Particle Motion

Part 1 of 7 — Position, Velocity, Acceleration

The Relationships

Key Interpretations

• Speed $= |v(t)|$ (always positive)
• Particle moves right when $v(t) > 0$
• Particle moves left when $v(t) < 0$
• Particle speeds up when $v$ and $a$ have the same sign
• Particle slows down when $v$ and $a$ have opposite signs

Position, Velocity, Acceleration 🎯

A particle moves along a line with $s(t) = t^3 - 6t^2 + 9t + 1$.

Key Takeaways — Part 1

1. Differentiate position to get velocity, differentiate again for acceleration
2. "At rest" means $v(t) = 0$
3. Direction of motion determined by sign of $v(t)$

Particle Motion

Part 2 of 7 — Displacement vs Total Distance

Displacement (Net Change)

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

Total Distance Traveled

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

Split the integral where $v(t) = 0$ (direction changes).

Worked Example

$v(t) = t^2 - 4$ on $[0, 3]$.

$v = 0$ at $t = 2$.

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

Total distance: $\int_0^2 |t^2-4|\,dt + \int_2^3 |t^2-4|\,dt$ $= \int_0^2 (4-t^2)\,dt + \int_2^3 (t^2-4)\,dt = \frac{16}{3} + \frac{7}{3} = \frac{23}{3}$

Displacement vs Distance 🎯

Key Takeaways — Part 2

1. Displacement can be negative (net change in position)
2. Total distance is always positive (use $|v(t)|$)
3. Split at points where $v(t) = 0$ (direction changes)

Particle Motion

Part 3 of 7 — Speed and Speeding Up/Slowing Down

Speed vs Velocity

• Velocity has direction (can be negative)
• Speed $= |v(t)|$ (always non-negative)
• Maximum speed occurs at an endpoint or where $\frac{d}{dt}|v(t)| = 0$

Speeding Up vs Slowing Down

Key: Same sign = speeding up. Opposite signs = slowing down.

Speed Analysis 🎯

$v(t) = t^2 - 5t + 4 = (t-1)(t-4)$, $a(t) = 2t - 5$

Key Takeaways — Part 3

1. Speed = $|v(t)|$
2. Same sign of $v$ and $a$ = speeding up
3. Opposite signs = slowing down

Particle Motion

Part 4 of 7 — Position from Velocity

Given $v(t)$, Find $s(t)$

$s(t) = s(0) + \int_0^t v(\tau)\,d\tau$

Worked Example

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

$s(2) = 5 + \int_0^2 (3t^2 - 2)\,dt = 5 + [t^3 - 2t]_0^2 = 5 + (8-4) = 9$

Finding Position 🎯

Key Takeaways — Part 4

1. Position = initial position + displacement
2. Maximum/minimum position occurs when $v = 0$ (direction change)

Particle Motion

Part 5 of 7 — Velocity from Acceleration

Given $a(t)$, Find $v(t)$

$v(t) = v(0) + \int_0^t a(\tau)\,d\tau$

Free Fall Model

Near Earth's surface: $a(t) = -g = -9.8$ m/s$^2$ (or $-32$ ft/s$^2$)

$v(t) = v_0 - gt$

$s(t) = s_0 + v_0 t - \frac{1}{2}gt^2$

Acceleration to Velocity 🎯

Key Takeaways — Part 5

1. Integrate acceleration to find velocity
2. Free fall: $a = -g$ (constant), giving quadratic position

Particle Motion

Part 6 of 7 — AP-Style Workshop

AP-Style Motion Problems 🎯

A particle moves with $v(t) = 2\sin(t)$ for $t \geq 0$.

Workshop Complete!

Particle Motion — Review

Part 7 of 7 — Final Assessment

Final Assessment 🎯

Particle Motion — Complete! ✅

You have mastered:

• ✅ Position, velocity, acceleration relationships
• ✅ Displacement vs total distance
• ✅ Speed analysis (speeding up/slowing down)
• ✅ Finding position from velocity and acceleration