College Board AP 2024 (BC, style)-style practice: Section II (free response, calculator). A particle moves with velocity v(t) = t - 2, t^2 - 4 for 0 t 3, starting at r(0) = 1, 1. (a) Find the position at t = 3. (b) Set up the integral for the total distance traveled.

A 4-point planar-motion problem. (a) (2 points) x(3) = 1 + _0^3 (t-2)\,dt = 1 + [(t^2)/(2) - 2t]_0^3 = 1 + ((9)/(2) - 6) = 1 - (3)/(2) = -(1)/(2). y(3) = 1 + _0^3 (t^2 - 4)\,dt = 1 + [(t^3)/(3) - 4t]_0^3 = 1 + (9 - 12) = -2. Position -(1)/(2), -2. (b) (2 points) Distance = _0^3sqrt((t-2)^2 + (t^2 - 4)^2)\,dt (evaluate numerically).

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How do you solve planar motion problems using parametric and vector-valued functions?

Topic 9.6 Solving Motion Problems Using Parametric and Vector-Valued Functions: find position, velocity, speed, acceleration, displacement and distance for a particle moving in the plane (BC).

A focused answer to AP Calculus BC Topic 9.6, solving planar motion problems with parametric and vector-valued functions, finding position, velocity, speed, acceleration, displacement and total distance, with worked examples.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-04

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

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Quick answer

For a particle with position $\mathbf{r}(t) = \langle x(t), y(t)\rangle$ :
velocity $\mathbf{v} = \langle x', y'\rangle$ (differentiate);
speed $= |\mathbf{v}| = \sqrt{(x')^2 + (y')^2}$ (magnitude);
acceleration $\mathbf{a} = \langle x'', y''\rangle$ ;
displacement over $[a, b]$ $= \int_a^b \mathbf{v}\,dt$ (a vector);
total distance $= \int_a^b |\mathbf{v}|\,dt$ (a scalar);
position $\mathbf{r}(b) = \mathbf{r}(a) + \int_a^b \mathbf{v}\,dt$ . The exam tests whether you pick the right tool for the noun in the question, especially the vector-versus-scalar split.

Jump to a section

What this topic is asking
The toolkit at a glance
A worked full-motion problem
Total distance traveled: the calculator workhorse
Finding when the particle is momentarily at rest or fastest
A worked distance setup

What this topic is asking

The College Board (Topic 9.6, BC only) is the synthesis topic of the parametric/vector strand: a single particle moving in the plane, and a question that may ask for any of position, velocity, speed, acceleration, displacement, or total distance. It combines the differentiation of Topic 9.4, the integration of Topic 9.5, and the arc length of Topic 9.3 into one problem type, the 2D analogue of the straight-line motion of Unit 4.

The toolkit at a glance

Key fact

Given a starting point and either position, velocity, or acceleration, every other quantity follows:

Want	From position $\mathbf{r}$	From velocity $\mathbf{v}$
Velocity	$\mathbf{r}'(t)$	given
Acceleration	$\mathbf{r}''(t)$	$\mathbf{v}'(t)$
Speed	$\|\mathbf{r}'(t)\|$	$\|\mathbf{v}(t)\|$
Position at $b$	given	$\mathbf{r}(a) + \int_a^b\mathbf{v}\,dt$
Displacement	$\mathbf{r}(b) - \mathbf{r}(a)$	$\int_a^b\mathbf{v}\,dt$
Distance	---	$\int_a^b\|\mathbf{v}\|\,dt$

Differentiate to go toward acceleration; integrate (with initial conditions) to go toward position.

A worked full-motion problem

Speed, displacement, and a position

A particle moves with velocity $\mathbf{v}(t) = \langle 3t^2,\, 2t\rangle$ , starting at $\mathbf{r}(1) = \langle 0,\, 5\rangle$ . Find the speed at $t = 1$ , the displacement on $[1, 2]$ , and the position at $t = 2$ .

step 1 Speed at t = 1

$\mathbf{v}(1) = \langle 3, 2\rangle$ , so speed $= \sqrt{3^2 + 2^2} = \sqrt{13}$ .

step 2 Displacement on [1, 2]

$\int_1^2\langle 3t^2, 2t\rangle\,dt = \left\langle [t^3]_1^2,\ [t^2]_1^2\right\rangle = \langle 8 - 1,\ 4 - 1\rangle = \langle 7,\, 3\rangle$ .

step 3 Position at t = 2

$\mathbf{r}(2) = \mathbf{r}(1) + \langle 7, 3\rangle = \langle 0, 5\rangle + \langle 7, 3\rangle = \langle 7,\, 8\rangle$ .

step 4 Sanity check

Both velocity components are positive on $[1, 2]$ , so the particle moves up and to the right, consistent with position increasing in both coordinates.

