🎯⭐ INTERACTIVE LESSON

Particle Motion

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Particle Motion - Complete Interactive Lesson

Part 1: Position, Velocity, Acceleration

Particle Motion

Part 1 of 7 — Position, Velocity, and Acceleration

Topic Overview

Part	Topic
1	Position, velocity, acceleration
2	Speed & direction of motion
3	Displacement vs. total distance
4	Position from velocity (integration)
5	Acceleration & velocity from integrals
6	AP-style workshop
7	Comprehensive assessment

The Motion Hierarchy

$\boxed{s(t) \xrightarrow{\text{derivative}} v(t) \xrightarrow{\text{derivative}} a(t)}$

Function	Symbol	Relationship
Position	$s(t)$ or $x(t)$	Given or found by integrating $v$
Velocity	$v (t) = s^{'} (t)$

Worked Example

$s(t) = t^3 - 6t^2 + 9t + 2$ . Find and .

$v(t) = s'(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3)$

Key Fact: Velocity is signed (direction matters). Speed is $|v(t)|$ (always non-negative).

Practice — Finding v and a 🎯

Connect the concepts. 🔍

Calculate. ✍️

Key Takeaways — Part 1

$v(t) = s'(t)$ : velocity is the derivative of position
$a(t) = v'(t) = s''(t)$ : acceleration is the derivative of velocity

Part 2: Displacement vs Total Distance

Particle Motion

Part 2 of 7 — Speed & Direction of Motion

Speed vs. Velocity

$\boxed{\text{Speed} = |v(t)|}$

Concept	Formula	Always positive?
Velocity

Part 3: Speed and Speeding Up/Slowing Down

Particle Motion

Part 3 of 7 — Displacement vs. Total Distance

Two Different Quantities

$\boxed{\text{Displacement} = \int_a^b v(t)\,dt = s(b) - s(a)}$

Part 4: Position from Velocity

Particle Motion

Part 4 of 7 — Position from Velocity (Integration)

Finding Position from Velocity

$\boxed{s(t) = s(t_0) + \int_{t_0}^{t} v(\tau)\,d\tau}$

Part 5: Velocity from Acceleration

Particle Motion

Part 5 of 7 — Acceleration & Velocity from Integrals

Finding Velocity from Acceleration

$\boxed{v(t) = v(t_0) + \int_{t_0}^{t} a(\tau)\,d\tau}$

Part 6: AP-Style Workshop

Particle Motion

Part 6 of 7 — AP-Style Free-Response Workshop

AP FRQ Patterns for Particle Motion

Part	Typical Prompt	Key Setup
(a)	"When is the particle at rest?"	Solve $v(t) = 0$
(b)	"Find total distance on $[a,b]$ "	$\int_a^b

Part 7: Final Assessment

Particle Motion

Part 7 of 7 — Comprehensive Assessment

Formula Reference

Relationship	Formula	Notes
Velocity from position	$v(t) = s'(t)$	Derivative
Acceleration from velocity

\boxed{a(t) \xrightarrow{\text{integral}} v(t) \xrightarrow{\text{integral}} s(t)}

a(t) = v'(t) = 6t - 12 = 6(t-2)

Velocity has sign (direction); speed is

|v(t)|

$v(t)$	Direction	Meaning
$v(t) > 0$	Moving right (or up)	Position increasing
$v(t) < 0$	Moving left (or down)	Position decreasing
$v(t) = 0$	At rest	Possibly changing direction

Speeding Up vs. Slowing Down

$\boxed{\text{Speeding up: } v \text{ and } a \text{ have the SAME sign}}$ $\boxed{\text{Slowing down: } v \text{ and } a \text{ have OPPOSITE signs}}$

$v(t)$	$a(t)$	Speed is...
$+$	$+$	Increasing
$-$	$-$	Increasing
$+$	$-$	Decreasing
$-$	$+$	Decreasing

$s(t) = t^3 - 6t^2 + 9t$ . When is the particle speeding up on $[0, 4]$ ?

$v(t) = 3(t-1)(t-3)$ , $a(t) = 6(t-2)$

$v > 0$ on $(0,1)$ and $(3,4)$ ; $v < 0$ on $(1,3)$
$a > 0$ on $(2,4)$ ; $a < 0$ on $(0,2)$

Same sign: $(1,2)$ (both negative) and $(3,4)$ (both positive).

Speeding up on $(1, 2) \cup (3, 4)$ .

Practice — Speed & Direction 🎯

Classify the motion. 🔍

Find the speed. ✍️

Key Takeaways — Part 2

Speed $= |v(t)|$ , always non-negative
$v > 0$ : right/up; $v < 0$ : left/down
Speeding up: $v$ and $a$ same sign
Slowing down: $v$ and $a$ opposite signs
Direction change: $v(t)$ changes sign

$\boxed{\text{Total Distance} = \int_a^b |v(t)|\,dt}$

Quantity	Formula	Sign	Meaning
Displacement	$\int_a^b v(t)\,dt$	Can be negative	Net change in position
Total distance	$\int_a^b	v(t)	,dt$

$v(t) = t^2 - 4$ on $[0, 3]$ . Find displacement and total distance.

