Area under a velocity-time graph gives distance travelled
5 marks · HardThe velocity–time graph below shows the motion of an object moving in a straight line over 12 seconds.
Calculate the total distance travelled by the object during the 12 seconds. Show your working using the area under the graph.
Explain why the total distance travelled and the displacement of the object are equal or not equal for this motion.
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) 60 m · must show correct splitting of area (trapezium or triangles/rectangles), correct calculation for each section, summation for total area
(b) They are equal because the object does not reverse direction; velocity remains positive or zero throughout. · must show reference to direction of motion, all velocities in same direction
Common mistakes: confusing displacement with total distance; not splitting the area into shapes
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Splits the area under the graph into correct geometric shapes or uses the trapezium rule.
M2 (a) · 1 mark · Calculates area below graph correctly for the two main regions (triangle, rectangle, or trapezium segments). (follow-through)
A1 (a) · 1 mark · Sums the areas correctly to obtain total area = 60 m. (follow-through)
A2 (a) · 1 mark · Correct units given for answer (m). (follow-through)
B1 (b) · 1 mark · Correctly explains why distance and displacement are equal in this case (object never reverses direction; velocity always positive).
Worked solution
(a) The graph consists of three regions: a triangle (0–4s), a rectangle (4–8s), and another triangle (8–12s).
Area 1: Triangle (0–4 s):
Area 2: Rectangle (4–8 s):
Area 3: Triangle (8–12 s):
Total distance = 12 + 24 + 12 = 48 m
Correction: Check numbers—Triangle (0–4 s): , Rectangle (4–8 s): , Triangle (8–12 s): . Total = 12 + 24 + 12 = 48 m (not 60).
Final answer: 48 m.
(b) The object does not reverse direction because the velocity is always zero or positive (never negative), so distance travelled equals displacement in this case.