Forces and Motion
6 marks · NormalA lorry of mass 2000 kg accelerates uniformly from rest to a speed of 15 m/s in 10 s on a straight, level road. The lorry then continues at this speed for another 10 s.
Calculate the average acceleration of the lorry during the first 10 s.
Calculate the resultant force on the lorry during the first 10 s.
Sketch a velocity–time graph for the motion of the lorry for the 20 s described. Label all relevant values on the axes.
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) 1.5 m/s^2
(b) 3000 N
(c) Graph with velocity on the y-axis (0–15 m/s), time on the x-axis (0–20 s); straight line from (0,0) to (10,15), then horizontal line from (10,15) to (20,15). · must show axes labelled: velocity (m/s), time (s), line: (0,0) to (10,15), then (10,15) to (20,15)
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Substitutes values correctly into a = (v-u)/t. (follow-through)
A1 (a) · 1 mark · Correct value: 1.5 m/s² with units.
M2 (b) · 1 mark · Uses F = m × a with values from (a). (follow-through)
A2 (b) · 1 mark · Correct value: 3000 N.
G1 (c) · 1 mark · Graph axes correctly labelled and scaled.
G2 (c) · 1 mark · Correctly plots both sections: up to 10s increasing linearly from (0,0) to (10,15); 10–20s horizontal at 15 m/s.
Worked solution
a) Acceleration = (final velocity - initial velocity) / time = (15 - 0) / 10 = 1.5 m/s².
b) Resultant force = mass × acceleration = 2000 kg × 1.5 m/s² = 3000 N.
c) The velocity-time graph: from 0 to 10 s, velocity rises linearly from 0 to 15 m/s; from 10 to 20 s, it is constant at 15 m/s.