The problems are on the calculus kinematics worksheet. Try them on your own first, then compare with the answers below.
Quick answer: In calculus-based kinematics, velocity is the derivative of position and acceleration is the derivative of velocity. These worked answers correct the 20 problems of the AP Physics C worksheet on calculus kinematics, from velocity and acceleration from a position function to numerical integration from data: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the derivative or integral, the use of initial conditions and the final value with its unit. If your result is off, check the constants of integration and whether you used distance where displacement was asked.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:M Unit 1
Problem 1 Velocity and acceleration from a position function
1. Firstly, v = dx/dt = 10t − 3.2. v(2) = 20 − 3 = 17 m/s. a = dv/dt = 10, so a = 10 m/s² at all times.
Problem 2 Derivatives of a cubic
1. Firstly, v = 6t², a = 12t.2. v(3) = 6 × 9 = 54 m/s, a(3) = 12 × 3 = 36 m/s².
Problem 3 Displacement from a velocity function
1. Firstly, x = ∫3t² dt = t³.2. x(4) − x(0) = 64 m.
3. v avg = 64 / 4 = 16 m/s.
Problem 4 Integrating a simple acceleration
1. Firstly, v = 2 + ∫6t dt = 2 + 3t².2. v(3) = 2 + 27 = 29 m/s.
Problem 5 Units of the constants
1. Firstly, v = 4.5t² + 4.0 = 4.5(4.0) + 4.0 = 22 m/s.2. a = 9.0t = 18 m/s².
3. Finally, 4.5 is in m/s³ and 4.0 is in m/s, so v ends in m/s.
Problem 6 Area under a v–t graph
1. Firstly, area = ½(4)(8) + (6)(8) + ½(4)(8) = 16 + 48 + 16 = 80 m.2. a = 8 / 4 = 2 m/s².
3. a = −8 / 4 = −2 m/s².
Problem 7 When is the particle at rest?
1. Firstly, v = 3t² − 12t + 9 = 3(t − 1)(t − 3), so it is at rest at t = 1 s and t = 3 s.2. x(1) = 4 m and x(3) = 0 m.
3. x(4) = 64 − 96 + 36 = 4 m, so displacement = 4 m. The path is 0 → 4 → 0 → 4 m, so distance = 4 + 4 + 4 = 12 m.
Problem 8 Average and instantaneous velocity
1. Firstly, (18 − 2) / 2 = 8 m/s.2. v = 4t = 8 m/s at t = 2 s.
3. Finally, for a uniformly accelerating motion v is linear in t, so the average velocity over an interval equals the instantaneous velocity at the midpoint.
Problem 9 Two integrations
1. Firstly, v = 1 + ∫(12t² − 4) dt = 4t³ − 4t + 1; x = ∫v dt = t⁴ − 2t² + t.2. v(2) = 32 − 8 + 1 = 25 m/s; x(2) = 16 − 8 + 2 = 10 m.
Problem 10 Thrown ball with integration
1. Firstly, v = 15 − 9.81t; y = 15t − 4.905t².2. v = 0 at t = 15 / 9.81 = 1.53 s.
3. y = 15(1.529) − 4.905(1.529)² = 11.5 m.
Problem 11 Area under an a–t graph
1. Firstly, area = 3 × 4 + ½(4)(3) = 12 + 6 = 18 m/s.2. v = 2.0 + 18 = 20 m/s.
Problem 12 Speeding up or slowing down
1. Firstly, v = 3t² − 12t + 9 and a = 6t − 12. At 0.5 s: v = 3.75 m/s, a = −9 m/s². At 2.5 s: v = −2.25 m/s, a = +3 m/s². At 4.0 s: v = 9 m/s, a = 12 m/s².2. Then opposite signs at 0.5 s and 2.5 s, so slowing; same signs at 4.0 s, so speeding up.
Problem 13 Exponential velocity
1. Firstly, v = 12e⁻¹ = 4.41 m/s.2. a = dv/dt = −0.5 × 12e^(−0.5t) = −0.5v. ✓
3. x = ∫0^∞ 12e^(−0.5t) dt = 12 / 0.5 = 24 m.
Problem 14 Acceleration depends on position
1. Firstly, ∫v dv = ∫−9x dx gives v²/2 = −4.5x² + 18, so v² = 36 − 9x².2. v = 0 when x² = 4, so x = 2.0 m.
3. v² = 36 − 9 = 27, so v = 5.2 m/s.
Problem 15 Braking car
1. Firstly, v = 20 − 0.4t².2. 20 = 0.4t², so t = √50 = 7.07 s.
3. x = 20t − (0.4/3)t³ = 141.4 − 47.1 = 94.3 m.
Problem 16 Oscillating particle
1. Firstly, v = 1.2 cos(4.0t); a = −4.8 sin(4.0t).2. Then maxima: 1.2 m/s and 4.8 m/s².
3. at t = 0.50, 4.0t = 2.0 rad: x = 0.30(0.909) = 0.273 m; v = 1.2(−0.416) = −0.499 m/s; a = −4.8(0.909) = −4.36 m/s². Check: −16 × 0.273 = −4.37 m/s² ✓. a = −ω²x.
Problem 17 Drag proportional to speed squared
1. Firstly, v dv/dx = −kv² gives dv/v = −k dx, so ln v = −kx + C and with v(0) = v₀, v = v₀e^(−kx).2. v = 25e^(−0.020 × 50) = 25e⁻¹ = 9.2 m/s.
Problem 18 Acceleration that fades away
1. Firstly, v = ∫6.0e^(−t/4) dt = 24(1 − e^(−t/4)); as t → ∞ the limit is 24 m/s.2. v(4) = 24(1 − 0.368) = 15.2 m/s.
3. x = 24[t + 4e^(−t/4)] from 0 to 4 = 24(1.4715) = 35.3 m.
Problem 19 Catching up
1. Firstly, 2t² = 18 + 6t gives t² − 3t − 9 = 0, so t = (3 + √45) / 2 = 4.85 s.2. x = 2(4.854)² = 47.1 m.
3. v_A = 4t = 19.4 m/s against v_B = 6 m/s, so A is moving 13.4 m/s faster than B.
Problem 20 Numerical integration from data
1. Firstly, area = ½(0 + 2.1) + ½(2.1 + 3.8) + ½(3.8 + 5.0) + ½(5.0 + 5.6) = 1.05 + 2.95 + 4.40 + 5.30 = 13.7 m.2. Then a ≈ (5.0 − 2.1) / 2 = 1.45 m/s².
3. Finally, the velocity rises more slowly each second, so the slope of the v–t graph falls, as when a force reduces with speed.
