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Rectilinear Motion from Antiderivatives

Reference: Stewart §3.9 • Chapter: 3 • Section: 9

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From Physics to Calculus

A ball is thrown upward. You know gravity causes constant downward acceleration of $-32$ ft/s². Can you predict where the ball will be in 3 seconds?

This is exactly what antiderivatives do: given how something is changing (acceleration), you can work backwards to find velocity, and from velocity to position.

The relationships are beautifully simple:

Velocity is the derivative of position: $v = s'$
Acceleration is the derivative of velocity: $a = v'$

Reading these backwards:

Position is the antiderivative of velocity
Velocity is the antiderivative of acceleration

Prerequisite Map

Quick Reference

Property	Value
Concept	Antiderivatives
Chapter	3.9
Difficulty	Intermediate
Time	~25 minutes

Key Concepts

The Position-Velocity-Acceleration Chain

For an object moving along a line (rectilinear motion):

Quantity	Symbol	Relationship
Position	$s(t)$	(base quantity)
Velocity	$v(t)$	$v = \frac{ds}{dt} = s'(t)$
Acceleration	$a(t)$	$a = \frac{dv}{dt} = v'(t) = s''(t)$

Going Backwards (Antidifferentiation)

Given	Find	Method
$a(t)$	$v(t)$	$v(t) = \int a(t)\,dt$
$v(t)$	$s(t)$	$s(t) = \int v(t)\,dt$
$a(t)$	$s(t)$	Antidifferentiate twice

Each antidifferentiation requires an initial condition to determine the constant.

Constant Acceleration (Free Fall)

Near Earth's surface, objects in free fall have constant acceleration:

$$a(t) = -g$$

where $g \approx 9.8$ m/s² (or 32 ft/s²). The negative sign indicates downward.

Starting from constant acceleration:

$$v(t) = \int -g\,dt = -gt + C_1$$

If $v(0) = v_0$ (initial velocity), then $C_1 = v_0$: $$v(t) = -gt + v_0$$

$$s(t) = \int (-gt + v_0)\,dt = -\frac{1}{2}gt^2 + v_0 t + C_2$$

If $s(0) = s_0$ (initial position), then $C_2 = s_0$:

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

This is the famous kinematic equation from physics, derived purely from calculus!

Key Relationships for Constant Acceleration

Formula	What it gives
$s(t) = s_0 + v_0 t + \frac{1}{2}at^2$	Position at time $t$
$v(t) = v_0 + at$	Velocity at time $t$
$v^2 = v_0^2 + 2a(s - s_0)$	Velocity without time

Interpreting Signs

Sign	Position ($s$)	Velocity ($v$)	Acceleration ($a$)
Positive	Above origin / to the right	Moving upward / rightward	Speeding up (if same sign as $v$)
Negative	Below origin / to the left	Moving downward / leftward	Slowing down (if opposite sign to $v$)

Key insight: An object slows down when velocity and acceleration have opposite signs.

Practice Problems

Level 1 Given Velocity, Find Position

A particle moves along a line with velocity $v(t) = 4t - 6$ m/s. If the particle is at position $s = 2$ m when $t = 0$, find its position at time $t$.

Thought Process

Position is the antiderivative of velocity. Antidifferentiate $v(t)$, then use $s(0) = 2$ to find $C$.

Show Answer

Step 1: Antidifferentiate velocity: $$s(t) = \int (4t - 6)\,dt = 2t^2 - 6t + C$$

Step 2: Use $s(0) = 2$: $$s(0) = 0 - 0 + C = 2 \quad \Rightarrow \quad C = 2$$

Step 3: Position function: $$s(t) = 2t^2 - 6t + 2 \text{ meters}$$

Level 2 Given Acceleration, Find Velocity

A car has acceleration $a(t) = 3$ m/s² (constant). If the car starts from rest, find its velocity after 4 seconds.

Thought Process

"Starts from rest" means $v(0) = 0$. Antidifferentiate acceleration to get velocity, apply the initial condition, then evaluate at $t = 4$.

Show Answer

Step 1: Antidifferentiate acceleration: $$v(t) = \int 3\,dt = 3t + C$$

Step 2: Use $v(0) = 0$ (starts from rest): $$v(0) = 0 + C = 0 \quad \Rightarrow \quad C = 0$$ $$v(t) = 3t$$

Step 3: Velocity at $t = 4$: $$v(4) = 3(4) = 12 \text{ m/s}$$

Level 3 From Acceleration to Position

A particle has acceleration $a(t) = 6t$ m/s². Its initial velocity is $v(0) = -4$ m/s and its initial position is $s(0) = 5$ m. Find $s(t)$.

