Dynamics
Newton's second law treated as what it actually is — a differential equation, \(F = m\dfrac{dv}{dt}\) — rather than a single algebra formula for one case. Constant force reduces to the kinematics already covered; forces that depend on time or velocity don't, and solving those is the entire reason this unit needs calculus at all. The velocity-dependent case (linear drag) is this course's first real differential equation, and its solution — an object approaching a terminal velocity it never quite reaches — is a genuinely different kind of motion than anything constant acceleration can produce.
What you'll learn
- Write Newton's second law as a differential equation and recognize constant-force motion as the special case matching Unit 101's kinematics
- Solve for velocity and position given a time-dependent force F(t), by dividing by mass and integrating
- Set up the differential equation for a velocity-dependent force (linear drag), and verify a given solution by substitution
- Explain terminal velocity as the velocity at which net force — and therefore acceleration — reaches zero, without the object ever truly stopping
- Apply Newton's third law and F=ma to systems of connected objects, unchanged from the algebra-based treatment
1. Newton's Second Law as a Differential Equation
Newton's second law is usually written \(F=ma\), but \(a\) is itself \(dv/dt\) — so the law is really \(F = m\dfrac{dv}{dt}\), a differential equation relating force to how velocity changes over time. When \(F\) is constant, dividing by \(m\) gives a constant \(a\), and integrating reproduces exactly the kinematics from Unit 101. Nothing new happens there. The interesting cases are when \(F\) isn't constant — when it depends on time, on velocity, or on position — because then finding the motion means actually solving a differential equation, not just plugging into a formula.
2. Time-Dependent Forces
When \(F(t)\) genuinely depends on time, \(a(t) = F(t)/m\) is just as genuine a function of time — and finding \(v(t)\) and \(x(t)\) is exactly the integration process from Unit 101, applied to whatever \(a(t)\) this new force produces. No new technique is needed here, just the recognition that the force itself is the thing that's changing.
3. Velocity-Dependent Forces and Terminal Velocity
A falling object with linear air resistance experiences two forces: gravity, pulling it down at a constant \(mg\), and drag, pushing back against its motion with a force proportional to speed, \(bv\). Newton's second law becomes \(m\dfrac{dv}{dt} = mg - bv\), or \(\dfrac{dv}{dt} = g - \dfrac{b}{m}v\) — and this is a genuinely different kind of equation than anything in Unit 101 or the constant/time-dependent cases above, because the unknown function \(v(t)\) appears on both sides. This can't be solved by direct integration of a known function of \(t\); it has to be solved as a differential equation (by separating variables), or — the more common approach in a first course — verified by substituting a proposed solution and checking that it satisfies the equation.
The solution is \(v(t) = \dfrac{mg}{b}\left(1-e^{-bt/m}\right)\). As \(t\) grows, \(e^{-bt/m}\to 0\), so \(v(t)\) approaches — but never exactly reaches — the terminal velocity \(v_{terminal}=mg/b\), the speed at which gravity and drag exactly balance and acceleration drops to zero.
4. Newton's Third Law and Connected-Object Systems
None of the calculus above changes how Newton's third law or multi-object systems work — every force still comes in an equal-and-opposite pair acting on two different objects, and solving a connected system (blocks on strings, pulleys) still means writing \(F=ma\) for each object separately and solving the resulting equations together. The only difference from an algebra-based course is that if any of those forces happen to depend on time or velocity, the resulting equations are the same differential equations covered above, applied per object instead of to a single mass.
Key equations
- F = m·dv/dt = m·d²x/dt² — Newton's second law in its actual differential form — the starting point for every problem in this unit.
- a(t) = F(t)/m → v(t) = v₀ + ∫a(t)dt — Time-dependent force: divide by mass to get a(t), then integrate exactly as in Unit 101.
- dv/dt = g − (b/m)v — Falling object with linear air resistance — a genuine differential equation in v, not solvable by direct integration of a known function of t.
- v(t) = (mg/b)(1 − e^(−bt/m)) — The solution to the drag equation above, for an object released from rest — approaches but never reaches mg/b.
- v_terminal = mg/b — The velocity at which net force (and so acceleration) is exactly zero — where gravity and drag balance.