Movement and Position
6 marks · NormalThe table shows the motion of a ball thrown vertically upwards. Air resistance is negligible.
| Time (s) | Velocity (m/s) |
|---|---|
| 0.0 | 24.0 |
| 1.0 | 14.2 |
| 2.0 | 4.4 |
| 3.0 | -5.4 |
| 4.0 | -15.2 |
Plot the velocity–time graph for the ball using the data given. Label both axes.
Determine the acceleration acting on the ball.
Calculate the distance the ball travels in the first 2 seconds.
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 of velocity against time, points plotted: (0,24), (1,14.2), (2,4.4), (3,–5.4), (4,–15.2). Axes: x = time (s), y = velocity (m/s). · must show Correct axes, points
(b) –9.8 m/s² (a = (–15.2–24.0)/4 = –39.2/4 = –9.8 m/s²) m/s² · must show Uses gradient formula
(c) 28.6 m (Area = (1/2) × (24.0+4.4)×2 = 28.6 m) m · must show Working and correct area method
Common mistakes: Forgetting negative velocity on way down; Mixing up signs for acceleration
Show mark scheme and worked solution
Marking points
B1 (a) · 2 marks · Plots points with correct axes and labels.
M1 (b) · 1 mark · Finds gradient: (change in velocity)/(change in time).
A1 (b) · 1 mark · Correct answer: –9.8 m/s². (follow-through)
M2 (c) · 1 mark · Area method: (1/2)(v₀+v)×t.
A2 (c) · 1 mark · Correct answer: 28.6 m. (follow-through)
Worked solution
a) Draw graph, plot each point, time on x-axis, velocity on y-axis.