Physics Applications of Derivatives
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 3.4: “Derivatives as Rates of Change” |
| Book URL | https://openstax.org/details/books/calculus-volume-1 |
Freely available and openly licensed.
Try This First: Watch a Quantity Change
A ball is thrown upward. Its height (in meters) at time $t$ seconds is $s(t) = 20t - 5t^2$.
Before any computation:
- Predict: at what time does the ball reach its highest point?
- Predict: is the ball moving upward or downward at $t = 1$? At $t = 3$?
Now compute $s'(t)$ and check your predictions. The derivative tells you whether the ball is rising or falling, and exactly how fast, at any moment.
Quantity-First Framing
When an object moves in a straight line, three quantities describe its motion:
- Position $s(t)$: where the object is at time $t$ (measured in meters, kilometers, etc. along a number line).
- Velocity $v(t) = s'(t)$: how fast the position is changing -- and in what direction. Positive velocity means moving in the positive direction; negative velocity means moving backward.
- Acceleration $a(t) = v'(t) = s''(t)$: how fast the velocity is changing. Positive acceleration means speeding up in the positive direction (or slowing down if currently moving in the negative direction).
The derivative connects each layer to the next: velocity is the derivative of position, and acceleration is the derivative of velocity.
Prerequisite Check
Before this lesson, make sure you can do all of the following:
Quick Reference
| Quantity | Notation | How to find it |
|---|---|---|
| Position | $s(t)$ | Given |
| Velocity | $v(t) = s'(t)$ | Differentiate $s$ |
| Acceleration | $a(t) = v'(t) = s''(t)$ | Differentiate $v$ (or differentiate $s$ twice) |
| Speed | $|v(t)|$ | Absolute value of velocity |
Sign interpretation:
| $v(t) > 0$ | Object moves in positive direction |
|---|---|
| $v(t) < 0$ | Object moves in negative direction |
| $v(t) = 0$ | Object momentarily at rest |
| $v(t)$ and $a(t)$ same sign | Speeding up |
| $v(t)$ and $a(t)$ opposite sign | Slowing down |
Key Concepts
1. Position, Velocity, Acceleration: The Full Picture
Example 1. A particle moves along a straight line with position $s(t) = t^3 - 6t^2 + 9t$ meters, $t \geq 0$.
Predict first: Sketch a rough graph of $s$ or describe its behavior from the formula.
Step 1: Velocity. \[ v(t) = s'(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3). \]
$v(t) = 0$ at $t = 1$ and $t = 3$. These are moments when the particle is at rest.
Step 2: Direction.
| Interval | Sign of $v(t)$ | Direction |
|---|---|---|
| $0 < t < 1$ | $+$ (test $t=0.5$: $3 \cdot 0.25 - 6 + 9 > 0$) | Positive |
| $1 < t < 3$ | $-$ (test $t=2$: $12 - 24 + 9 < 0$) | Negative (backward) |
| $t > 3$ | $+$ | Positive again |
The particle moves right, stops at $t=1$, turns around, moves left, stops at $t=3$, turns around, and moves right again.
Step 3: Acceleration. \[ a(t) = v'(t) = 6t - 12. \]
$a(t) = 0$ at $t = 2$ (inflection in velocity). For $t < 2$: $a < 0$ (velocity decreasing). For $t > 2$: $a > 0$ (velocity increasing).
Step 4: Speed and speeding up/slowing down.
At $t = 0.5$: $v = 3(0.25) - 6 + 9 = 3.75 > 0$, $a = 3 - 12 = -9 < 0$. Opposite signs: slowing down.
At $t = 4$: $v = 48 - 48 + 9 = 9 > 0$, $a = 24 - 12 = 12 > 0$. Same sign: speeding up.
2. Two Representations: Graph and Table
For $s(t) = 20t - 5t^2$ (the ball from Try This First):
$v(t) = 20 - 10t$. $v(t) = 0$ at $t = 2$: highest point. $s(2) = 40 - 20 = 20$ m.
$a(t) = -10$ m/s$^2$ (constant, the acceleration due to gravity near Earth’s surface).
