The questions are on the suvat equations test. Take it under real conditions before reading the mark scheme.
Quick answer: The suvat equations connect displacement, initial velocity, final velocity, acceleration and time for motion in a straight line with constant acceleration. This mark scheme gives the model answers and the marks for the 8 questions of the Physics 11-12 test on suvat equations, from choosing equations to van overtaking, with the equivalent answers to accept.
Marks are given for correct equations, correct substitutions and correct answers with units. Follow-through marks apply when an earlier error is carried into a later part correctly. The marking notes show where each mark is earned.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: A Level 3.4, IB A.1
Question 1 Choosing equations (2 marks)
- v = u + at.
- s = ½(u + v)t.
Marking: 1 mark for each correct equation.
Question 2 Cyclist (3 marks)
- v = u + at = 2.0 + 0.80 × 10 = 10 m/s.
- s = ut + ½at² = 2.0 × 10 + 0.5 × 0.80 × 100 = 20 + 40 = 60 m.
Marking: (a) 1 mark; (b) 1 mark for the correct equation and substitution, 1 mark for the answer.
Question 3 Canyon drop (3 marks)
- s = ½gt², so t = √(2 × 80 ÷ 9.81) = √16.31 = 4.04 s.
- v = gt = 9.81 × 4.04 = 39.6 m/s.
Marking: (a) 1 mark for the rearranged equation, 1 for the time; (b) 1 mark.
Question 4 Bus braking (3 marks)
- a = (0 − 15) ÷ 5.0 = −3.0 m/s².
- Distance = area under the line = ½ × 5.0 × 15 = 37.5 m.
Marking: (a) 1 mark; (b) 1 mark for using the area, 1 mark for the answer.
Question 5 Thrown ball (3 marks)
- v² = u² + 2as with v = 0: s = 144 ÷ (2 × 9.81) = 7.34 m.
- Time up = 12 ÷ 9.81 = 1.223 s, so total = 2.45 s.
Marking: (a) 1 mark for the equation, 1 for the value; (b) 1 mark.
Question 6 Top of the flight (2 marks)
The velocity is zero but the acceleration is still 9.81 m/s² downwards. Because the acceleration is not zero, the velocity keeps changing and the ball starts to fall.Marking: 1 mark for both values; 1 mark for the explanation.
Question 7 Limits of the equations (2 marks)
The equations need constant acceleration. As the parachutist speeds up, air resistance increases, so the resultant force and the acceleration change continuously.Marking: 1 mark for constant acceleration needed; 1 mark for air resistance changing the acceleration.
Question 8 Van overtaking (2 marks)
v² = u² + 2as: a = (625 − 25) ÷ 200 = 3.0 m/s². t = (25 − 5.0) ÷ 3.0 = 6.7 s.Marking: 1 mark for the acceleration; 1 mark for the time.
