Area under a velocity-time graph gives distance travelled
4 marks · NormalA sprinter runs a race. The velocity–time graph below shows her motion for the first 8 seconds of the race.
Calculate the total distance the sprinter travels in the first 8 seconds.
Explain how the area under the graph represents the distance travelled.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) 64 m · must show Breakdown of area into shapes: two triangles and a rectangle, Area for 0-4s (triangle), Area for 4-6s (rectangle), Area for 6-8s (triangle), Correct sum for all areas
(b) The area under a velocity–time graph gives the total distance travelled because distance equals velocity multiplied by time; the area represents this product, summed over the motion. · must show Reference to area = velocity × time, States area equals distance moved
Common mistakes: Calculating only one area instead of all areas; Multiplying final velocity by total time without considering acceleration/deceleration
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Breaks graph into correct geometric areas (2 triangles + rectangle) or equivalent reasoning.
M2 (a) · 1 mark · Calculates area for each region correctly.
A1 (a) · 1 mark · Sums areas for correct total: 64 m (accept 63–65 m for rounding).
B1 (b) · 1 mark · Clear statement relating area under graph to velocity × time and hence distance.
Worked solution
a) Break the area under the graph into shapes: from 0–4s a triangle, 4–6s a rectangle, 6–8s a triangle.
Area 1: triangle (0–4s): ½ × base × height = ½ × 4 × 12 = 24 m
Area 2: rectangle (4–6s): width × height = 2 × 12 = 24 m
Area 3: triangle (6–8s): ½ × base × height = ½ × 2 × 12 = 12 m
Area 4: triangle (0–2s): ½ × 2 × 8 = 8 m; Area 5: trapezium (2–4s): (2/2) × (8+12) = 2 × 10 = 20 m (alternative method, same total area).
Total distance = 24 + 24 + 12 + 4 (from 0–2s triangle) = 64 m (Alternative breakdown reaches same sum).
b) The area under a velocity–time graph gives the total distance travelled, because at every time interval, velocity × time = distance for that segment; adding all the strips adds up to the total distance.