Topic 8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals: use integrals to find velocity, position, displacement and total distance.

What this topic is asking

The College Board (Topic 8.2) is the integral side of motion. Since acceleration is the derivative of velocity and velocity the derivative of position, integrating moves the other way: from acceleration to velocity, from velocity to position. You must compute displacement and total distance, and tell them apart.

Recovering velocity and position

A worked motion computation

Why total distance needs absolute value

When the velocity changes sign, the particle reverses direction, and the forward and backward motions partly cancel in the displacement (signed integral). Total distance counts every meter travelled regardless of direction, so it integrates the speed $|v(t)|$. In practice you find where $v$ changes sign, integrate $v$ over each subinterval, then add the magnitudes. Integrating $|v|$ directly is awkward, so the standard method is exactly this split-and-add-magnitudes approach. The error of reporting the displacement when the question asks for total distance (or vice versa) is one of the most penalized in all of AP Calculus motion.

Position needs the initial condition

To find an actual position (not just displacement), you need a starting position. The formula $s(t) = s(t_0) + \int_{t_0}^t v$ shows that the integral gives only the change; the constant $s(t_0)$ comes from the given initial condition. So "find the particle's position at $t = 5$" requires both the displacement integral and the stated $s(0)$. Likewise recovering velocity from acceleration needs an initial velocity. Reading the problem for these initial values, and adding them in, completes the answer. Carrying units throughout and writing a contextual sentence finishes a full-credit free response.

Speeding up versus slowing down

A subtle motion question asks whether the particle is speeding up or slowing down at an instant, which depends on the signs of velocity and acceleration together. The particle speeds up when $v$ and $a$ have the same sign (both pushing the same direction) and slows down when they have opposite signs. This is distinct from the direction of motion (the sign of $v$ alone) and from where the particle turns around (where $v$ changes sign). Because acceleration is the derivative of velocity, you find $a$ by differentiating $v$, then compare its sign with that of $v$ at the instant. Keeping these three questions, direction, turning, and speeding up, separate prevents a common tangle on motion free responses.

The symmetry with the differentiation side

This topic mirrors the Unit 4 motion topic, run in the opposite direction. There, differentiation took position to velocity to acceleration; here, integration takes acceleration back to velocity back to position. The two perspectives reinforce each other: the Fundamental Theorem says that integrating the velocity recovers the net change in position, which is exactly undoing the differentiation that defined velocity as the rate of change of position. Recognizing motion-with-integrals as the inverse of motion-with-derivatives lets you check answers across the two units, for example confirming that differentiating a recovered position function returns the given velocity.