Table:
| $t$ (s) | $s(t)$ (m) | $v(t)$ (m/s) | $a(t)$ (m/s$^2$) | Motion |
|---|---|---|---|---|
| 0 | 0 | 20 | $-10$ | Rising, decelerating |
| 1 | 15 | 10 | $-10$ | Rising, slowing |
| 2 | 20 | 0 | $-10$ | At peak |
| 3 | 15 | $-10$ | $-10$ | Falling, speeding up |
| 4 | 0 | $-20$ | $-10$ | Back at ground, fast |
Graph description. $s(t)$ is a downward parabola peaking at $(2, 20)$. $v(t) = 20 - 10t$ is a line crossing zero at $t = 2$. $a(t) = -10$ is a horizontal line.
Translation prompt. From the table: at $t = 3$, $s = 15$ and $v = -10$. What does it mean for position to be positive (15 m) while velocity is negative ($-10$ m/s)? The ball is above the ground (positive position) but moving downward (negative velocity).
3. Speed vs. Velocity
Velocity is a signed quantity: positive or negative depending on direction. Speed is $|v(t)|$ -- always non-negative, measuring how fast the object moves without regard to direction.
Example 2. In Example 1, at $t = 0.5$: $v(0.5) = 3.75$ m/s. Speed $= 3.75$ m/s.
At $t = 2$: $v(2) = 3(4) - 24 + 9 = -3$ m/s. Speed $= 3$ m/s.
Same speed, but opposite direction.
4. Ask Why: Why is Acceleration the Second Derivative?
Acceleration measures how the velocity is changing, just as velocity measures how the position is changing. Because velocity is the first derivative of position, the derivative of velocity is the second derivative of position. In Leibniz notation: \[ a = \frac{dv}{dt} = \frac{d}{dt}\left(\frac{ds}{dt}\right) = \frac{d^2s}{dt^2}. \]
This chain of derivatives appears throughout physics: just as position $\to$ velocity $\to$ acceleration, in electrical circuits charge $\to$ current $\to$ rate of current change.
5. Multiple Valid Paths
To determine whether a particle is speeding up or slowing down at $t = t_0$:
Path 1 (sign comparison): compute $v(t_0)$ and $a(t_0)$; if same sign, speeding up; if opposite sign, slowing down.
Path 2 (speed derivative): compute $\dfrac{d}{dt}|v(t)|$ at $t_0$; if positive, speeding up; if negative, slowing down. (This requires careful handling at $v = 0$.)
Both paths are valid. Path 1 is faster and recommended for MATH161; Path 2 is more general.
Common Misconception: rate-as-fixed-number
A particle with position $s(t) = t^3 - 6t^2 + 9t$ does not move at a fixed speed. The velocity $v(t) = 3t^2 - 12t + 9$ changes at every moment.
A common error: computing $v(1) = 0$ and concluding “the particle is not moving” -- but this is only true at $t = 1$. At $t = 0.5$, the velocity is $3.75$ m/s. At $t = 1.5$, it is $-2.25$ m/s. The derivative gives the instantaneous rate, which varies continuously, not a single answer that applies to all of $t$.
Common Errors
| Error | Specific example | Correction |
|---|---|---|
| Confusing position with velocity | “At $t = 3$, $s = 0$ so the particle is at rest” | $s = 0$ means the particle is at the origin; $v = 0$ means it is at rest |
| Confusing speed with velocity | “Velocity is $-3$ m/s, speed is $-3$ m/s” | Speed is $|v| = 3$ m/s; it is never negative |
| Wrong criterion for speeding up | “Velocity is increasing, so particle is speeding up” | Speeding up requires velocity and acceleration to have the same sign; if $v < 0$ and $a < 0$, the particle is speeding up in the negative direction |
Common Misconceptions
when $s(t) = 0$, the particle is at rest.
This is the height-vs-slope error. For a particle with position $s(t) = t^3 - 6t^2 + 9t$, evaluating $s(3) = 27 - 54 + 27 = 0$ means the particle is at the origin, not that it has stopped moving. The particle is at rest only when $v(t) = s'(t) = 0$. At $t = 3$, the velocity $v(3) = 27 - 36 + 9 = 0$ happens to also be zero, but these are two separate facts: one about position ($s = 0$) and one about velocity ($v = 0$). At $t = 0$, for example, $s(0) = 0$ but $v(0) = 9 \neq 0$, so the particle is at the origin but moving.
