Integrating Acceleration

You’ll be able to

Recover velocity from acceleration and position from velocity by integration
Use initial conditions to fix the constant of integration
Derive the constant-acceleration kinematic equations as integrals of a = constant

Integration reverses differentiation

If differentiating position gives velocity, then integrating acceleration must give velocity back, and integrating velocity must give position. Integration is the accumulation of change: the area under an a–t graph over an interval is the change in velocity across it, and the area under a v–t graph is the change in position.

Building motion back up by integration

v(t) = v₀ + ∫ a dt x(t) = x₀ + ∫ v dt

Each indefinite integral introduces a constant; v₀ and x₀ are those constants, set by the initial conditions.

The constant of integration is a physical initial condition

An indefinite integral is only known up to an added constant C, because many functions share the same derivative. Physically that constant is the value at t = 0. Integrate a(t) and the leftover constant is exactly the initial velocity v₀; integrate v(t) and the constant is the initial position x₀. Without a starting condition, the motion is undetermined.

Worked example

A particle starts at x₀ = 0 with velocity v₀ = 2 m/s and undergoes acceleration a(t) = 6t (SI units). Find v(t) and x(t), then find the position at t = 3 s.

1.Integrate acceleration: v(t) = ∫ 6t dt = 3t² + C₁.
2.Apply v(0) = 2: 3(0)² + C₁ = 2, so C₁ = 2 and v(t) = 3t² + 2.
3.Integrate velocity: x(t) = ∫ (3t² + 2) dt = t³ + 2t + C₂.
4.Apply x(0) = 0: C₂ = 0, so x(t) = t³ + 2t.
5.Evaluate at t = 3: x(3) = 3³ + 2(3) = 27 + 6 = 33 m.

Answer: v(t) = 3t² + 2, x(t) = t³ + 2t; at t = 3 s the position is 33 m

Where the kinematic equations come from

When acceleration is constant, these integrals become the equations you memorized in an algebra-based course. Integrate a = constant once to get v = v₀ + at; integrate that to get x = x₀ + v₀t + ½at². They are not separate magic formulas — they are just the special case of ∫a dt and ∫v dt when a does not depend on time.

Constant-a kinematics as integral results

v = v₀ + at x = x₀ + v₀t + ½at²

Valid ONLY when a is constant. If a depends on t, integrate a(t) directly instead.

Checkpoint

A particle starts from rest (v₀ = 0) with acceleration a(t) = 6t (SI units). What is its velocity at t = 2 s?

Checkpoint

You integrate a known a(t) and obtain v(t) = (an expression) + C, where C is an unknown constant. What determines the value of C?

Watch out

Never apply v = v₀ + at or x = x₀ + v₀t + ½at² when acceleration varies with time. If a(t) is anything but a constant, those shortcuts are wrong — go back to integrating a(t) directly.

Answer the 2 checkpoints as you read.