Instantaneous Velocity

7 min read

Reference: Stewart §2.1

What Does “Speed Right Now” Mean?

Your car’s speedometer shows 60 mph. But what does that mean? You’re not traveling for an hour; you’re looking at the dial at one instant. How can we define speed at a single moment in time?

Here’s the insight: average speed over shorter and shorter time intervals gives better and better approximations of your “true” speed at an instant. As the time interval shrinks to zero, the average speed approaches the instantaneous velocity.

This is the same limiting process we used for tangent lines, and it’s no coincidence. The instantaneous velocity is the slope of the position-time graph.

Prerequisite Map

Quick Reference

Property	Value
Concept	Tangent and Velocity Problems
Chapter	1.4
Difficulty	Intermediate
Time	~20 minutes

Key Concepts

Average vs. Instantaneous Velocity

Type	Formula	Meaning
Average velocity	$\displaystyle v_{\text{avg}} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}$	Total displacement ÷ total time
Instantaneous velocity	$\displaystyle v(t) = \lim_{h \to 0} \frac{s(t+h) - s(t)}{h}$	Speed at a single moment

Here $s(t)$ is the position function: it tells you where the object is at time $t$.

The Definition

The instantaneous velocity at time $t = a$ is:

$$\boxed{v(a) = \lim_{h \to 0} \frac{s(a+h) - s(a)}{h}}$$

or equivalently:

$$\boxed{v(a) = \lim_{t \to a} \frac{s(t) - s(a)}{t - a}}$$

This is exactly the same as the tangent slope formula, just with physical names:

$s$ instead of $f$
$t$ instead of $x$
“velocity” instead of “slope”

Visual Interpretation

  position s(t)
        │
        │              ·
        │           ·
        │        ·       · curve s(t)
        │     ·    ╱
        │   ·    ╱  tangent line
        │ ·    ╱
        │·   ╱
        └─────────────────── time t
              ↑
           t = a

  The slope of the tangent = instantaneous velocity at t = a

The slope of the position-time graph at any point equals the instantaneous velocity at that time.

Why This Connection Matters

Geometric View	Physical View
Curve $y = f(x)$	Position $s(t)$
Point $(a, f(a))$	Time $t = a$ and position $s(a)$
Secant slope	Average velocity
Tangent slope	Instantaneous velocity

The same limit appears in both contexts. This unification is the power of calculus.

Common Position Functions

Motion Type	Position Function	Example
Free fall (near Earth)	$s(t) = \frac{1}{2}gt^2$	$s(t) = 4.9t^2$ meters
Constant velocity	$s(t) = vt + s_0$	$s(t) = 5t + 2$
Vertical throw	$s(t) = -\frac{1}{2}gt^2 + v_0 t + s_0$	$s(t) = -4.9t^2 + 20t$

Here $g \approx 9.8 \text{ m/s}^2$ (or $32 \text{ ft/s}^2$) is gravitational acceleration.

Units Analysis

If position $s(t)$ is in meters and time $t$ is in seconds, then:

$$v = \frac{s(t+h) - s(t)}{h} = \frac{\text{meters}}{\text{seconds}} = \text{m/s}$$

Velocity always has units of distance ÷ time.

Common misconception

average velocity over a short interval equals instantaneous velocity.

This is the rate-as-fixed-number error. The average velocity $\frac{s(3.001) - s(3)}{0.001}$ is not the instantaneous velocity at $t = 3$ -- it is an approximation. For $s(t) = t^2$, this average is $\frac{(3.001)^2 - 9}{0.001} = \frac{9.006001 - 9}{0.001} = 6.001$. The instantaneous velocity at $t = 3$ is exactly $6$ m/s, found by the limit. A smaller $h$ gives a better approximation, but no nonzero $h$ gives the exact value. The exact instantaneous velocity is the limit, not any one average -- and it is constant at the single instant, even though it came from an infinite process.

Common misconception

“instantaneous velocity” is a contradiction because velocity requires a time interval.

This is the limit-as-unreachable-barrier error. The objection is philosophically natural: a ratio $\frac{\Delta s}{\Delta t}$ becomes $\frac{0}{0}$ at a single instant. But the limit process avoids dividing by zero by defining instantaneous velocity as the value the averages APPROACH, not the value at $h = 0$. The limit $\lim_{h \to 0} \frac{s(3+h) - s(3)}{h}$ is a well-defined number because the expression $\frac{s(3+h) - s(3)}{h}$ simplifies to a function that approaches $6$ as $h$ approaches $0$. Instantaneous velocity is not the result of dividing by zero; it is the result of a limit that produces a precise, finite number.

Practice Problems

Level 1 Average vs. Instantaneous

A ball’s position is given by $s(t) = t^2$ meters, where $t$ is in seconds.

(a) Find the average velocity from $t = 1$ to $t = 3$.

