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Rectilinear Motion

Reference: Stewart §2.7

What Can Calculus Tell Us About Moving Objects?

A particle moves along a line. You know its position at every moment in time: $s(t)$. From this single function, calculus can tell you:

How fast is it moving? (velocity)
Is it speeding up or slowing down? (speed analysis)
Which direction is it going? (sign of velocity)
When does it stop and turn around? (when $v = 0$)
How far does it travel in total? (not the same as displacement!)

This is rectilinear motion: motion along a straight line. The derivative transforms position into velocity, and the derivative of velocity gives acceleration.

Prerequisite Map

Quick Reference

Property	Value
Concept	Rates of Change
Chapter	2.7
Difficulty	Intermediate
Time	~25 minutes

Key Concepts

The Motion Hierarchy

Quantity	Symbol	Meaning	How to Find
Position	$s(t)$	Where the particle is	Given
Velocity	$v(t)$	Rate of change of position	$v(t) = \frac{ds}{dt} = s'(t)$
Acceleration	$a(t)$	Rate of change of velocity	$a(t) = \frac{dv}{dt} = s''(t)$

Each quantity is the derivative of the one above it:

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

Interpreting Velocity

The velocity $v(t) = s'(t)$ tells you:

Condition	Meaning
$v(t) > 0$	Moving in the positive direction (right/up)
$v(t) < 0$	Moving in the negative direction (left/down)
$v(t) = 0$	Particle is at rest (momentarily stopped)

The speed is $\vert v(t)\vert $, which is velocity without the direction.

When Does It Change Direction?

A particle changes direction when velocity changes sign. This happens when:

$v(t) = 0$ (the particle stops momentarily), AND
$v(t)$ changes from positive to negative (or vice versa)

    s(t)
      │       ╭──╮
      │      ╱    ╲
      │     ╱      ╲    ← Maximum: v = 0, changing direction
      │    ╱        ╲
      │   ╱          ╲
      └──────────────────── t
           ↑
       v > 0      v < 0
       (moving    (moving
        right)     left)

Speeding Up vs. Slowing Down

This is the trickiest concept! The particle is:

Condition	Behavior	Why?
$v$ and $a$ have same sign	Speeding up	Acceleration pushes in direction of motion
$v$ and $a$ have opposite signs	Slowing down	Acceleration opposes motion

Example:

$v = 5$, $a = 2$: Both positive → speeding up (going right, accelerating right)
$v = -5$, $a = -2$: Both negative → speeding up (going left, accelerating left)
$v = 5$, $a = -2$: Opposite signs → slowing down (going right, braking)
$v = -5$, $a = 2$: Opposite signs → slowing down (going left, braking)

Total Distance vs. Displacement

Displacement = final position − initial position = $s(b) - s(a)$

Total distance = sum of all distances traveled, counting every meter

If a particle goes right 4 meters, then left 3 meters:

Displacement = $4 - 3 = 1$ meter (net change)
Total distance = $4 + 3 = 7$ meters (odometer reading)

To find total distance:

Find when $v(t) = 0$ (direction changes)
Calculate $\vert s(t_i) - s(t_{i-1})\vert $ for each interval
Add them up

Practice Problems

Level 1 Finding Velocity and Acceleration

A particle moves along a line with position $s(t) = t^3 - 6t^2 + 9t$ meters, where $t \geq 0$ is in seconds.

(a) Find the velocity function $v(t)$.

(b) Find the acceleration function $a(t)$.

Thought Process

Velocity is the derivative of position. Acceleration is the derivative of velocity.

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

$v(t) = s'(t)$: use power rule on each term
$a(t) = v'(t)$: differentiate again

Show Answer

(a) Velocity: $$v(t) = s'(t) = \frac{d}{dt}(t^3 - 6t^2 + 9t) = 3t^2 - 12t + 9$$

(b) Acceleration: $$a(t) = v'(t) = \frac{d}{dt}(3t^2 - 12t + 9) = 6t - 12$$

So $\boxed{v(t) = 3t^2 - 12t + 9}$ and $\boxed{a(t) = 6t - 12}$.

Level 2 When Is the Particle at Rest?

Using $v(t) = 3t^2 - 12t + 9$ from Level 1:

(a) When is the particle at rest?

(b) Find the position of the particle at each rest time.

Thought Process

(a) At rest means $v(t) = 0$. Solve $3t^2 - 12t + 9 = 0$.

Factor: $3(t^2 - 4t + 3) = 3(t-1)(t-3) = 0$

(b) Plug the rest times back into $s(t) = t^3 - 6t^2 + 9t$.

Show Answer

(a) Set $v(t) = 0$: $$3t^2 - 12t + 9 = 0$$ $$3(t^2 - 4t + 3) = 0$$ $$3(t - 1)(t - 3) = 0$$ $$t = 1 \text{ or } t = 3$$

The particle is at rest at $\boxed{t = 1 \text{ s and } t = 3 \text{ s}}$.

