AP Calculus BC — Unit 9.6
Solving Motion Problems using Parametric and Vector-Valued Functions
Key Concepts Packet + Worked Examples + Interactive MCQ Practice (36 Questions)
Table of Contents
- Position, Velocity, Acceleration (Vector View)
- Speed, Distance Traveled, and Displacement
- Resting, Moving Forward/Backward, and Turning Points
- Average vs. Instantaneous Velocity
- Distance Traveled from Speed (and When It Works)
- Calculator Expectations + Common AP Traps
- Unit 9.6 Multiple-Choice Practice (36 Questions)
- Answer Key
Unit 9.6 Key Concepts Packet
1) Position, Velocity, Acceleration (Vector View)
Core definitions
If \( \mathbf{r}(t) \) is position, then
\( \mathbf{v}(t)=\mathbf{r}'(t) \) (velocity),
\( \mathbf{a}(t)=\mathbf{v}'(t)=\mathbf{r}''(t) \) (acceleration).
- Velocity points in the direction of motion.
- Acceleration measures how velocity changes (speeding up, slowing down, changing direction).
2) Speed, Distance Traveled, and Displacement
Key quantities
Speed: \( \|\mathbf{v}(t)\| \)
Distance traveled on \([a,b]\): \( \displaystyle \int_a^b \|\mathbf{v}(t)\|\,dt \)
Displacement: \( \mathbf{r}(b)-\mathbf{r}(a) \)
Distance traveled on \([a,b]\): \( \displaystyle \int_a^b \|\mathbf{v}(t)\|\,dt \)
Displacement: \( \mathbf{r}(b)-\mathbf{r}(a) \)
- Distance traveled is always nonnegative (it accumulates motion).
- Displacement is a vector (it only compares start and end).
- In 1D, if \(x(t)\) is position, distance traveled is \( \int_a^b |x'(t)|dt \).
3) Resting, Moving Forward/Backward, and Turning Points
Most common AP motion checks
At rest when \( \mathbf{v}(t)=\mathbf{0} \) (or in 1D, \(x'(t)=0\)).
In 1D, moving right when \(x'(t)>0\), moving left when \(x'(t)<0\).
In 1D, moving right when \(x'(t)>0\), moving left when \(x'(t)<0\).
- A turning point (1D) happens when velocity changes sign (not just when it equals zero).
- In vectors, “changing direction” can mean the velocity vector changes direction even if speed is constant.
4) Average vs. Instantaneous Velocity
Vector average velocity
Average velocity on \([a,b]\):
\( \displaystyle \mathbf{v}_{avg}=\frac{\mathbf{r}(b)-\mathbf{r}(a)}{b-a} \).
Instantaneous velocity is \( \mathbf{v}(t)=\mathbf{r}'(t) \).
- Average speed is generally \( \frac{\text{distance traveled}}{b-a} \).
- Average velocity uses displacement (can be zero even if the particle moved a lot).
5) Distance Traveled from Speed (and When It Works)
Distance traveled (vector motion)
\( \displaystyle \text{Distance}=\int_a^b \|\mathbf{r}'(t)\|\,dt \)
where
\( \|\mathbf{r}'(t)\|=\sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2} \).
- Do not confuse this with \( \left\|\int_a^b \mathbf{r}'(t)\,dt\right\| \), which is magnitude of displacement.
- Distance traveled is an integral of a magnitude (so it accumulates all movement).
6) Calculator Expectations + Common AP Traps
- AP commonly asks for velocity, speed, acceleration, distance traveled, and displacement.
- Trap: Reporting \( \mathbf{r}'(t) \) when the question asks for \( \|\mathbf{r}'(t)\| \).
- Trap: Using displacement when distance traveled is required.
- Important: If you ever see raw backslashes instead of formatted math, it means MathJax did not typeset that section—this page forces typesetting on load and when solutions open.
Worked Examples + Notes
Example 1 — Average velocity vs. average speed (1D)
A particle has position \(x(t)=t^3-6t^2+9t\) on \([0,4]\). Find (i) average velocity and (ii) distance traveled on \([0,4]\).
Step 1: Compute displacement: \(x(4)-x(0)=(64-96+36)-0=4\).
Step 2: Average velocity: \( \frac{4}{4-0}=1 \).
Step 3: Velocity: \(x'(t)=3t^2-12t+9=3(t-1)(t-3)\).
Step 4: Speed is \( |x'(t)| \). Break at sign changes \(t=1,3\), then compute \( \int_0^4 |x'(t)|dt \) piecewise.
Conclusion: Average velocity uses displacement; distance traveled uses \( \int |x'(t)|dt \).
Example 2 — Vector distance traveled
A particle has position \( \mathbf{r}(t)=\langle t,\ t^2\rangle \) on \([0,2]\). Find the distance traveled.
Step 1: Velocity \( \mathbf{r}'(t)=\langle 1,\ 2t\rangle \).
Step 2: Speed \( \|\mathbf{r}'(t)\|=\sqrt{1^2+(2t)^2}=\sqrt{1+4t^2} \).
Step 3: Distance traveled \( \displaystyle \int_0^2 \sqrt{1+4t^2}\,dt \).
Conclusion: Distance traveled is \( \int \|\mathbf{r}'(t)\|dt \), not the magnitude of displacement.
Example 3 — When is the particle at rest?
Let \( \mathbf{r}(t)=\langle t^2-4t,\ t^3-3t\rangle \). Find times when the particle is at rest.
Step 1: Velocity \( \mathbf{v}(t)=\mathbf{r}'(t)=\langle 2t-4,\ 3t^2-3\rangle \).
Step 2: At rest means \( \mathbf{v}(t)=\mathbf{0} \), so solve \(2t-4=0\Rightarrow t=2\) and \(3t^2-3=0\Rightarrow t=\pm 1\).
Conclusion: No common solution, so the particle is never at rest.
Example 4 — Speed vs. velocity at a time
Suppose \( \mathbf{v}(2)=\langle -3,\ 4\rangle \). Find the speed at \(t=2\).
Step 1: Speed is the magnitude: \( \|\mathbf{v}(2)\|=\sqrt{(-3)^2+4^2}=5 \).
Conclusion: Speed is \(5\).
Note: This page forces MathJax typesetting on load and when solution panels open.
Unit 9.6 Multiple-Choice Practice (36 Questions)
Answer Key