(b) Is this the same as the instantaneous velocity at $t = 2$? (You don’t need to calculate the instantaneous velocity, just answer conceptually.)

Thought Process

(a) Average velocity = $\frac{s(3) - s(1)}{3 - 1}$

(b) Think about whether “average over an interval” equals “exact value at the midpoint.” For a curved graph, these are generally different.

Show Answer

(a) $v_{\text{avg}} = \frac{s(3) - s(1)}{3 - 1} = \frac{9 - 1}{2} = \frac{8}{2} = 4$ m/s

(b) No. The average velocity over $[1, 3]$ is not the same as the instantaneous velocity at $t = 2$.

For a function that curves (like $t^2$), the average over an interval doesn’t equal the value at the midpoint. (In fact, the instantaneous velocity at $t = 2$ is $4$ m/s as well, but this is a coincidence for this particular function; it won’t hold in general.)

Level 2 Compute Instantaneous Velocity

A particle moves along a line with position $s(t) = t^2 + 3t$ meters. Find the instantaneous velocity at $t = 2$ seconds.

Thought Process

Use the limit definition: $$v(2) = \lim_{t \to 2} \frac{s(t) - s(2)}{t - 2}$$

Compute $s(2) = 4 + 6 = 10$
Write $s(t) - s(2) = (t^2 + 3t) - 10 = t^2 + 3t - 10$
Factor: look for factors of $-10$ that sum to $3$... that’s $(t+5)(t-2)$
Cancel $(t-2)$ and evaluate the limit

Show Answer

First, $s(2) = 2^2 + 3(2) = 4 + 6 = 10$.

$$v(2) = \lim_{t \to 2} \frac{(t^2 + 3t) - 10}{t - 2} = \lim_{t \to 2} \frac{t^2 + 3t - 10}{t - 2}$$

Factor: $t^2 + 3t - 10 = (t + 5)(t - 2)$

$$v(2) = \lim_{t \to 2} \frac{(t+5)(t-2)}{t-2} = \lim_{t \to 2} (t + 5) = 7 \text{ m/s}$$

The instantaneous velocity at $t = 2$ is $\boxed{7 \text{ m/s}}$.

Level 3 Free Fall Velocity

A stone is dropped from a bridge, and its height above the ground is given by $s(t) = 100 - 4.9t^2$ meters, where $t$ is time in seconds.

(a) Find the instantaneous velocity at $t = 3$ seconds.

(b) Interpret the sign of your answer.

Thought Process

(a) Use the limit definition with $a = 3$: $$v(3) = \lim_{h \to 0} \frac{s(3+h) - s(3)}{h}$$

Expand $s(3+h) = 100 - 4.9(3+h)^2$ and simplify.

(b) Think about what negative velocity means physically: is the stone going up or down?

Show Answer

(a) First, $s(3) = 100 - 4.9(9) = 100 - 44.1 = 55.9$ m.

Using the $h$ form: $$v(3) = \lim_{h \to 0} \frac{[100 - 4.9(3+h)^2] - 55.9}{h}$$

Expand $(3+h)^2 = 9 + 6h + h^2$: $$100 - 4.9(9 + 6h + h^2) = 100 - 44.1 - 29.4h - 4.9h^2 = 55.9 - 29.4h - 4.9h^2$$

So: $$v(3) = \lim_{h \to 0} \frac{(55.9 - 29.4h - 4.9h^2) - 55.9}{h} = \lim_{h \to 0} \frac{-29.4h - 4.9h^2}{h}$$

$$= \lim_{h \to 0} (-29.4 - 4.9h) = -29.4 \text{ m/s}$$

(b) The negative sign indicates the stone is moving downward (in the direction of decreasing height). After 3 seconds, it’s falling at 29.4 m/s.

The instantaneous velocity at $t = 3$ is $\boxed{-29.4 \text{ m/s}}$ (downward).

Level 4 When Does It Stop?

A ball is thrown upward from ground level with position $s(t) = 20t - 5t^2$ meters.

(a) Find the instantaneous velocity as a function of time (find $v(t)$ for general $t$).

(b) At what time does the ball reach its maximum height? (When is $v(t) = 0$?)

Thought Process

(a) Use the limit definition with general $t$: $$v(t) = \lim_{h \to 0} \frac{s(t+h) - s(t)}{h}$$

Expand and simplify to get a formula in terms of $t$.

(b) The ball stops momentarily at the top of its arc, which is when $v(t) = 0$.

Show Answer

(a) $$s(t+h) = 20(t+h) - 5(t+h)^2 = 20t + 20h - 5(t^2 + 2th + h^2)$$ $$= 20t + 20h - 5t^2 - 10th - 5h^2$$

$$s(t+h) - s(t) = (20t + 20h - 5t^2 - 10th - 5h^2) - (20t - 5t^2)$$ $$= 20h - 10th - 5h^2 = h(20 - 10t - 5h)$$

$$v(t) = \lim_{h \to 0} \frac{h(20 - 10t - 5h)}{h} = \lim_{h \to 0} (20 - 10t - 5h) = 20 - 10t$$

So $\boxed{v(t) = 20 - 10t}$ m/s.

