Forces and Motion
5 marks · NormalA trolley of mass 1.5 kg is pulled along a straight track by a constant force. The trolley accelerates from rest to a velocity of 2.8 m/s in 4.0 s. A student draws the velocity–time graph below for the motion:
Calculate the acceleration of the trolley.
Calculate the resultant (unbalanced) force on the trolley.
State two differences between the information shown on a velocity–time graph and a distance–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.
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(a) 0.70 m/s^2 · must show a = (v-u)/t
(b) 1.1 N · must show F=ma with calculated a
(c) Velocity–time graph shows acceleration (gradient); distance–time graph shows speed (gradient) and not acceleration directly. · must show One reference to what is shown by each graph
Common mistakes: Using wrong gradient for acceleration; Not converting time units; Misapplying F=ma
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Correctly substitutes v, u, and t into the acceleration formula.
A1 (a) · 1 mark · Correct acceleration calculation to 2 sig. fig.
M2 (b) · 1 mark · Uses F=ma with calculated acceleration and correct mass. (follow-through)
A2 (b) · 1 mark · Evaluates force to 2 sig. fig. (follow-through)
B1 (c) · 1 mark · States two differences correctly between the two graph types.
Worked solution
a) Acceleration, a = (v – u) / t = (2.8 – 0) / 4.0 = 0.70 m/s^2
b) Force, F = m × a = 1.5 kg × 0.70 m/s² = 1.05 N ≈ 1.1 N
c) A velocity–time graph's gradient shows acceleration; a distance–time graph's gradient shows speed. A velocity–time graph shows changes in velocity (including negative/zero), while a distance–time graph does not.