Leveled Practice
Level 1 -- Computing Velocity and Acceleration
Problem 1. A particle’s position is $s(t) = 2t^3 - 9t^2 + 12t$ meters. Find $v(t)$ and $a(t)$. At what times is the particle at rest?
Show answer
$v(t) = 6t^2 - 18t + 12 = 6(t-1)(t-2)$.
At rest when $v = 0$: $t = 1$ s and $t = 2$ s.
$a(t) = 12t - 18$.
Problem 2. For the particle in Problem 1, determine whether it is speeding up or slowing down at $t = 1.5$.
Show answer
$v(1.5) = 6(2.25) - 27 + 12 = 13.5 - 27 + 12 = -1.5$ m/s (negative; moving in negative direction).
$a(1.5) = 18 - 18 = 0$ m/s$^2$.
Acceleration is zero: neither speeding up nor slowing down at $t = 1.5$. The speed is at a local minimum.
Level 2 -- Full Analysis
Problem 3. A ball is dropped from rest from a height of 80 m. Its position (measured downward) is $s(t) = 5t^2$ meters.
(a) Find the velocity when it hits the ground.
(b) Predict the answer first using energy conservation: $v = \sqrt{2gh}$ where $g = 10$ m/s$^2$, $h = 80$ m. Compare to the calculus answer.
Show answer
(a) Ground: $s(t) = 80$, so $5t^2 = 80$, $t = 4$ s. $v(t) = 10t$; $v(4) = 40$ m/s.
(b) $v = \sqrt{2 \cdot 10 \cdot 80} = \sqrt{1600} = 40$ m/s. Same answer. The calculus and energy approaches agree.
Level 3 -- Low-Floor-High-Ceiling Extension
Problem 4 (Extension).
(a) (Floor) For $s(t) = 20t - 5t^2$, find the total distance traveled from $t = 0$ to $t = 4$.
(b) (Mid) Show that for any constant-acceleration motion $s(t) = s_0 + v_0 t + \frac{1}{2}at^2$, the velocity is $v(t) = v_0 + at$ and acceleration is $a$ (constant). Interpret: what does $s_0$ represent?
(c) (Ceiling) Jerk is defined as $j(t) = a'(t) = s'''(t)$. For the polynomial $s(t) = t^3 - 6t^2 + 9t$, find $j(t)$ and interpret: what does constant jerk mean geometrically for the acceleration graph?
Show answer
(a) Ball rises from $t=0$ to $t=2$ (when $v=0$) and falls from $t=2$ to $t=4$.
Distance up: $s(2) - s(0) = 20 - 0 = 20$ m. Distance down: $s(2) - s(4) = 20 - 0 = 20$ m. Total: 40 m.
(b) $v(t) = v_0 + at$, $a(t) = a$ (constant). $s_0 = s(0)$ is the initial position.
(c) $j(t) = s'''(t) = 6$ (constant). Constant jerk means the acceleration graph is a line with slope 6 -- the acceleration changes at a constant rate. In a car, constant jerk means the driver presses the gas pedal at a steady rate (so acceleration builds linearly).
Mastery Checklist
You have mastered this skill when you can do all of the following without referring to notes:
Mental Model
Position, velocity, and acceleration form a chain. Position tells you where; velocity tells you the direction and rate of change of where; acceleration tells you the direction and rate of change of that rate.
The derivative connects each level to the next. Speed up this derivative chain and you climb from position to velocity to acceleration. Every physical motion problem is a question about somewhere along this chain: where is the object? How fast? In what direction? Is it speeding up or slowing down?
The sign of velocity tells you direction. The sign of acceleration, relative to velocity, tells you whether the object is speeding up (same sign) or slowing down (opposite sign). These two pieces of sign information are what connects the algebraic derivative to the physical motion.
Connections
Within Calculus I (MATH161)
- Higher derivatives: Position, velocity, and acceleration is the primary example of a chain of higher derivatives. The pattern continues: jerk ($s'''$), snap ($s^{(4)}$), etc.
- Optimization: The highest point in the ball example occurs where $v = 0$ -- this is a critical point. Optimization (Ch. 3) generalizes this idea.
- Related rates: When multiple quantities change simultaneously (position and angle, for instance), related-rate techniques extend these ideas.
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