Non-Constant Acceleration & Separation of Variables

You’ll be able to

Recognize when the constant-acceleration kinematic equations do not apply
Solve motion problems in which acceleration depends on time, velocity or position
Use separation of variables to integrate a velocity-dependent acceleration

The kinematic equations are a special case, not a law

Equations such as v = v₀ + at and x = x₀ + v₀t + ½at² are not fundamental. They are what you get by integrating a = dv/dt twice when a is constant. The moment acceleration varies — with time, with velocity, or with position — those formulas are simply wrong, and the only correct move is to return to the definitions and integrate. This is the single most reliable way Physics C separates itself from Physics 1, and it is worth recognizing the trigger instantly.

Three cases, three techniques

If a = a(t), integrate directly with respect to time: v = v₀ + ∫a dt. If a = a(v) — drag is the standard case — separate the variables: dv/dt = a(v) becomes dv/a(v) = dt, and you integrate each side over its own variable. If a = a(x), use the chain-rule identity a = v dv/dx, which converts the problem into v dv = a(x) dx and eliminates time entirely. Choosing the right one of these three is most of the work.

The identity that removes time

a = dv/dt = (dv/dx)(dx/dt) = v dv/dx

Use this whenever acceleration is given as a function of position and the question does not ask about time. It turns a differential equation in t into one relating v and x directly.

Linear drag as the standard example

A body moving through fluid at low speed feels resistance proportional to velocity: F = −bv, so a = −(b/m)v = −kv. Separating gives dv/v = −k dt, and integrating from v₀ gives v(t) = v₀e^(−kt). The velocity decays exponentially and never reaches exactly zero, yet the total distance is finite: integrating v(t) from zero to infinity gives v₀/k. A body that never quite stops still travels a bounded distance, which is a result worth being surprised by once.

Worked example

A boat of mass m moving at v₀ = 20 m/s shuts off its engine and experiences a = −kv with k = 0.50 s⁻¹. Find its speed at t = 2.0 s and the total distance it coasts.

1.The acceleration depends on velocity, so separate: dv/v = −k dt.
2.Integrate: ln(v/v₀) = −kt, so v(t) = v₀e^(−kt).
3.At t = 2.0 s: v = 20e^(−0.50 × 2.0) = 20e^(−1) ≈ 7.4 m/s.
4.Total distance is ∫₀^∞ v₀e^(−kt) dt = v₀/k = 20/0.50 = 40 m.

Answer: v(2.0 s) = 20e⁻¹ ≈ 7.4 m/s; the boat coasts a total of v₀/k = 40 m

Checkpoint

A particle has acceleration a = −4x, where x is its position. The most efficient route to a relation between v and x is to —

Watch out

The most common lost point on this material is using v² = v₀² + 2aΔx when a is not constant. Before writing any kinematic equation, check whether the problem statement gives acceleration as a number or as a function. If it is a function of anything, integrate.

Checkpoint

For linear drag a = −kv with initial speed v₀, the total distance traveled before the object stops is —

Answer the 2 checkpoints as you read.