In planar motion, integrating velocity over time gives a vector result that captures net change in position, connecting derivatives, accumulation, and initial conditions in a central BC motion model.

Displacement from Velocity

When a particle moves in the plane, its position is often written as $\mathbf{r}(t)=\langle x(t),y(t)\rangle$. Its velocity vector is $\mathbf{v}(t)=\mathbf{r}'(t)=\langle x'(t),y'(t)\rangle$. Since integration undoes differentiation, the definite integral of velocity over a time interval tells how far the position vector changes over that interval, not how much ground the particle travels along its path.

A velocity–time graph with shaded signed area illustrates that displacement is the accumulation of velocity over time. Regions above the $t$-axis contribute positive displacement, while regions below contribute negative displacement, reinforcing that the definite integral measures net change rather than total travel. Source

Displacement is a vector, so direction matters. A positive $x$-component means the particle ends farther to the right than it started, while a negative $x$-component means it ends farther to the left. The same idea applies to the $y$-component. Because displacement records only starting and ending position, opposite motions can cancel each other.

This piecewise velocity–time graph uses rectangular regions to show how signed area accumulates across time intervals. The positive block and negative block demonstrate how direction reversals subtract from the total, so the final displacement reflects net area rather than total distance traveled. Source

In AP Calculus BC, this integral is usually evaluated component by component. If $\mathbf{v}(t)=\langle v_x(t),v_y(t)\rangle$, then $\int_a^b \mathbf{v}(t),dt=\left\langle \int_a^b v_x(t),dt,\int_a^b v_y(t),dt \right\rangle$. Each component of displacement comes from accumulating the corresponding component of velocity. This matters because the particle may move in one direction horizontally while moving in another direction vertically. Units also follow the meaning of the integral: if velocity is measured in meters per second and time in seconds, displacement is measured in meters.

Determining Position from Displacement

Once displacement is known, it can be combined with a starting location to find where the particle is at a later time. This is the main use of displacement in motion problems: velocity describes how position changes, and the definite integral packages that change over a chosen interval.

To determine an actual position, you need an initial condition, such as the particle’s position at time $t=a$. Velocity alone describes change, but it does not identify a unique location. Many different particles could have the same velocity function and still be in different places.

This formula shows the relationship among position, velocity, and displacement. Starting position plus net change equals ending position. In component form, if $\mathbf{r}(a)=\langle x_0,y_0\rangle$, then the final position is found by adding the displacement components to $x_0$ and $y_0$. You can also think of this as an application of the Fundamental Theorem of Calculus: because $\mathbf{r}'(t)=\mathbf{v}(t)$, any position function is an antiderivative of the velocity vector, and the initial condition selects the correct one.

Interpreting the Result

A displacement vector should always be read geometrically.

It tells how the endpoint compares with the starting point, not what route the particle followed in between. A particle can move forward, reverse direction, and still have a small displacement if the final position ends close to the initial position.

Several common interpretations are important:

• If $\int_a^b \mathbf{v}(t),dt=\langle 0,0\rangle$, the particle’s final position is the same as its initial position over that interval.

• If only one component of the displacement is zero, the particle ends with the same coordinate in that direction but may still have changed position overall.

• A nonzero displacement does not mean the particle moved in a straight line.

• A large displacement component indicates a substantial net change in that coordinate, even if the particle changed direction during the interval.

These interpretations help on free-response and multiple-choice questions, where you may need to explain the meaning of a vector answer rather than just compute it.

Common AP Expectations and Pitfalls

On AP problems, the velocity vector may be given directly, or it may be embedded in a word problem with component functions. In either case, the key idea is to integrate the vector over the stated time interval and then use any given initial position correctly.

Keep these points in mind:

• Use a definite integral for displacement over an interval. An indefinite integral gives a family of position functions and is not the same as displacement.

• Keep the answer as a vector or ordered pair unless the problem clearly asks for something else.

• Apply initial conditions after integrating when the question asks for position rather than displacement.

• Watch the interval carefully. Changing the limits changes the sign and meaning of the displacement.

• Integrate each component separately. Do not try to combine vector components into a single scalar expression.

• Include units when provided. Displacement and position have the same units as the coordinate system.

• Do not confuse net change with total travel. A particle can travel a long distance and still have small displacement if its motion partly cancels.

When interpreting or checking an answer, ask whether the vector points from the starting location to the ending location. That viewpoint keeps the calculus connected to the geometry of motion.