🎯⭐ 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, Acceleration

The Relationships

Function	Symbol	Relationship
Position	$s(t)$	Given or found by integrating $v$
Velocity	$v(t)$	$v(t) = s'(t)$
Acceleration	$a(t)$	$a(t) = v'(t) = s''(t)$

Key Interpretations

Speed $= |v(t)|$ (always positive)
Particle moves right when $v(t) > 0$
Particle moves left when $v(t) < 0$

Position, Velocity, Acceleration 🎯

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

Key Takeaways — Part 1

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

Part 2: Displacement vs Total Distance

Particle Motion

Part 2 of 7 — Displacement vs Total Distance

Displacement (Net Change)

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

Part 3: Speed and Speeding Up/Slowing Down

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 $\frac{d}{dt}|v(t)| = 0$

Part 4: Position from Velocity

Particle Motion

Part 4 of 7 — Position from Velocity

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

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

Part 5: Velocity from Acceleration

Particle Motion

Part 5 of 7 — Velocity from Acceleration

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

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

Part 6: AP-Style Workshop

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!

Part 7: Final Assessment

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

Particle speeds up when

v

and

a

have the same sign

Particle slows down when

v

and

a

have opposite signs

Total Distance Traveled

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

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

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

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

Speeding Up vs Slowing Down

$v(t)$	$a(t)$	Particle is...
$+$	$+$	Speeding up (moving right, accelerating right)
$-$	$-$	Speeding up (moving left, accelerating left)
$+$	$-$	Slowing down
$-$	$+$	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

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

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

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

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

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

Acceleration to Velocity 🎯

Key Takeaways — Part 5

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

Particle Motion

Total Distance Traveled

Worked Example

Key Takeaways — Part 2

Speeding Up vs Slowing Down

Key Takeaways — Part 3

Worked Example

Key Takeaways — Part 4

Free Fall Model

Key Takeaways — Part 5

Particle Motion

Particle Motion - Complete Interactive Lesson

Part 1: Position, Velocity, Acceleration

Particle Motion

The Relationships

Key Interpretations

Key Takeaways — Part 1

Part 2: Displacement vs Total Distance

Particle Motion

Displacement (Net Change)

Part 3: Speed and Speeding Up/Slowing Down

Particle Motion

Speed vs Velocity

Part 4: Position from Velocity

Particle Motion

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

Part 5: Velocity from Acceleration

Particle Motion

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

Part 6: AP-Style Workshop

Particle Motion

Workshop Complete!

Part 7: Final Assessment

Particle Motion — Review

Particle Motion — Complete! ✅

Total Distance Traveled

Worked Example

Key Takeaways — Part 2

Speeding Up vs Slowing Down

Key Takeaways — Part 3

Worked Example

Key Takeaways — Part 4

Free Fall Model

Key Takeaways — Part 5

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

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