Rates of Change in Context: Motion, Velocity and Acceleration

Quick answer

A derivative is a rate of change, measured in output units per input unit, so f′(3) = −2 liters per minute says the quantity is falling at 2 liters each minute at that moment. For motion along a line, velocity is the derivative of position and acceleration is the derivative of velocity. The sign of velocity gives the direction; speed is its absolute value. An object speeds up when velocity and acceleration have the same sign and slows down when their signs differ.

What you'll learn

Interpret a derivative as a rate of change, with units
Find velocity and acceleration from a position function
Decide when an object moves left or right, and when it speeds up or slows down
Find total distance traveled as opposed to displacement

A derivative is a rate, with units

If $V(t)$ is the volume of water in a tank, in liters, after $t$ minutes, then

V'(t) = \lim_{h \to 0} \frac{V(t + h) - V(t)}{h}

is a change in liters divided by a change in minutes. Its units are liters per minute. A statement such as $V'(10) = -3$ reads, in context: at $10$ minutes, the volume is decreasing at $3$ liters per minute.

Every good interpretation names the time, the quantity, whether it is increasing or decreasing, and the rate with its units.

Position, velocity and acceleration

For an object moving along a line with position $s(t)$ :

Quantity	Formula	Meaning
velocity	$v(t) = s'(t)$	rate of change of position; its sign is the direction
speed	$\lvert v(t) \rvert$	how fast, regardless of direction
acceleration	$a(t) = v'(t) = s''(t)$	rate of change of velocity

The object moves in the positive direction when $v > 0$ , in the negative direction when $v < 0$ , and is momentarily at rest when $v = 0$ . A change of direction happens where $v$ changes sign.

s(t) = t³ − 6t² + 9t

Position of a particle

Why speeding up depends on two signs

Speed is $\lvert v \rvert$ , the size of the velocity. Acceleration pushes the velocity toward positive values when $a > 0$ and toward negative values when $a < 0$ .

If the object is moving in the positive direction ( $v > 0$ ) and $a > 0$ , the push is along the motion, so the speed grows. If $v < 0$ and $a < 0$ , the push is again along the motion — toward more negative velocity — so the speed grows too. When the signs differ, the push opposes the motion and the speed shrinks. An object speeds up when velocity and acceleration have the same sign, and slows down when they have opposite signs. Negative acceleration alone does not mean slowing down.

Higher derivatives

Differentiating again gives the second derivative, $f''$ or $\tfrac{d^2y}{dx^2}$ , the rate of change of the rate of change. Acceleration is the second derivative of position. The third derivative, $f'''$ , and beyond follow the same pattern.

Worked examples

Example 2 — Particle motion

A particle has position $s(t) = t^3 - 6t^2 + 9t$ for $t \ge 0$ . When is it at rest, and when is it speeding up?

$v(t) = 3t^2 - 12t + 9 = 3(t - 1)(t - 3)$ , which is $0$ at $t = 1$ and $t = 3$ . So $v > 0$ on $[0, 1)$ , $v < 0$ on $(1, 3)$ and $v > 0$ after $3$ .

$a(t) = 6t - 12$ , which is negative before $t = 2$ and positive after.

Interval	$v$	$a$	Motion
$0 < t < 1$	$+$	$-$	forward, slowing down
$1 < t < 2$	$-$	$-$	backward, speeding up
$2 < t < 3$	$-$	$+$	backward, slowing down
$t > 3$	$+$	$+$	forward, speeding up

Common mistakes

Practice problems

$P(t)$ is a town’s population, $t$ years after 2020, and $P'(5) = 120$ . Interpret this.

Answer

In 2025 the population is increasing at $120$ people per year.

Full solution

$P'$ has units of people per year. A positive value means the population is growing at that moment.
A particle has position $s(t) = t^2 - 8t + 3$ . Find $v(t)$ and $a(t)$ .

Answer

$v(t) = 2t - 8$ and $a(t) = 2$

Full solution

Differentiate once for velocity and again for acceleration.
For the particle in problem 2, when does it change direction?

Answer

At $t = 4$

Full solution

$v(t) = 2t - 8$ changes sign from negative to positive at $t = 4$ .
For the particle in problem 2, is it speeding up or slowing down at $t = 2$ ?

Answer

Slowing down

Full solution

$v(2) = -4 < 0$ and $a(2) = 2 > 0$ . The signs differ, so it is slowing down.
A particle has velocity $v(t) = -3$ and acceleration $a(t) = -1$ at some moment. Is it speeding up?

Answer

Yes

Full solution

Both are negative. The acceleration points along the motion, so the speed increases.
Find the second derivative of $f(x) = x^4 - 2x^3 + x$ .

Answer

$f''(x) = 12x^2 - 12x$

Full solution

$f'(x) = 4x^3 - 6x^2 + 1$ , then $f''(x) = 12x^2 - 12x$ .
A stone is dropped from $144$ feet, so $h(t) = 144 - 16t^2$ . When does it land, and how fast is it going?

Answer

At $t = 3$ seconds, at $96$ feet per second

Full solution

$144 - 16t^2 = 0$ gives $t = 3$ . $v(t) = -32t$ , so $v(3) = -96$ : a speed of $96$ feet per second, downward.
Position $s(t) = t^2 - 4t$ for $0 \le t \le 5$ . Find the displacement and the total distance.

Answer

Displacement $5$ ; distance $13$

Full solution

$v = 2t - 4$ turns around at $t = 2$ . $s(0) = 0$ , $s(2) = -4$ , $s(5) = 5$ . Displacement: $5 - 0 = 5$ . Distance: $4 + 9 = 13$ .
$C(x)$ is the cost, in dollars, of producing $x$ chairs, and $C'(100) = 45$ . What does this mean, approximately?

Answer

The 101st chair costs about 45 dollars to make.

Full solution

$C'(100)$ is the rate in dollars per chair at a production level of $100$ . Producing one more chair adds about $C'(100) \cdot 1 = 45$ dollars.
A student says a particle with $a(t) < 0$ for all $t$ must always be slowing down. Give a counterexample.

Hint

What if the velocity is negative as well?

Answer

$s(t) = -t^2$ for $t > 0$ : $a = -2$ , yet the particle speeds up.

Full solution

$v(t) = -2t$ , which is negative for $t > 0$ , and its speed $2t$ grows.

Velocity and acceleration are both negative, so the particle speeds up in the negative direction.

Frequently asked questions

What are the units of a derivative?

Output units per input unit. If V is in liters and t in minutes, V′(t) is in liters per minute.

How are position, velocity and acceleration related?

Velocity is the derivative of position, v = s′, and acceleration is the derivative of velocity, a = v′ = s″.

What is the difference between velocity and speed?

Velocity has a sign that gives the direction of motion. Speed is the absolute value of velocity, so it is never negative.

When is an object speeding up?

When its velocity and acceleration have the same sign. When the signs differ, it is slowing down.

How is total distance different from displacement?

Displacement is final position minus starting position. Total distance adds up every stretch traveled, counting backward motion as distance too.

What to learn next

For teachersPrint problems onlyTick to print a blank worksheet; leave it clear to print the lesson with the full answer key.

Quick answer

What you'll learn

A derivative is a rate, with units

Position, velocity and acceleration

Why speeding up depends on two signs

Higher derivatives

Worked examples

Common mistakes

Practice problems

Frequently asked questions

What are the units of a derivative?

How are position, velocity and acceleration related?

What is the difference between velocity and speed?

When is an object speeding up?

How is total distance different from displacement?

What to learn next

More Calculus lessons