The problems are on the terminal velocity worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Terminal velocity is the steady speed at which a resistive force balances the weight, so the net force is zero. These worked answers correct the 20 problems of the AP Physics C worksheet on terminal velocity, from drag direction to power at terminal speed: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the equation, the substitution and the result. Where a derivation is asked for, the integration steps are written out. Use g = 9.81 m/s² and give final values to two or three significant figures, as the data allow.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:M Unit 2
Problem 1 Drag direction
1. The net force equals mass times acceleration: m dv/dt = ΣF, and ΣF includes the drag term.2. Drag always points opposite to the velocity, so it slows the object relative to the fluid.
Problem 2 Units of coefficients
1. b = F/v, so N ÷ (m/s) = kg·m/s² ÷ m/s = kg/s.2. c = F/v², so N ÷ (m²/s²) = kg/m.
Problem 3 Syrup terminal speed
At terminal speed mg = bvt, so vt = mg/b = (0.20 × 9.81)/0.50 = 3.9 m/s.Problem 4 Time constant
1. τ = m/b = 0.20/0.50 = 0.40 s.2. v = vt(1 − e−1) = 0.632 vt, so about 63 %.
Problem 5 Paper cone
m = 4.0 × 10⁻³ kg. vt = √(mg/c) = √[(4.0 × 10⁻³ × 9.81)/(2.0 × 10⁻³)] = √19.62 = 4.4 m/s.Problem 6 Zero acceleration, nonzero speed
1. Zero acceleration means the velocity is constant, not zero. The object keeps moving at vt.2. The drag force equals the weight in magnitude, so the net force is zero.
Problem 7 Speed in oil
1. vt = mg/b = 0.981/0.50 = 1.962 m/s; τ = m/b = 0.20 s.2. t/τ = 1.5, so v = 1.962(1 − e−1.5) = 1.962 × 0.7769 = 1.5 m/s.
Problem 8 Acceleration history
1. At t = 0 the speed is zero, so there is no drag and a = g = 9.81 m/s².2. a = g e−t/τ = 9.81 × e−1 = 9.81 × 0.368 = 3.61 m/s².
3. As v rises, drag bv rises, so the net force mg − bv falls.
Problem 9 Time to reach 90 %
1. 1 − e−t/τ = 0.90, so e−t/τ = 0.10 and t = τ ln 10.2. t = 0.20 × 2.303 = 0.46 s.
Problem 10 Reading a graph
1. The speed levels off at 2.0 m/s.2. 63 % of 2.0 m/s is 1.26 m/s, reached at 0.25 s, so τ = 0.25 s.
3. b = mg/vt = (0.040 × 9.81)/2.0 = 0.20 kg/s. Check: m/τ = 0.040/0.25 = 0.16 kg/s, close given the graph reading.
Problem 11 Parachute constant
1. At constant speed cv² = mg, so c = mg/v² = (75 × 9.81)/36 = 20.4, which rounds to 20 kg/m.2. F = cv² = 20.4 × 9.0 = 1.8 × 10² N.
Problem 12 Doubling the mass
1. vt = mg/b, so it doubles: factor 2.2. vt = √(mg/c), so it rises by √2: factor 1.4.
3. Drag in the quadratic model grows as v², so a speed increase of only √2 is needed to balance double the weight.
Problem 13 Spot the sign error
1. Drag opposes the downward velocity, so it acts upward and must be negative when down is positive.2. m dv/dt = mg − bv.
3. With +bv the acceleration would grow without limit as the ball speeds up: there would be no terminal speed, which is unphysical.
Problem 14 Verify the solution
1. dv/dt = vt(1/τ) e−t/τ = (mg/b)(b/m) e−t/τ = g e−t/τ.2. Left side: m dv/dt = mg e−t/τ. Right side: mg − bv = mg − b(mg/b)(1 − e−t/τ) = mg e−t/τ. The two sides agree, and v(0) = 0.
Problem 15 Distance fallen
1. x = ∫ v dt = vt[t + τ e−t/τ] − vtτ = vtt − vtτ(1 − e−t/τ).2. At t = τ: x = vtτ e−1 = 1.5 × 0.30 × 0.3679 = 0.17 m.
Problem 16 Boat coasting to rest
1. m dv/dt = −bv gives dv/v = −(b/m) dt; integrating, ln(v/v₀) = −bt/m.2. τ = m/b = 8.0 s; v = 6.0 e−1.5 = 6.0 × 0.2231 = 1.3 m/s.
3. x = ∫₀^∞ v dt = v₀τ = 6.0 × 8.0 = 48 m.
Problem 17 Skydiver with tanh
1. gt/vt = 29.43/30 = 0.981; tanh(0.981) = 0.754; v = 23 m/s.2. v = gt = 29 m/s.
3. Drag has already reduced the speed by about 7 m/s: at 23 m/s the drag is (23/30)² = 59 % of the weight.
Problem 18 Ball thrown upward
1. a = g + (c/m)v² = 9.81 + 2.0 × 10⁻³ × 625 = 11.1 m/s² (downward).2. vt = √(g/(c/m)) = √(9.81/0.0020) = 70 m/s.
3. With k = c/m, separate: v dv/(g + kv²) = −dx. Integrating from v₀ to 0: (1/2k) ln[(g + kv₀²)/g] = H. So H = (1/2k) ln(1 + kv₀²/g) = 250 × ln(1.1274) = 250 × 0.1199 = 30 m.
Problem 19 Block on an oiled ramp
1. Taking down the slope as positive: m dv/dt = mg sin 30° − bv.2. At vt, bvt = mg sin 30°, so vt = (2.0 × 9.81 × 0.50)/5.0 = 2.0 m/s.
3. τ = m/b = 2.0/5.0 = 0.40 s.
Problem 20 Power at terminal speed
1. The kinetic energy is constant, so all the energy lost by gravity each second is transferred to the air by the drag force: P = FDvt = mg vt.2. P = 80 × 9.81 × 62.6 = 4.9 × 10⁴ W = 49 kW.
