For p(t) = t, 2t, find the position at t = 5. [1 point]

If the displacement over 4 seconds is 8, -4, what is the average velocity? [1 point]

College Board AP 2025 (style)-style practice: Section II (free response, calculator allowed). A particle has position p(t) = t^2, 4t. (a) Find the displacement vector from t = 1 to t = 3. (b) Find the average velocity over [1, 3].

A 4-point question on displacement and average velocity. (a) Displacement (2 points): p(1) = 1, 4 and p(3) = 9, 12. The displacement is p(3) - p(1) = 9 - 1, 12 - 4 = 8, 8. (b) Average velocity (2 points): average velocity is the displacement divided by the elapsed time, (1)/(3 - 1) 8, 8 = (1)/(2) 8, 8 = 4, 4.

United StatesPrecalculusSyllabus dot point

What is a vector-valued function, and how does it describe position and motion in the plane?

Topic 4.9 Vector-Valued Functions: interpret a vector-valued function whose output is a position vector, and relate it to parametric motion and velocity.

A focused answer to AP Precalculus Topic 4.9, covering vector-valued functions whose output is a position vector, their equivalence to parametric functions, how to evaluate position at a time, and how average velocity is the displacement vector over time.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

A vector-valued function takes a parameter $t$ and returns a vector, usually the position vector $\mathbf{p}(t) = \langle x(t), y(t) \rangle$ of a moving point. It is exactly a parametric function written with vector notation: the components $x(t)$ and $y(t)$ are the same. To find the position at a time, evaluate each component. The displacement from $t_1$ to $t_2$ is the vector difference $\mathbf{p}(t_2) - \mathbf{p}(t_1)$ , pointing from the earlier position to the later one. The average velocity over $[t_1, t_2]$ is the displacement divided by the elapsed time, $\frac{\mathbf{p}(t_2) - \mathbf{p}(t_1)}{t_2 - t_1}$ , a vector whose direction is the direction of net travel and whose magnitude is the average speed along that straight-line displacement. Vector-valued functions package planar motion so that position, displacement and velocity are all vectors.

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What this topic is asking
Position as a vector
Displacement between two times
Average velocity
How this ties the unit together
Try this

What this topic is asking

The College Board (Topic 4.9) wants you to interpret a vector-valued function, whose input is a parameter (often time) and whose output is a vector, typically a position vector. This is the same idea as a parametric function, written in vector notation. You evaluate the position at a time, find the displacement vector between two times, and compute average velocity.

Position as a vector

So everything from Topics 4.1 and 4.2 applies: evaluate componentwise to find where the point is at a given time.

Displacement between two times

Displacement is net change in position, not total distance travelled along the path; a point that loops back can have a small displacement despite a long journey.

Average velocity

This is the vector version of the average-rate-of-change idea: change in position over change in time, with the change in position now a vector.

Position, displacement and average velocity

A particle has position $\mathbf{p}(t) = \langle 3t, t^2 - 1 \rangle$ . Find its position at $t = 2$ and $t = 4$ , the displacement, and the average velocity over $[2, 4]$ .

step 1 Evaluate the positions

$\mathbf{p}(2) = \langle 3 \cdot 2, 2^2 - 1 \rangle = \langle 6, 3 \rangle$ . $\mathbf{p}(4) = \langle 3 \cdot 4, 4^2 - 1 \rangle = \langle 12, 15 \rangle$ .

step 2 Find the displacement

$\mathbf{p}(4) - \mathbf{p}(2) = \langle 12 - 6, 15 - 3 \rangle = \langle 6, 12 \rangle$ .

step 3 Divide by the elapsed time

Elapsed time $4 - 2 = 2$ . Average velocity $= \frac{1}{2}\langle 6, 12 \rangle = \langle 3, 6 \rangle$ .

step 4 Interpret

The average velocity $\langle 3, 6 \rangle$ points up and to the right: over $[2, 4]$ the particle moves, on average, $3$ units right and $6$ units up per unit of time. Its direction is the direction of the displacement.

How this ties the unit together

A point worth stating once is that the vector-valued function is the synthesis of the unit so far: it is a parametric function (Topic 4.1) used to model motion (Topic 4.2), with its rates of change (Topic 4.3) now expressed as vectors. Displacement is a vector difference (Topic 4.8), and average velocity is that displacement scaled by the reciprocal of elapsed time (scalar multiplication). Seeing position, displacement and velocity as one chain of vector operations, each built from the last, is what makes the motion picture coherent and sets up the matrix transformations of Topics 4.10 to 4.14, where vectors are the objects that matrices act on.

Try this

Q1. For $\mathbf{p}(t) = \langle t, 2t \rangle$ , find the position at $t = 5$ . [1 point]

Cue. $\langle 5, 10 \rangle$ , evaluating each component.

Q2. If the displacement over $4$ seconds is $\langle 8, -4 \rangle$ , what is the average velocity? [1 point]

Cue. Divide by the time: $\frac{1}{4}\langle 8, -4 \rangle = \langle 2, -1 \rangle$ .

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2025 (style)1 marksSection I, Part B (multiple choice, calculator allowed). A vector-valued function gives position

\mathbf{p}(t) = \langle 2t, t^2 \rangle

. What is the position vector at

t = 3

? (A)

\langle 6, 9 \rangle

(B)

\langle 6, 6 \rangle

(C)

\langle 5, 9 \rangle

(D)

\langle 9, 6 \rangle

Show worked answer →

The correct answer is (A), $\langle 6, 9 \rangle$ .

Evaluate each component at $t = 3$ : the first is $2(3) = 6$ and the second is $3^2 = 9$ . The position vector is $\langle 6, 9 \rangle$ . A vector-valued function works just like a parametric function, with the output written as a vector.

AP 2025 (style)4 marksSection II (free response, calculator allowed). A particle has position

\mathbf{p}(t) = \langle t^2, 4t \rangle

. (a) Find the displacement vector from

t = 1

t = 3

. (b) Find the average velocity over

[1, 3]

Show worked answer →

A 4-point question on displacement and average velocity.

(a) Displacement (2 points): $\mathbf{p}(1) = \langle 1, 4 \rangle$ and $\mathbf{p}(3) = \langle 9, 12 \rangle$ . The displacement is $\mathbf{p}(3) - \mathbf{p}(1) = \langle 9 - 1, 12 - 4 \rangle = \langle 8, 8 \rangle$ .
(b) Average velocity (2 points): average velocity is the displacement divided by the elapsed time, $\frac{1}{3 - 1}\langle 8, 8 \rangle = \frac{1}{2}\langle 8, 8 \rangle = \langle 4, 4 \rangle$ .

Related dot points

Sources & how we know this

AP Precalculus Course and Exam Description — College Board (2023)