Motion Graphs

• Distance–time graph — slope gives speed; horizontal line = rest; straight sloping line = uniform speed.
• Velocity–time graph — slope gives acceleration; area under graph gives displacement.
• Reading graphs for uniform and non-uniform motion.
• NCERT graphical interpretation of motion.

1. Distance–time (s–t) — slope = Δs/Δt = speed; steeper slope → higher speed.
2. Horizontal s–t line — object at rest (speed = 0).
3. Curved s–t line — changing speed (non-uniform motion).
4. Velocity–time (v–t) — slope = acceleration; horizontal line → uniform velocity (a = 0).
5. Area under v–t graph — displacement (rectangle + triangle for trapezoid cases).
6. Diagram (text) — v–t: triangle from 0 to 10 m/s in 5 s → area = ½ × 5 × 10 = 25 m.
7. NCERT Ch. 8 — compare graphs of walker, cyclist, and car in city traffic.
8. Real world — speedometer vs odometer; cruise control gives flat v–t (approx.).

Finding displacement from a velocity–time graph (uniform acceleration 0 to 8 m/s in 4 s)

Step 1 — Graph is a triangle: base = 4 s, height = 8 m/s.
Step 2 — Area = ½ × base × height = ½ × 4 × 8 = 16 m.
Step 3 — Displacement = 16 m in direction of motion.
Step 4 — Average velocity = 16/4 = 4 m/s (= (u + v)/2 for uniform accel from rest).
Step 5 — Acceleration = slope = 8/4 = 2 m/s².

• Reading distance from v–t graph as slope instead of area.
• Misconception: steep s–t graph means acceleration (s–t slope is speed, not acceleration).
• Confusing s–t and v–t axes.
• Negative slope on v–t always means slowing (depends on sign convention).
• Assuming curved v–t means non-accelerated motion (curvature means changing acceleration).

• What does the slope of a distance–time graph represent?
• What does the area under a velocity–time graph represent?
• How is rest shown on a distance–time graph?
• A v–t graph is a horizontal line at 5 m/s for 10 s. Find displacement.

Open the Practice tab for graded questions on Motion Graphs.