Interpreting Motion Graphs
Overview
This is a self-contained, no-technology substitute packet in which students learn to read motion graphs (kinematics). Working alone with a pencil (a calculator is allowed), students read a short original reference on position–time graphs (slope = velocity), velocity–time graphs (slope = acceleration, area = displacement), and the formulas speed = distance ÷ time and acceleration = Δv ÷ Δt; study two original labeled graphs; read values off the graphs; compute an average velocity from a position–time segment and an acceleration from a velocity–time segment; match described motions to the correct graph; fix a common graph-reading error; and write an evidence-based explanation (CER) of an object's motion. It is a paper investigation with no lab materials and no hazards. Runs in a ~50-minute standard period; a ~90-minute block adds computing displacement as the area under a velocity–time graph.
At a glance
Course: Physics (Grades 9–12)
Subject: Science
Time: about 50 minutes standard, or a ~90-minute block (block adds an area/displacement extension)
Materials: printed packet and a pencil; a calculator is allowed (not required)
Work mode: independent
Product: graph-value calculations and a claim–evidence–reasoning explanation (CER)
Standards (provisional): 19 TAC Chapter 112 (Physics) — describing motion with position–time and velocity–time graphs, and calculating velocity and acceleration. Provisional — pending educator verification against the current official TAC source; a specific section number is intentionally not asserted. Standards are paraphrased, not quoted. This packet does not claim formal alignment to the TEKS.
Accessible version of the student activity
The full student activity is reproduced below in plain, screen-reader-friendly HTML. It reflows on phones and at 200% zoom. Write your answers on the printed packet.
Start (5 minutes) — Notice & Wonder
- Look at Graph A (a position–time graph). Without doing math, write one thing you notice and one thing you wonder about how the line changes as time goes on.
- On a graph of an object's position over time, what do you think a flat (horizontal) line means the object is doing? Give your best first idea.
Build (8–12 minutes) — Read the science
A position–time (p–t) graph has time on the horizontal axis and position (meters) on the vertical axis; the slope of the line equals the object's velocity. A steeper line means a faster speed, and a flat line means the object is stopped (velocity 0). A velocity–time (v–t) graph has time on the horizontal axis and velocity (m/s) on the vertical axis; here the slope is the acceleration, and the area under the line is the displacement.
Key formulas: speed = distance ÷ time (m/s); average velocity = Δposition ÷ Δtime; acceleration = Δv ÷ Δt (m/s²); displacement = area under a v–t line (rectangle area = velocity × time; triangle area = ½ × base × height).
Word bank
- position
- where an object is, measured as its distance from a starting point (meters, m).
- velocity
- how fast position is changing, with a direction (m/s); the slope of a position–time graph.
- acceleration
- how fast velocity is changing (m/s²); the slope of a velocity–time graph.
- slope
- rise ÷ run; velocity on a p–t graph, acceleration on a v–t graph.
- displacement
- the overall change in position (meters); the area under a velocity–time line.
- Using Graph A, describe what Object A is doing during segment 2 (4 s to 7 s), and say what the slope of that segment equals.
Apply (20–25 minutes) — Use the graphs
Read values straight off the gridlines; show work and keep units. The plotted values are:
| Time (s) | Position (m) | Segment |
|---|---|---|
| 0 | 0 | start |
| 4 | 40 | end of segment 1 (rise) |
| 7 | 40 | end of segment 2 (flat) |
| 10 | 100 | end of segment 3 (steeper rise) |
| Time (s) | Velocity (m/s) | Segment |
|---|---|---|
| 0 | 10 | start of segment 1 (flat) |
| 4 | 10 | end of segment 1 / start of rise |
| 8 | 30 | end of segment 2 (rise) |
| 10 | 30 | end of segment 3 (flat) |
- Read Graph A: position of Object A at t = 0 s, 4 s, 7 s, and 10 s.
- Average velocity from Graph A segment 1 (0 s to 4 s): use average velocity = Δposition ÷ Δtime; show the formula, numbers with units, and the answer in m/s.
- Find the average velocity of Graph A segment 3 (7 s to 10 s), then state which is faster (segment 1 or segment 3) and how the steepness of the line tells you.
- Acceleration from Graph B segment 2 (4 s to 8 s): use acceleration = Δv ÷ Δt; show the formula, numbers with units, and the answer in m/s².
- Match the motion to the graph — write A or B: a) "stopped in the middle, then speeds up and covers ground faster than before"; b) "steady 10 m/s, then steadily speeds up to 30 m/s, then holds 30 m/s."
- Error analysis: a student said segment 2 of Graph A is "flat and high up, so the object is moving very fast and steadily." Explain what the student confused and give the correct interpretation of segment 2 (including the velocity value).
Explain (7–10 minutes) — Claim, Evidence, Reasoning
Question 10. Using Graph A, write a claim describing Object A's motion across all three segments, support it with two pieces of numeric evidence (position and time values from Graph A), and explain your reasoning using the idea that the slope of a position–time graph equals velocity.
Sentence stems you may use: "Object A first…, then…, and finally…"; "One piece of evidence is that at t = ___ s the position is ___ m."; "A second piece of evidence is…"; "This shows the velocity because the slope of a position–time graph equals…"
Close (5 minutes) — ACE
- Articulate: explain what the slope means on a position–time graph versus a velocity–time graph.
- Connect: point to one segment where the velocity is not changing and name the values.
- Extend: describe a new three-step real-life trip and what its position–time graph would look like.
Continue (optional) — Early finisher & block extension
Early finisher: sketch your own labeled position–time graph with at least one stopped (flat) segment and one segment steeper than another, then name the fastest segment.
Block extension (~+30 min): using Graph B, compute Object B's total displacement as the area under the velocity–time line (rectangles plus a triangle), then explain why area under a v–t graph equals distance traveled. (Total displacement = 180 m.)
Turn in
Hand in the whole packet with your name, class period, and date, with questions 1–10 and the ACE box answered. Include the optional sketch and block calculation if you did them.