Kinematics

v = s'(t), a = v'(t), displacement = ∫ v dt.

The idea

Kinematics describes motion along a line, and calculus links its three quantities: position, velocity, and acceleration.

A particle travelling along a line has a position $s(t)$ at each time $t$. Its velocity is the rate of change of position — a derivative: $v(t) = s'(t).$ Velocity carries a sign: a negative value means the particle is moving backwards along the line. The instants where $v(t) = 0$ are the instants it is momentarily at rest, and only there can it reverse direction.

The acceleration is the rate of change of velocity: $a(t) = v'(t)$.

Integration runs the chain the other way. Velocity is the rate at which position accumulates, so integrating it recovers the change in position: $s(b) - s(a) = \int_{a}^{b} v(t)\,dt.$ This number is the displacement, the net change in position, and it is signed: a trip out and back contributes nothing. Distance travelled is the different quantity $\int_{a}^{b} |v(t)|\,dt$, which counts every stretch as positive and allows no cancellation. The two agree only when the particle never turns around.

Ways to work on it

• Walkthrough. Position, velocity, acceleration, and rest points.
• Practice. Velocity from a position function.
• Hardest. Displacement via a definite integral; distance vs displacement.

Not sure where to start? Take the ten-question placement test.