Movement and Position
6 marks · NormalA student uses a toy car to investigate acceleration down a track. The results are shown in the table.
| Time (s) | Velocity (m/s) |
|---|---|
| 0.0 | 0.0 |
| 1.0 | 2.0 |
| 2.0 | 4.0 |
| 3.0 | 6.0 |
| 4.0 | 8.0 |
Plot a velocity–time graph for the results above. Label each axis.
Determine the gradient of the velocity–time graph. State what this gradient represents.
Calculate the distance travelled by the car in the first 4 seconds using your graph or the area method.
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 time on x-axis (0–4 s), velocity on y-axis (0–8 m/s), straight line through points (0,0), (1,2), (2,4), (3,6), (4,8). · must show Axes labelled and points correctly plotted
(b) Gradient = 2 m/s². The gradient represents the car's acceleration. m/s² · must show States gradient is 2 m/s² and represents acceleration
(c) 16 m. Area under the graph = (1/2) × base × height = (1/2)×4×8 = 16 m. m · must show Method and correct answer
Must show: Axes labelled time (s) and velocity (m/s), Points plotted accurately
Also accept: Area under graph calculated using triangles or rectangles
Common mistakes: Mixing up axes; Not identifying what gradient means
Show mark scheme and worked solution
Marking points
B1 (a) · 2 marks · Axes correctly labelled, data points accurately plotted, straight line drawn.
M1 (b) · 1 mark · Correctly calculates gradient: (8-0)/(4-0)=2 m/s².
B2 (b) · 1 mark · States gradient represents acceleration.
M2 (c) · 1 mark · Finds area under graph: (1/2)×base×height or sums rectangles/triangles.
A1 (c) · 1 mark · Correct distance calculated: 16 m. (follow-through)
Worked solution
a) The velocity–time graph is a straight line from (0,0) up to (4,8). The axes are labelled as required.
b) The gradient is 2 m/s², which is the acceleration of the car.
c) Area under the graph = (1/2) × 4 × 8 = 16 m.