The questions are on the calculus kinematics test. Take it under real conditions before reading the mark scheme.
Quick answer: In calculus-based kinematics, velocity is the derivative of position and acceleration is the derivative of velocity. This mark scheme gives the model answers and the marks for the 8 questions of the AP Physics C test on calculus kinematics, from derivatives of a position function to simple harmonic motion, with the equivalent answers to accept.
Marks are shown for each step. Method marks are given for a correct derivative or integral even if the arithmetic slips, but final answers need units.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:M Unit 1
Question 1 Derivatives of a position function (3 marks)
a. v = 3t² − 12t + 9, a = 6t − 12.b. 3(t − 1)(t − 3) = 0, so at rest at t = 1 s and t = 3 s.
Marking: each derivative (1); both times (1).
Question 2 Integrating acceleration (2 marks)
a. v = 2 + ∫6t dt = 2 + 3t².b. v(3) = 2 + 27 = 29 m/s.
Marking: integral with constant (1); value (1).
Question 3 Reading a velocity graph (3 marks)
a. Area = ½(4)(8) + 6(8) + ½(4)(8) = 80 m.b. a = slope = 8 / 4 = 2 m/s².
Marking: correct areas (1); total with unit (1); slope (1).
Question 4 Exponential slowing (3 marks)
a. v = 12e⁻¹ = 4.41 m/s.b. x = ∫0^∞ 12e^(−0.50t) dt = 12 / 0.50 = 24 m.
Marking: value (1); integral (1); result with unit (1).
Question 5 Acceleration as a function of position (3 marks)
a. ∫v dv = ∫−9x dx gives v² = 36 − 9x². Set v = 0: x² = 4, so x = 2.0 m.b. v² = 27, so v = 5.2 m/s.
Marking: separation and integration (1); largest x (1); speed (1).
Question 6 Why constant-acceleration equations fail (2 marks)
a. The equation is derived by integrating a constant acceleration; here a changes with time, so it does not apply.b. The area under an a–t graph is the change in velocity.
Marking: one mark each part.
Question 7 Speeding up or slowing down (2 marks)
a. At 2.5 s, v = −2.25 m/s and a = +3 m/s², opposite signs, so the speed is decreasing.b. At 4.0 s, v = 9 m/s and a = 12 m/s², same sign, so the speed is increasing.
Marking: one mark each part.
Question 8 Simple harmonic motion (2 marks)
a. v = 1.2 cos(4.0t), so the maximum speed is 1.2 m/s.b. a = −4.8 sin(4.0t), so the maximum acceleration is 4.8 m/s².
Marking: one mark each part.
