1.2 Kinematic Relationships in One Dimension

Calculus links position, velocity, acceleration

Velocity is the derivative of position with respect to time: v = dx/dt. Acceleration is the derivative of velocity: a = dv/dt. Since AP Physics C uses calculus, you can differentiate position functions to get velocity, and differentiate velocity to get acceleration — no need for constant-acceleration shortcuts if given x(t).

Integration goes the other way

Given acceleration as a function of time, integrate a(t) to find v(t), and integrate v(t) to find x(t), using initial conditions to solve for constants of integration. Graphically, the area under an a-t graph gives change in velocity, and the area under a v-t graph gives displacement.

Constant acceleration equations still apply

When acceleration is constant, the kinematic equations (v = v₀ + at, x = x₀ + v₀t + ½at², v² = v₀² + 2a(x − x₀)) are shortcuts derived from calculus. They only work when a is constant — for variable acceleration, you must integrate or differentiate directly.

Key terms

v(t) in terms of x(t)

v(t) = dx/dt — velocity is the derivative of position with respect to time.

a(t) in terms of v(t)

a(t) = dv/dt — acceleration is the derivative of velocity with respect to time.

Finding v(t) from a(t)

Integrate a(t) with respect to time and add the initial velocity constant: v(t) = ∫a(t)dt + v₀.

Area under a-t graph

Represents the change in velocity, Δv.

When to use constant-acceleration kinematic equations

Only valid when acceleration is constant; otherwise use derivatives/integrals of the actual functions.

Try an AP-style question

a(t) = 2t, with x(0) = 0 and v(0) = 0. What is x at t = 3 s?

1. 27 m
2. 9 m
3. 6 m
4. 18 m