40.8

Question 1

Using the synthetic falling-object graph above, calculate the distance travelled and average speed from 0.0–0.1 s and from 0.3–0.4 s.

Question 2

A bar chart compares material categories A, B and C. Why avoid implying values between A and B?

Question 3

Explain why the falling-object graph is not a straight line. State two assumptions and what would be required to compare it with a real falling object.

Teacher answer key — 3 questions

Model responses identify the essential ideas. Accept scientifically correct alternatives that address the question; wording need not match exactly.

  1. First: (0.05 − 0)/0.10 = 0.5 m/s over 0.05 m. Last: (0.80 − 0.45)/0.10 = 3.5 m/s over 0.35 m. These are interval-average speeds. The final interval covers seven times the distance in the same time.
  2. The categories are not a continuous numerical variable; an intermediate category need not exist. Separate bars/points preserve that structure.
  3. The object covers more distance in equal time intervals, so its speed increases. The model assumes release from rest and negligible air resistance (also accept constant acceleration 10 m/s²). Comparison needs a safe teacher-approved recording with calibrated distance/time, matching start conditions and an evaluation of air resistance. Synthetic points are not primary observations.