Short on time? This page is the recap. For the derivations, the graph and two worked examples, use the full study guide.
Recap
- A resistive force depends on velocity and points opposite to it. Course model: F_r = −kv, with k in kg/s.
- Newton’s second law, m dv/dt = ΣF, then contains v on both sides: a differential equation.
- Solve by separation of variables: v terms on one side, t on the other, then integrate from the initial state (0, v₀) to a general state (t, v).
- The solutions are exponential with time constant τ = m/k. They approach asymptotes set by the forces and initial conditions.
- Terminal velocity is reached when the constant force and the resistive force balance: ΣF = 0.
Key relationships
| Situation (axis) | Equation | Velocity | Acceleration | Asymptote |
|---|---|---|---|---|
| Coasting (+x along v₀) | m dv/dt = −kv | v₀ e^(−t/τ) | −(v₀/τ) e^(−t/τ) | v → 0; x → x₀ + mv₀/k |
| Falling from rest (+y down) | m dv/dt = mg − kv | v_T(1 − e^(−t/τ)) | g e^(−t/τ) | v → v_T = mg/k |
| Thrown down at v₀ (+y down) | m dv/dt = mg − kv | v_T + (v₀ − v_T) e^(−t/τ) | −((v₀ − v_T)/τ) e^(−t/τ) | v → v_T |
Useful numbers: after one τ an approach is 63% complete; after 5τ it is over 99% complete. The time to halve in coasting is τ ln 2.
Assumptions
- The drag is exactly proportional to v (a model; real drag on fast objects is closer to v²).
- k stays constant: same shape, same fluid.
- Near Earth’s surface, g = 9.8 m/s². Buoyancy is ignored.
Mistakes to avoid
- Constant-acceleration equations. The acceleration changes with v. Integrate instead.
- Wrong sign for drag. It opposes the velocity, not “up”. On the way up, drag acts down.
- Mismatched limits. Pair v₀ with t = 0 in the two integrals.
- “No force at terminal velocity.” Two forces act; the net force is zero.
- “a = 0 at release.” From rest, a = g at t = 0; a → 0 as v → v_T.
- Losing the constant in ln. ln v − ln v₀ = ln(v/v₀); exponentiate the whole side.
Quick self-check
- A 0.30 kg object with k = 1.5 kg/s falls from rest. Find v_T and τ. (v_T = 0.30 × 9.8 ÷ 1.5 ≈ 2.0 m/s; τ = 0.20 s)
- An object coasts with v₀ = 4.0 m/s and τ = 2.0 s. Find v at 2.0 s and the total coasting distance. (4.0 e^(−1) ≈ 1.5 m/s; v₀τ = 8.0 m)
- A falling object has reached half its terminal speed. What is its acceleration? (Drag = mg/2, so a = g/2 = 4.9 m/s²)
Next: practice questions.