(b) Positions at rest times: $$s(1) = 1 - 6 + 9 = 4 \text{ m}$$ $$s(3) = 27 - 54 + 27 = 0 \text{ m}$$

At $t = 1$, the particle is at position $4$ m. At $t = 3$, the particle is back at the origin.

Level 3 Direction and Speed Analysis

For the particle with $s(t) = t^3 - 6t^2 + 9t$, $v(t) = 3t^2 - 12t + 9$, $a(t) = 6t - 12$:

(a) When is the particle moving in the positive direction? Negative direction?

(b) When is the particle speeding up? Slowing down?

Thought Process

(a) Positive direction when $v(t) > 0$. Use the factored form $v(t) = 3(t-1)(t-3)$ and sign analysis.

(b) Speeding up when $v$ and $a$ have the same sign. Find when $a = 0$ (at $t = 2$), then check signs in each interval.

Show Answer

(a) We have $v(t) = 3(t-1)(t-3)$.

Sign analysis:

For $0 \leq t < 1$: $(t-1) < 0$, $(t-3) < 0$ → $v(t) > 0$ (positive × negative × negative)
For $1 < t < 3$: $(t-1) > 0$, $(t-3) < 0$ → $v(t) < 0$
For $t > 3$: $(t-1) > 0$, $(t-3) > 0$ → $v(t) > 0$

Positive direction: $0 \leq t < 1$ and $t > 3$ Negative direction: $1 < t < 3$

(b) We have $a(t) = 6t - 12 = 6(t - 2)$, so $a = 0$ at $t = 2$.

For $t < 2$: $a(t) < 0$
For $t > 2$: $a(t) > 0$

Now compare signs of $v$ and $a$:

Interval	$v$	$a$	Same sign?	Behavior
$0 \leq t < 1$	$+$	$-$	No	Slowing down
$1 < t < 2$	$-$	$-$	Yes	Speeding up
$2 < t < 3$	$-$	$+$	No	Slowing down
$t > 3$	$+$	$+$	Yes	Speeding up

Speeding up: $1 < t < 2$ and $t > 3$ Slowing down: $0 \leq t < 1$ and $2 < t < 3$

Level 4 Total Distance Traveled

Find the total distance traveled by the particle with $s(t) = t^3 - 6t^2 + 9t$ during the first 5 seconds.

Thought Process

The particle changes direction at $t = 1$ and $t = 3$.

Find $s(0)$, $s(1)$, $s(3)$, $s(5)$
Calculate distance in each interval where motion is one direction
Add up the absolute values

Show Answer

Position values: $$s(0) = 0$$ $$s(1) = 1 - 6 + 9 = 4$$ $$s(3) = 27 - 54 + 27 = 0$$ $$s(5) = 125 - 150 + 45 = 20$$

Distance traveled in each interval:

$[0, 1]$: Moving positive (right), from $s = 0$ to $s = 4$ $$d_1 = \vert s(1) - s(0)\vert = \vert 4 - 0\vert = 4 \text{ m}$$

$[1, 3]$: Moving negative (left), from $s = 4$ to $s = 0$ $$d_2 = \vert s(3) - s(1)\vert = \vert 0 - 4\vert = 4 \text{ m}$$

$[3, 5]$: Moving positive (right), from $s = 0$ to $s = 20$ $$d_3 = \vert s(5) - s(3)\vert = \vert 20 - 0\vert = 20 \text{ m}$$

Total distance: $$d_1 + d_2 + d_3 = 4 + 4 + 20 = \boxed{28 \text{ m}}$$

Note: The displacement is only $s(5) - s(0) = 20$ m, but the total distance is 28 m because the particle backtracked.

Level 5 Complete Motion Analysis

A ball is thrown upward with position $s(t) = -5t^2 + 30t + 10$ meters, where $t \geq 0$.

(a) Find velocity and acceleration as functions of time.

(b) What is the maximum height reached?

(d) With what speed does it hit the ground?

(e) Describe the motion: when is the ball going up? Down? Speeding up? Slowing down?

Thought Process

(a) Differentiate $s(t)$ for velocity; differentiate again for acceleration.

(b) Maximum height occurs when $v(t) = 0$. Find that time, then evaluate $s(t)$.

(d) Speed at impact = $\vert v(t)\vert $ where $t$ is the impact time.

(e) Going up when $v > 0$; down when $v < 0$. For speeding up/slowing down, compare signs of $v$ and $a$.