(b) Set $v(t) = 0$: $$20 - 10t = 0 \implies t = 2 \text{ seconds}$$

The ball reaches a maximum height of $\boxed{20 \text{ meters}}$ at $t = 2$ seconds.

Level 5 Velocity from Data

A car’s position $s(t)$ (in meters) is recorded at several times:

$t$ (sec)	0	0.5	1.0	1.5	2.0	2.5	3.0
$s(t)$ (m)	0	2.0	6.5	12.0	18.5	26.0	34.5

(a) Estimate the instantaneous velocity at $t = 1.5$ seconds using secant lines from both sides.

(b) The position function is actually $s(t) = t^2 + 2t$. Verify your estimate by computing the exact instantaneous velocity at $t = 1.5$.

Thought Process

(a) To estimate velocity at $t = 1.5$:

Compute the secant slope from $t = 1.0$ to $t = 1.5$ (approaching from left)
Compute the secant slope from $t = 1.5$ to $t = 2.0$ (approaching from right)
Average these or note they should be close

(b) Use the limit definition or pattern recognition. For $s(t) = t^2 + 2t$, we found in Level 4 style that $v(t) = 2t + 2$.

Show Answer

(a) Left secant (from $t = 1.0$ to $t = 1.5$): $$v_{\text{left}} = \frac{12.0 - 6.5}{1.5 - 1.0} = \frac{5.5}{0.5} = 11 \text{ m/s}$$

Right secant (from $t = 1.5$ to $t = 2.0$): $$v_{\text{right}} = \frac{18.5 - 12.0}{2.0 - 1.5} = \frac{6.5}{0.5} = 13 \text{ m/s}$$

Average estimate: $\frac{11 + 13}{2} = 12$ m/s

(We could also use closer data points if available for better accuracy.)

(b) For $s(t) = t^2 + 2t$, compute: $$v(t) = \lim_{h \to 0} \frac{(t+h)^2 + 2(t+h) - (t^2 + 2t)}{h}$$

Expanding: $$= \lim_{h \to 0} \frac{t^2 + 2th + h^2 + 2t + 2h - t^2 - 2t}{h} = \lim_{h \to 0} \frac{2th + h^2 + 2h}{h}$$ $$= \lim_{h \to 0} (2t + h + 2) = 2t + 2$$

At $t = 1.5$: $$v(1.5) = 2(1.5) + 2 = 3 + 2 = 5 \text{ m/s}$$

The formula and the table describe different functions: $s(t) = t^2 + 2t$ gives $s(1.5) = 1.5^2 + 2(1.5) = 2.25 + 3 = 5.25$, while the table records $12.0$. The table values do not fit $s(t) = t^2 + 2t$, so they come from a separate data set built only for the part (a) estimate. Use the formula for part (b): with $s(t) = t^2 + 2t$, the exact instantaneous velocity is $v(1.5) = 2(1.5) + 2 = 5$ m/s.

The exact instantaneous velocity at $t = 1.5$ is $\boxed{5 \text{ m/s}}$ (assuming $s(t) = t^2 + 2t$).

(Note: The table data was constructed separately and doesn’t exactly match. In real applications, estimate from data first, then verify with the formula.)

Mastery Checklist

I can distinguish between average velocity and instantaneous velocity
I can compute instantaneous velocity using the limit definition
I understand that instantaneous velocity = slope of position graph
I can interpret the sign of velocity (positive = forward, negative = backward)
I can find when an object stops (when $v(t) = 0$)
I can estimate velocity from a table of data

Mental Model

The Shrinking Stopwatch:

Imagine you have a magical stopwatch. You measure how far you travel in 1 second, then 0.1 seconds, then 0.01 seconds, then 0.001 seconds...

Each measurement gives an average speed over that interval. As the interval shrinks toward zero, your “average” speed over that tiny interval becomes your exact speed at that instant.

The limit captures what happens when the stopwatch interval becomes infinitely small.

Connections

Looking back:

Secant Lines gave us the difference quotient (average velocity)
Tangent Slope is the same limit with different names

Looking ahead:

The derivative $f'(t)$ formalizes this as “the instantaneous rate of change”
Acceleration is the rate of change of velocity (the derivative of velocity)
All rates of change in physics, economics, biology use this same concept

The Big Picture: The tangent problem (geometry) and velocity problem (physics) are mathematically identical. This is why calculus unifies so many fields: the same tools work everywhere rates of change appear.

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Tangent Slope via Limits	Skills Index	Derivative Definition

Last updated: 2026-01-22