Displacement: $\int_0^3 (t^2-4)\,dt = \left[\frac{t^3}{3}-4t\right]_0^3 = (9-12) - 0 = -3$

Total distance: $v(t) = 0$ at $t = 2$ . Split at $t = 2$ : $\int_0^2 |t^2-4|\,dt + \int_2^3 |t^2-4|\,dt = \int_0^2 (4-t^2)\,dt + \int_2^3 (t^2-4)\,dt$ $= \left[4t - \frac{t^3}{3}\right]_0^2 + \left[\frac{t^3}{3} - 4t\right]_2^3 = \frac{16}{3} + \frac{7}{3} = \frac{23}{3}$

AP Tip: Displacement can be negative (particle ends left of start). Total distance is always positive. AP loves asking for both in the same problem.

Practice — Displacement & Distance 🎯

Distinguish displacement and distance. 🔍

Calculate total distance. ✍️

Key Takeaways — Part 3

Displacement $= \int_a^b v(t)\,dt$ (net change, can be negative)
Total distance $= \int_a^b |v(t)|\,dt$ (always positive)
Split the integral at points where $v(t) = 0$
Displacement $= 0$ means the particle returned to start

This combines the initial condition $s(t_0)$ with the displacement $\int v\,dt$ .

$v(t) = 3t^2 - 2$ and $s(0) = 5$ . Find $s(t)$ .

$s(t) = s(0) + \int_0^t (3\tau^2 - 2)\,d\tau = 5 + [\tau^3 - 2\tau]_0^t = 5 + t^3 - 2t$

$\boxed{s(t) = t^3 - 2t + 5}$

Position at a Specific Time

$v(t) = 6t - 4$ , $s(1) = 3$ . Find $s(4)$ .

$s(4) = s(1) + \int_1^4 (6t-4)\,dt = 3 + [3t^2 - 4t]_1^4$ $= 3 + (48-16) - (3-4) = 3 + 32 + 1 = 36$

Key Fact: You don't need to find $s(t)$ as a formula — just compute the definite integral and add the initial position.

Practice — Position from Velocity 🎯

Build the position function. 🔍

Find position. ✍️

Key Takeaways — Part 4

$s(t) = s(t_0) + \int_{t_0}^t v(\tau)\,d\tau$
Initial condition + displacement gives position
You can find $s$ at a specific time without finding $s(t)$ as a formula
AP FRQs often give $v(t)$ and initial position, ask for $s$ at another time

Full Chain: $a \to v \to s$

Given	To Find	Formula
$a(t)$ and $v(t_0)$	$v(t)$	$v(t_0) + \int_{t_0}^t a\,d\tau$
$v(t)$ and $s(t_0)$	$s(t)$	$s (_{}$
$a(t)$ , $v(t_0)$ , $s(t_0)$

$a(t) = 6t$ , $v(0) = -4$ , $s(0) = 1$ . Find $s(t)$ .

Step 1: $v(t) = -4 + \int_0^t 6\tau\,d\tau = -4 + 3t^2$

Step 2: $s(t) = 1 + \int_0^t (-4 + 3\tau^2)\,d\tau = 1 - 4t + t^3$

$\boxed{s(t) = t^3 - 4t + 1}$

When Does the Particle Change Direction?

$v(t) = 3t^2 - 4 = 0$ at $t = 2/\sqrt{3}$ . Check sign change: $v(0) = -4 < 0$ , $v(2) = 8 > 0$ . Direction changes at $t = 2/\sqrt{3}$ .

Practice — Integration 🎯

Build from acceleration. 🔍

Find velocity. ✍️

Key Takeaways — Part 5

$v(t) = v(t_0) + \int_{t_0}^t a\,d\tau$
Integrate twice to go from $a$ to $s$
Each integration adds a constant (initial condition)
Direction changes when $v(t)$ changes sign

AP Tip: In table-based problems, use the given values with Riemann sums or trapezoidal approximations.

A particle moves along the $x$ -axis with velocity $v(t) = t^2 - 5t + 4$ for $t \ge 0$ and position $s(0) = 2$ .

(a) When is the particle at rest?

$v(t) = t^2 - 5t + 4 = (t-1)(t-4) = 0$ at $t = 1$ and $t = 4$ .

(b) Total distance traveled on $[0,6]$

Sign analysis: $v > 0$ on $[0,1)$ , $v < 0$ on $(1,4)$ , $v > 0$ on $(4,6]$ .

