Extension — displacement from the v–t graph

• (Not stated in the book text, but needed for graph questions like "find the total distance covered".) The area between the v–t graph and the time axis equals the displacement (distance for motion in one direction).
• For the graph above:

• AB: ½ × 10 × 25 = 125 m
• BC: 10 × 25 = 250 m
• CD: ½ × 10 × 25 = 125 m
• Total distance = 500 m; average velocity = 500/30 = 16.7 m/s.

For constant acceleration, the following results (all follow from the straight-line graph) are useful:

• v = u + at (this is just the definition of acceleration, a = (v − u)/t),
• s = ½(u + v)t (area of a trapezoid),
• s = ut + ½at², v² = u² + 2as.

Example. A car starts from rest and accelerates at 2 m/s² for 10 s. Then v = 0 + 2 × 10 = 20 m/s and the distance s = ½ × 2 × 10² = 100 m (check: area of the triangle ½ × 10 × 20 = 100 m).