Total distance traveled: the calculator workhorse

The most frequently tested piece of Topic 9.6 is total distance traveled, $\int_a^b\sqrt{(x'(t))^2 + (y'(t))^2}\,dt$ . This integrand rarely has an elementary antiderivative, so it lives on the calculator-active sections, where you write the integral with correct limits and let the calculator produce the number. The setup must use the speed (magnitude of velocity), not the velocity components themselves; integrating a signed component gives a displacement coordinate, not distance. A reliable habit is to write "distance $= \int_a^b\sqrt{(x')^2 + (y')^2}\,dt$ " as a template, then drop in the derivatives. The marks are in the correct integrand and limits.

Finding when the particle is momentarily at rest or fastest

Some problems ask for instants when the particle is at rest or moving fastest or slowest. The particle is at rest where the velocity vector is zero, meaning $x'(t) = 0$ and $y'(t) = 0$ at the same $t$ , which is more restrictive than in 1D. To find extreme speed, treat the speed $\sqrt{(x')^2 + (y')^2}$ as a function of $t$ and use the usual critical-point analysis (or, on a calculator, examine the graph of speed). Note that a single component being zero does not stop the particle; it only means the motion is momentarily purely horizontal or vertical. Distinguishing "velocity is zero" (both components) from "a velocity component is zero" (motion along one axis) is a frequent point of confusion.

A worked distance setup

Setting up total distance

A particle has position $\mathbf{r}(t) = \langle t^2,\, \ln(t+1)\rangle$ for $0 \le t \le 3$ . Set up the integral for the total distance it travels.

step 1 Find the velocity components

$x'(t) = 2t$ , $y'(t) = \frac{1}{t+1}$ .

step 2 Form the speed

$|\mathbf{v}(t)| = \sqrt{(2t)^2 + \left(\frac{1}{t+1}\right)^2} = \sqrt{4t^2 + \frac{1}{(t+1)^2}}$ .

step 3 Write the distance integral

\text{distance} = \int_0^3 \sqrt{4t^2 + \frac{1}{(t+1)^2}}\,dt.

step 4 Evaluate numerically

This integrand has no elementary antiderivative, so evaluate the definite integral on a calculator.

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 2021 (BC, style)1 marksSection I (multiple choice, calculator). A particle moves with velocity

\langle \cos t, \sin t\rangle

. Its speed at any time

t

is (A)

1

(B)

\cos t + \sin t

(C)

\sqrt{2}

(D)

0

Show worked answer →

The correct answer is (A), $1$ .

Speed $= \sqrt{\cos^2 t + \sin^2 t} = \sqrt{1} = 1$ for all $t$ . The particle moves at constant unit speed.

AP 2024 (BC, style)4 marksSection II (free response, calculator). A particle moves with velocity

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

for

0 \le t \le 3

, starting at

\mathbf{r}(0) = \langle 1, 1\rangle

. (a) Find the position at

t = 3

. (b) Set up the integral for the total distance traveled.

Show worked answer →

A 4-point planar-motion problem.

(a) (2 points) $x(3) = 1 + \int_0^3 (t-2)\,dt = 1 + \left[\frac{t^2}{2} - 2t\right]_0^3 = 1 + \left(\frac{9}{2} - 6\right) = 1 - \frac{3}{2} = -\frac{1}{2}$ . $y(3) = 1 + \int_0^3 (t^2 - 4)\,dt = 1 + \left[\frac{t^3}{3} - 4t\right]_0^3 = 1 + (9 - 12) = -2$ . Position $\langle -\frac{1}{2}, -2\rangle$ .
(b) (2 points) Distance $= \int_0^3\sqrt{(t-2)^2 + (t^2 - 4)^2}\,dt$ (evaluate numerically).

Related dot points

Sources & how we know this

AP Calculus AB and BC Course and Exam Description — College Board (2020)