$\int_0^1 (t^2-5t+4)\,dt = \left[\frac{t^3}{3} - \frac{5t^2}{2} + 4t\right]_0^1 = \frac{1}{3} - \frac{5}{2} + 4 = \frac{11}{6}$

$\int_1^4 (t^2-5t+4)\,dt = \left[\frac{t^3}{3} - \frac{5t^2}{2} + 4t\right]_1^4 = \left(\frac{64}{3} - 40 + 16\right) - \left(\frac{1}{3} - \frac{5}{2} + 4\right)$

$= \frac{64}{3} - 24 - \frac{11}{6} = \frac{128}{6} - \frac{144}{6} - \frac{11}{6} = -\frac{27}{6} = -\frac{9}{2}$

$\int_4^6 (t^2-5t+4)\,dt = \left[\frac{t^3}{3} - \frac{5t^2}{2} + 4t\right]_4^6 = \left(72 - 90 + 24\right) - \left(\frac{64}{3} - 40 + 16\right)$

$= 6 - \frac{64}{3} + 24 = 30 - \frac{64}{3} = \frac{26}{3}$

$\text{Total distance} = \frac{11}{6} + \frac{9}{2} + \frac{26}{3} = \frac{11}{6} + \frac{27}{6} + \frac{52}{6} = \frac{90}{6} = 15$

(c) Is speed increasing or decreasing at $t = 3$ ?

$v(3) = 9 - 15 + 4 = -2 < 0$ and $a(3) = 2(3) - 5 = 1 > 0$

Opposite signs $\Rightarrow$ speed is decreasing at $t = 3$ .

(d) Find $s(6)$ .

$s(6) = 2 + \int_0^6 (t^2-5t+4)\,dt = 2 + \left[\frac{t^3}{3} - \frac{5t^2}{2} + 4t\right]_0^6 = 2 + 72 - 90 + 24 = 8$

AP-style questions. 🎯

Analyze motion. 🔍

AP FRQ practice. ✍️

Key Takeaways — Part 6

AP FRQs test all aspects: rest, direction, distance, position
Always split integrals at $v(t) = 0$ for total distance
Speed increasing $\iff$ $v$ and $a$ same sign
Show every step and justify sign changes for full credit

a (t) = v^{'} (t) = s^{''} (t)

Mistake	Correction
Confusing displacement with total distance	Displacement is signed; total distance splits at $v = 0$
Forgetting to check sign of $v$ for direction	Always state direction (left/right or positive/negative)
Using $a > 0$ means "speeding up"	Speed increases only when $v$ and $a$ have the same sign
Missing initial conditions	Every antiderivative needs $+C$ or initial value
Not justifying sign changes	AP requires explicit sign analysis for direction change

Assessment — Set 1 🎯

Assessment — Set 2 🎯

Complete the analysis. 🔍

Final challenge. ✍️

Particle Motion — Complete! 🎓

Part	Topic	Status
1	Position, Velocity & Acceleration	✅
2	Speed & Direction of Motion	✅
3	Displacement vs. Total Distance	✅
4	Position from Velocity	✅
5	Acceleration & Velocity from Integrals	✅
6	AP-Style Free-Response Workshop	✅
7	Comprehensive Assessment	✅

You have completed the full Particle Motion unit. You should now be confident with all AP Calculus AB particle motion problems!

s (t_{0}) + \int_{t_{0}}^{t} v d τ

Particle Motion

Direction of Motion

Speeding Up vs. Slowing Down

Worked Example

Key Takeaways — Part 2

Worked Example

Key Takeaways — Part 3

Worked Example

Position at a Specific Time

Key Takeaways — Part 4

Full Chain: $a \to v \to s$

Worked Example

When Does the Particle Change Direction?

Key Takeaways — Part 5

Complete Worked FRQ

Key Takeaways — Part 6

Common AP Mistakes

Particle Motion — Complete! 🎓

Particle Motion

Particle Motion - Complete Interactive Lesson

Part 1: Position, Velocity, Acceleration

Particle Motion

Topic Overview

The Motion Hierarchy

Worked Example

Key Takeaways — Part 1

Part 2: Displacement vs Total Distance

Particle Motion

Speed vs. Velocity

Part 3: Speed and Speeding Up/Slowing Down

Particle Motion

Two Different Quantities

Part 4: Position from Velocity

Particle Motion

Finding Position from Velocity

Part 5: Velocity from Acceleration

Particle Motion

Finding Velocity from Acceleration

Part 6: AP-Style Workshop

Particle Motion

AP FRQ Patterns for Particle Motion

Part 7: Final Assessment

Particle Motion

Formula Reference

Direction of Motion

Speeding Up vs. Slowing Down

Worked Example

Key Takeaways — Part 2

Worked Example

Key Takeaways — Part 3

Worked Example

Position at a Specific Time

Key Takeaways — Part 4

Full Chain: a→v→sa \to v \to sa→v→s

Worked Example

When Does the Particle Change Direction?

Key Takeaways — Part 5

Complete Worked FRQ

Key Takeaways — Part 6

Common AP Mistakes

Particle Motion — Complete! 🎓

Full Chain: $a \to v \to s$