8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals

Integrals undo derivatives in motion

Since velocity is the derivative of position and acceleration is the derivative of velocity, integrating reverses this: integrating acceleration gives velocity, and integrating velocity gives position. Use the Fundamental Theorem of Calculus: v(t) = v(a) + ∫ₐᵗ a(x) dx, and s(t) = s(a) + ∫ₐᵗ v(x) dx. You need an initial condition to find the constant.

Total distance vs. displacement

Displacement over [a,b] is ∫ₐᵇ v(t) dt — it can be positive, negative, or zero. Total distance traveled is ∫ₐᵇ |v(t)| dt, found by splitting the interval where v(t) changes sign, computing each piece's absolute value, and summing. Distance is always ≥ 0 and accounts for direction changes.

Using given data

When v(t) or a(t) is given as a table, graph, or formula, apply the same integral relationships. The net change in position equals ∫ₐᵇ v(t) dt regardless of whether the particle moves back and forth. Always check for sign changes in v(t) to correctly separate distance calculations.

Key terms

Relationship between position, velocity, acceleration via integrals

v(t) = v(a) + ∫ₐᵗ a(x) dx; s(t) = s(a) + ∫ₐᵗ v(x) dx — integration reverses differentiation.

Displacement formula

∫ₐᵇ v(t) dt — the net change in position; can be positive, negative, or zero.

Total distance traveled formula

∫ₐᵇ |v(t)| dt — always nonnegative; requires splitting the interval where v(t) changes sign.

Why split the interval for distance?

Because v(t) may change sign (direction reversal); integrating |v(t)| requires evaluating each sign-consistent piece separately and adding absolute values.

Finding position from velocity with initial condition

s(t) = s(a) + ∫ₐᵗ v(x) dx, where s(a) is the known position at time a.

Try an AP-style question

v(t) = t² − 4 on [0, 3]. What is the particle's displacement?

1. −3
2. 23/3
3. 3
4. −16/3