Area under a velocity-time graph gives distance travelled
4 marks · HardA go-kart travels with constant velocity for 2 seconds, then slows down uniformly to a stop in the next 3 seconds.
The initial velocity is 8 m/s. Draw a labelled velocity–time graph for the go-kart’s motion over the 5 seconds.
Using your graph, calculate the total distance travelled by the go-kart during the 5 seconds. Show your working.
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) Graph with a horizontal line at 8 m/s from t = 0 to 2 s, then a straight line sloping down to 0 m/s at t = 5 s. Both axes labelled. · must show horizontal section from (0,8) to (2,8), straight sloped section from (2,8) to (5,0), axes correctly labelled
(b) 20 m · must show splitting area into rectangle (or trapezium/triangle), correct area calculation for each shape
Common mistakes: forgetting to use a triangle for deceleration part; not splitting areas correctly
Show mark scheme and worked solution
Marking points
B1 (a) · 1 mark · Graph shows horizontal line from (0,8) to (2,8), then straight line descending from (2,8) to (5,0).
B2 (a) · 1 mark · Axes labelled correctly (time/s and velocity/m/s); points and lines clearly plotted.
M1 (b) · 1 mark · Rectangular area for first 2 s (2 × 8 = 16 m) and triangular area for next 3 s ( m) identified.
A1 (b) · 1 mark · Sums areas correctly to obtain total distance: 16 + 12 = 28 m. (follow-through)
Worked solution
(a) The velocity–time graph starts at 8 m/s at t = 0, remains at 8 m/s until t = 2 s, then reduces uniformly to 0 m/s at t = 5 s. Axes are labelled.
(b) The distance for the first part is:
The slowing section is a triangle:
Total distance = 16 + 12 = 28 m