Particle Motion

Particle Motion

Part 1 of 7 — Position, Velocity, and Acceleration

Topic Overview

The Motion Hierarchy

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

Worked Example

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

$v(t) = s'(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3)$ $a(t) = v'(t) = 6t - 12 = 6(t-2)$

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
• At rest: $v(t) = 0$
• Velocity has sign (direction); speed is $|v(t)|$

Particle Motion

Part 2 of 7 — Speed & Direction of Motion

Speed vs. Velocity

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

Direction of Motion

Speeding Up vs. Slowing Down

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

Worked Example

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

Particle Motion

Part 3 of 7 — Displacement vs. Total Distance

Two Different Quantities

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

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

Worked Example

$v(t) = t^2 - 4$ 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

Particle Motion

Part 4 of 7 — Position from Velocity (Integration)

Finding Position from Velocity

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

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

Worked Example

$v(t) = 3t^2 - 2$ 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

Particle Motion

Part 5 of 7 — Acceleration & Velocity from Integrals

Finding Velocity from Acceleration

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

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

Worked Example

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

Particle Motion

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

AP FRQ Patterns for Particle Motion

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

Complete Worked FRQ

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

Particle Motion

Part 7 of 7 — Comprehensive Assessment

Formula Reference

Common AP Mistakes

Assessment — Set 1 🎯

Assessment — Set 2 🎯

Complete the analysis. 🔍

Final challenge. ✍️

Particle Motion — Complete! 🎓

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