Connecting Position, Velocity, and Acceleration Using Integrals

A particle's velocity can be positive, negative, or zero, and its sign tells you which way the particle is moving along the line. If you integrate v(t) directly over an interval, the positive stretches and negative stretches cancel against each other, and what you get is displacement: the net change in position, start to finish. That is why a particle can travel 20 feet of actual path and still end up back where it started, or even closer to the start than it began moving away from it. The positive distance it covered going one way gets wiped out by the negative distance it covered coming back.

Total distance asks a different question: how much ground did the particle actually cover, ignoring direction. To get that, you integrate |v(t)| instead of v(t). Taking the absolute value flips any interval where velocity is negative up to positive before you integrate, so that backward stretch adds to the total instead of subtracting from it. Mechanically, this means finding where v(t) changes sign, splitting the integral at those points, and integrating the negative pieces with a minus sign in front to make them positive contributions. If velocity never changes sign on the interval, there is nothing to split and nothing to flip, so displacement and total distance come out equal.

This topic is the reverse of what you did with position, velocity, and acceleration as derivatives earlier. There, you differentiated position to get velocity. Here, you integrate velocity to recover position or displacement, using the Fundamental Theorem of Calculus to undo that derivative. The same logic extends one level further: integrating acceleration recovers velocity, which is exactly why this skill sits right before area-between-curves and volume problems, where you are constantly choosing between a signed integral and an absolute-value version depending on what question is actually being asked.