The questions are on the kinematics test. Take it under real conditions before reading the mark scheme.
Quick answer: One-dimensional kinematics describes straight-line motion using position, velocity and acceleration, without asking what causes it. This mark scheme gives the model answers and the marks for the 8 questions of the AP Physics 1 test on kinematics, from scalars and vectors to footbridge, with the equivalent answers to accept.
The model answers give full working and a marking guide. Award method marks even when a numerical slip occurs, and allow any consistent sign convention. Do not penalise answers rounded to two significant figures when the data have two.
Level: AP Physics 1 (algebra-based) · Curriculum: AP P1 Unit 1
Question 1 Scalars and vectors (2 marks)
Scalar: speed (or distance). Vector: velocity (or acceleration). A scalar has size only, while a vector has size and direction.
Marking: 1 mark for a correct scalar and vector pair, 1 mark for the difference.
Question 2 Round trip (2 marks)
(a) Displacement = 300 − 120 = 180 m east; average velocity = 180/150 = 1.2 m/s east.
(b) Distance = 420 m; average speed = 420/150 = 2.8 m/s.
Marking: 1 mark for part (a) with direction, 1 mark for part (b).
Question 3 Reading a graph (2 marks)
(a) a = (10 − 0)/4 = 2.5 m/s².
(b) Area = ½ × 4 × 10 + 4 × 10 = 20 + 40 = 60 m.
Marking: 1 mark each part; accept the trapezium method.
Question 4 Using an equation (3 marks)
(a) v² = v₀² + 2aΔx: a = (27² − 12²)/(2 × 90) = (729 − 144)/180 = 3.25 m/s².
(b) t = (27 − 12)/3.25 = 4.6 s.
Marking: 1 mark for the equation, 1 mark for a, 1 mark for t.
Question 5 Stone into a canyon (3 marks)
(a) Δy = ½gt² = ½ × 9.81 × 3.0² = 44.1 ≈ 44 m.
(b) v = gt = 9.81 × 3.0 = 29 m/s.
Marking: 1 mark for the equation, 1 mark for the depth, 1 mark for the speed.
Question 6 Interpreting a curve (2 marks)
The slope increases steadily, so the velocity increases uniformly: the object starts at rest and moves with constant positive acceleration. The velocity-time graph is a straight line through the origin with positive slope.
Marking: 1 mark for constant acceleration from rest, 1 mark for the straight-line v–t graph.
Question 7 Experimental design (3 marks)
Drop the ball from several measured heights h and time each fall at least three times, taking the mean. Plot h (vertical) against t² (horizontal). Since h = ½gt², the graph is a straight line through the origin and g = 2 × slope.
Marking: 1 mark for varying and measuring h and t with repeats, 1 mark for plotting h against t², 1 mark for g = 2 × gradient.
Question 8 Footbridge (3 marks)
(a) Taking down as positive, Δy = v₀t + ½gt² = 6.0 × 1.8 + ½ × 9.81 × 1.8² = 10.8 + 15.9 = 26.7 ≈ 27 m.
(b) v = 6.0 + 9.81 × 1.8 = 6.0 + 17.7 = 23.7 ≈ 24 m/s.
Marking: 1 mark for the equation, 1 mark for the height, 1 mark for the speed.