Thought Process

This is a second-order problem. Antidifferentiate $a(t)$ to get $v(t)$, use $v(0) = -4$ to find the first constant. Then antidifferentiate $v(t)$ to get $s(t)$, use $s(0) = 5$ to find the second constant.

Show Answer

Step 1: Find velocity: $$v(t) = \int 6t\,dt = 3t^2 + C_1$$

Use $v(0) = -4$: $$v(0) = 0 + C_1 = -4 \quad \Rightarrow \quad C_1 = -4$$ $$v(t) = 3t^2 - 4$$

Step 2: Find position: $$s(t) = \int (3t^2 - 4)\,dt = t^3 - 4t + C_2$$

Use $s(0) = 5$: $$s(0) = 0 - 0 + C_2 = 5 \quad \Rightarrow \quad C_2 = 5$$

Step 3: Position function: $$s(t) = t^3 - 4t + 5 \text{ meters}$$

Level 4 Ball Thrown Upward

A ball is thrown upward from the top of a 96-foot building with an initial velocity of 80 ft/s. Taking $g = 32$ ft/s²:

(a) Find the position function $s(t)$. (b) When does the ball reach its maximum height? (c) What is the maximum height? (d) When does the ball hit the ground?

Thought Process

Use $a = -32$, $v(0) = 80$, $s(0) = 96$. Maximum height occurs when $v(t) = 0$. Ball hits ground when $s(t) = 0$.

Show Answer

(a) Position function:

$a(t) = -32$

$v(t) = -32t + C_1$, and $v(0) = 80$, so $C_1 = 80$: $$v(t) = -32t + 80$$

$s(t) = -16t^2 + 80t + C_2$, and $s(0) = 96$, so $C_2 = 96$: $$s(t) = -16t^2 + 80t + 96 \text{ feet}$$

(b) Maximum height time:

Maximum height when $v(t) = 0$: $$-32t + 80 = 0 \quad \Rightarrow \quad t = \frac{80}{32} = 2.5 \text{ seconds}$$

(c) Maximum height: $$s(2.5) = -16(2.5)^2 + 80(2.5) + 96 = -100 + 200 + 96 = 196 \text{ feet}$$

(d) Hitting the ground:

Solve $s(t) = 0$: $$-16t^2 + 80t + 96 = 0$$ $$t^2 - 5t - 6 = 0$$ $$(t - 6)(t + 1) = 0$$ $$t = 6 \text{ or } t = -1$$

Since $t > 0$, the ball hits the ground at $t = 6$ seconds.

Level 5 Variable Acceleration

A rocket has acceleration $a(t) = 60t$ ft/s² for the first 2 seconds (while fuel burns), then $a(t) = -32$ ft/s² afterward (free fall). If the rocket starts from rest at ground level:

(a) Find the velocity and position at $t = 2$ s (end of fuel burn). (b) Find the maximum height the rocket reaches. (c) Find when the rocket returns to the ground.

Thought Process

This is a piecewise problem. Solve the first phase ($0 \leq t \leq 2$) with initial conditions at $t = 0$. Then use the values at $t = 2$ as initial conditions for the second phase ($t > 2$).

Show Answer

(a) Phase 1 ($0 \leq t \leq 2$):

$a(t) = 60t$, $v(0) = 0$, $s(0) = 0$

$v(t) = 30t^2 + C_1$, $v(0) = 0 \Rightarrow C_1 = 0$ $$v(t) = 30t^2$$

$s(t) = 10t^3 + C_2$, $s(0) = 0 \Rightarrow C_2 = 0$ $$s(t) = 10t^3$$

At $t = 2$:

$v(2) = 30(4) = 120$ ft/s
$s(2) = 10(8) = 80$ ft

(b) Phase 2 ($t > 2$):

Let $\tau = t - 2$ (time since fuel burnout).