Show Answer

(a) Velocity: $$v(t) = s'(t) = -10t + 30$$

Acceleration: $$a(t) = v'(t) = -10 \text{ m/s}^2$$

(b) Maximum height when $v(t) = 0$: $$-10t + 30 = 0 \implies t = 3 \text{ s}$$

$$s(3) = -5(9) + 30(3) + 10 = -45 + 90 + 10 = 55 \text{ m}$$

Maximum height: $\boxed{55 \text{ m}}$

(c) Ground when $s(t) = 0$: $$-5t^2 + 30t + 10 = 0$$ $$5t^2 - 30t - 10 = 0$$ $$t^2 - 6t - 2 = 0$$ $$t = \frac{6 \pm \sqrt{36 + 8}}{2} = \frac{6 \pm \sqrt{44}}{2} = 3 \pm \sqrt{11}$$

Since $t \geq 0$, we take $t = 3 + \sqrt{11} \approx 6.32$ s.

Impact time: $\boxed{t = 3 + \sqrt{11} \approx 6.32 \text{ s}}$

(d) Velocity at impact: $$v(3 + \sqrt{11}) = -10(3 + \sqrt{11}) + 30 = -10\sqrt{11} \approx -33.2 \text{ m/s}$$

Impact speed: $\boxed{10\sqrt{11} \approx 33.2 \text{ m/s}}$

(e) Motion description:

Since $a = -10 < 0$ always (constant downward acceleration):

Going up: when $v > 0$, i.e., $0 \leq t < 3$
Going down: when $v < 0$, i.e., $t > 3$
Slowing down: when $v$ and $a$ have opposite signs, i.e., $0 \leq t < 3$ (going up while gravity pulls down)
Speeding up: when $v$ and $a$ have the same sign, i.e., $t > 3$ (falling with gravity)

Time Interval	Direction	Speed
$0 \leq t < 3$	Up	Slowing down
$t = 3$	At rest (peak)	---
$t > 3$	Down	Speeding up

CCI-Style Conceptual Questions

Level 3 Conceptual: Interpreting Graphs

The graph shows the velocity $v(t)$ of a particle:

   v(t)
    │
  2 │──────╲
    │       ╲
  0 │────────╲────────
    │         ╲      t
 -2 │          ╲────
    │
    └──────────────────
        1    2    3

At $t = 2$, the particle is:

(A) At rest and speeding up (B) Moving left and speeding up (C) Moving left and slowing down (D) Moving right and slowing down

Thought Process

At $t = 2$:

Read $v(2)$ from the graph: it's negative (below the axis)
The slope of $v$ is the acceleration: the line is decreasing, so $a < 0$
Negative $v$ means moving left
Negative $v$ and negative $a$: same sign → speeding up

Show Answer

(B) Moving left and speeding up

At $t = 2$: $v < 0$ (moving left) and $a = v' < 0$ (slope is negative).

Same sign means speeding up. The particle is moving left and getting faster.

Level 3 Conceptual: Distance vs. Displacement

A particle starts at position $s = 0$, moves to $s = 10$, then back to $s = 3$.

Which statement is correct?

(A) Displacement = 13, Distance = 3 (B) Displacement = 3, Distance = 13 (C) Displacement = 3, Distance = 17 (D) Displacement = 17, Distance = 3

Thought Process

Displacement = final position − initial position = where you ended up relative to where you started
Distance = total path length = how far you actually walked

Show Answer

(C) Displacement = 3, Distance = 17

Displacement = $3 - 0 = 3$ (ended 3 units from start)
Distance = $10 + 7 = 17$ (went 10 right, then 7 left)

Mastery Checklist

[ ] I can find velocity from position by differentiation: $v = ds/dt$
[ ] I can find acceleration from velocity: $a = dv/dt$
[ ] I know that $v(t) = 0$ means the particle is at rest
[ ] I can determine direction of motion from the sign of velocity
[ ] I understand speeding up vs. slowing down depends on whether $v$ and $a$ have the same sign
[ ] I can find total distance by breaking the motion into intervals where direction doesn't change
[ ] I can distinguish between displacement (net change) and total distance (odometer)

Mental Model

The Car Dashboard:

Your car has three gauges that mirror $s$, $v$, and $a$:

Odometer/GPS: Position $s(t)$, where you are
Speedometer: Speed $\vert v(t)\vert $, how fast you're going (always positive)
Feeling of push/pull: Acceleration $a(t)$, whether you are being pushed into your seat (speeding up) or toward the dashboard (braking)

When you press the gas while moving forward, $v > 0$ and $a > 0$ → speeding up. When you brake while moving forward, $v > 0$ and $a < 0$ → slowing down.

The key insight: "speeding up" means your speed is increasing, which happens when velocity and acceleration point the same way.

Connections

Looking back:

Average vs. Instantaneous Rate established that derivatives measure instantaneous rates
Instantaneous Velocity introduced velocity as a limit

Looking ahead:

Related Rates extends these ideas to connected quantities
Optimization finds when things are biggest or smallest
Integrals (Chapter 4) reverse this: given velocity, find position

The Big Picture: Rectilinear motion is the prototype for all rate-of-change problems. The relationship $s \to v \to a$ (differentiate to go forward) becomes the foundation for understanding derivatives in every context.

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Average vs. Instantaneous Rate	Skills Index	Interpreting Derivatives in Context

Last updated: 2026-01-22