Area under a velocity-time graph gives distance travelled
4 marks · HardA trolley starts from rest and accelerates steadily. After 2 s, its velocity is 4 m/s, and after a further 3 s, its velocity is 10 m/s.
Sketch a velocity–time graph for the motion of the trolley for the first 5 seconds.
Calculate the total distance travelled by the trolley in 5 seconds by finding the total area under the velocity–time graph.
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) Velocity–time graph with straight lines joining (0,0) to (2,4) to (5,10), axes labelled. · must show axes labelled, straight line from (0,0) to (2,4), straight line from (2,4) to (5,10)
(b) 21 m · must show correct method splitting the area under the graph, area calculations for each section
Common mistakes: incorrectly joining points with curves instead of straight lines for uniform acceleration; incorrect area calculation methods
Show mark scheme and worked solution
Marking points
B1 (a) · 1 mark · Axes labelled correctly; graph covers 0–5 s and velocity 0–10 m/s.
B2 (a) · 1 mark · Straight line from (0, 0) to (2, 4), followed by straight line from (2, 4) to (5, 10); points plotted correctly.
M1 (b) · 1 mark · Divides the area under the graph into correct geometric shapes (triangles/trapezia).
A1 (b) · 1 mark · Evaluates total area correctly to obtain 21 m. (follow-through)
Worked solution
(a) The graph should have axes labelled (time in s and velocity in m/s), starting at (0, 0), going to (2, 4) in a straight line, then from (2, 4) to (5, 10) in a straight line.
(b) Calculate the area under both sections:
Section 1 (0–2 s): area of triangle:
Section 2 (2–5 s): area of trapezium:
Total area = 4 + 17 = 21 m