$a = -32$, $v(\tau = 0) = 120$, $s(\tau = 0) = 80$

$v(\tau) = -32\tau + 120$

$s(\tau) = -16\tau^2 + 120\tau + 80$

Maximum height when $v = 0$: $$-32\tau + 120 = 0 \quad \Rightarrow \quad \tau = 3.75 \text{ s}$$

Maximum height: $$s(3.75) = -16(3.75)^2 + 120(3.75) + 80 = -225 + 450 + 80 = 305 \text{ feet}$$

(c) Return to ground:

Solve $s(\tau) = 0$: $$-16\tau^2 + 120\tau + 80 = 0$$ $$2\tau^2 - 15\tau - 10 = 0$$

Using the quadratic formula: $$\tau = \frac{15 \pm \sqrt{225 + 80}}{4} = \frac{15 \pm \sqrt{305}}{4}$$

Taking the positive root: $$\tau \approx \frac{15 + 17.46}{4} \approx 8.12 \text{ s}$$

Total time: $t = 2 + 8.12 = 10.12$ seconds

CCI-Style Conceptual Questions

Conceptual Speeding Up or Slowing Down?

At time $t = 3$, a particle has $v(3) = -4$ m/s and $a(3) = -2$ m/s². Is the particle speeding up or slowing down at this instant?

Thought Process

Compare the signs of velocity and acceleration. If they have the same sign, the object is speeding up. If opposite signs, slowing down.

Show Answer

Speeding up.

Both $v(3) = -4$ and $a(3) = -2$ are negative. When velocity and acceleration have the same sign, the object is speeding up.

The particle is moving leftward (negative velocity) and accelerating leftward (negative acceleration), so its speed is increasing.

Conceptual At the Top of Its Arc

A ball is thrown straight up. At the instant it reaches its maximum height, what are the values of its velocity and acceleration?

Thought Process

At maximum height, the ball momentarily stops (changes direction). What does this say about velocity? Does gravity "turn off" at this instant?

Show Answer

Velocity = 0, Acceleration = -g (about -32 ft/s² or -9.8 m/s²)

At maximum height, $v = 0$ because the ball momentarily stops before reversing direction.

But acceleration is still $-g$! Gravity doesn't turn off just because the ball is at its peak. The ball is still being pulled downward: that's why it starts falling immediately after.

Common misconception: Students often think $a = 0$ at the top. But if that were true, there would be no force to make the ball start falling!

Conceptual Two Balls Problem

Ball A is dropped from a height of 100 ft. One second later, Ball B is thrown downward from the same height with initial velocity 40 ft/s. Which ball hits the ground first?

Thought Process

Set up position functions for both balls. Ball A: $s_A(t) = 100 - 16t^2$. Ball B starts 1 second later with downward velocity. Find when each hits ground ($s = 0$) and compare.

Show Answer

Ball A: $s_A(t) = 100 - 16t^2$

Hits ground when $s_A = 0$: $100 = 16t^2$, so $t = 2.5$ s.

Ball B: Starts at $t = 1$ s with $v(1) = -40$ ft/s (downward).

Let $\tau = t - 1$ be time since Ball B was thrown. $$s_B(\tau) = 100 - 40\tau - 16\tau^2$$

Hits ground when $s_B = 0$: $$16\tau^2 + 40\tau - 100 = 0$$ $$\tau = \frac{-40 + \sqrt{1600 + 6400}}{32} = \frac{-40 + \sqrt{8000}}{32} \approx 1.55 \text{ s}$$

Ball B lands at $t = 1 + 1.55 = 2.55$ s.

Ball A hits first (at 2.5 s vs 2.55 s), but just barely!

Mastery Checklist

[ ] I understand that position, velocity, and acceleration are related by differentiation/antidifferentiation
[ ] I can find velocity from acceleration using antidifferentiation and initial conditions
[ ] I can find position from velocity using antidifferentiation and initial conditions
[ ] I can solve problems involving constant acceleration (free fall, $g = 32$ ft/s² or 9.8 m/s²)
[ ] I can determine when an object reaches maximum height ($v = 0$)
[ ] I can find when an object returns to a given position ($s = $ specified value)
[ ] I understand when an object is speeding up vs. slowing down based on signs of $v$ and $a$

Mental Model

Think of the position-velocity-acceleration chain as a ladder:

        d/dt          d/dt
Position → Velocity → Acceleration
    s(t)      v(t)       a(t)

        ∫ dt          ∫ dt
Position ← Velocity ← Acceleration

Going UP the ladder (differentiation): tells you rates of change Going DOWN the ladder (antidifferentiation): reconstructs the original quantities

Each step DOWN requires knowing where you "started" (initial condition), because information is lost when you go UP.

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Last updated: 2026-01-22