Movement and Position
6 marks · NormalA car is initially travelling at 15 m/s. It accelerates uniformly at 4 m/s² for a certain time.
Calculate the velocity of the car after 3 seconds.
Calculate the distance the car travels during these 3 seconds.
Sketch a velocity–time graph for the motion described above. Include numerical values for axes.
Explain the physical meaning of the gradient and the area under the graph you have drawn.
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) 27 m/s (v = 15 + 4 × 3 = 27 m/s) m/s · must show Correct formula used
(b) 63 m (s = ut + 0.5at² = 15×3 + 0.5×4×9 = 45 + 18 = 63 m) m · must show Formula, steps and final answer
(c) Velocity–time graph: time 0–3 s (x-axis), velocity 15–27 m/s (y-axis), straight line from (0,15) to (3,27); axes labelled. · must show Axes, endpoints, gradual rise
(d) Gradient = acceleration (4 m/s²); area under the graph = distance travelled (63 m). · must show Both gradient and area described
Common mistakes: Wrong calculation for distance
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Applies v = u + at correctly.
M2 (b) · 1 mark · Applies s = ut + 0.5at².
A1 (b) · 1 mark · Finds distance as 63 m. (follow-through)
B1 (c) · 2 marks · Draws labelled, correctly shaped v–t graph for interval, with endpoints.
B2 (d) · 1 mark · Explains both gradient (acceleration) and area (distance).
Worked solution
c) Graph: x-axis 0–3 s, y-axis 15–27 m/s, straight line. d) Gradient is 4 m/s² = acceleration; area under line = 63 m = distance.