Instantaneous Velocity from the Derivative

7 min read

Reference: Stewart §2.7

What Does Your Speedometer Actually Show?

When you glance at your car’s speedometer, it doesn’t show your average speed over the trip. It shows how fast you’re going right now - at this exact instant.

But what does “speed at an instant” even mean? An instant has no duration, so how can there be a rate of change?

This is precisely what the derivative captures. The instantaneous velocity is the limit of average velocities over shorter and shorter time intervals.

Prerequisite Map

Quick Reference

Property	Value
Section	Stewart 2.1
Course	MATH161
Difficulty	Intermediate
Time	~15 minutes

Key Concepts

Position, Velocity, and Speed

If an object moves along a straight line, its position function $s = f(t)$ tells us where it is at time $t$.

Quantity	Symbol	Meaning	Relationship
Position	$s = f(t)$	Where the object is	Given function
Velocity	$v(t) = f'(t)$	How fast and which direction	Derivative of position
Speed	$\\|v(t)\\|$	How fast (no direction)	Absolute value of velocity

Average vs. Instantaneous Velocity

Average velocity over $[a, a+h]$: $$v_{\text{avg}} = \frac{f(a+h) - f(a)}{h} = \frac{\text{change in position}}{\text{change in time}}$$

Instantaneous velocity at $t = a$: $$\boxed{v(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = f'(a)}$$

The instantaneous velocity is what the average velocity approaches as the time interval shrinks to zero.

Physical Interpretation

Position s(t)
    │
    │        ●───── At time t = a, position is s(a)
    │       /│
    │      / │ After time h, position is s(a+h)
    │     /  │
    │    /   │
    │   ●────┘
    │
    └──────────────────── t
        a    a+h

Average velocity = slope of this secant line
Instantaneous velocity = slope of tangent at t = a

Sign Convention

Velocity	Motion
$v(t) > 0$	Moving in positive direction (often “right” or “up”)
$v(t) < 0$	Moving in negative direction (often “left” or “down”)
$v(t) = 0$	Momentarily at rest (may be changing direction)

Free Fall Example

A common position function for objects falling under gravity (in meters, seconds):

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

where $g \approx 9.8$ m/s² on Earth, $s_0$ is initial height, and $v_0$ is initial velocity.

For an object dropped from rest at height $h$: $$s(t) = h - 4.9t^2$$

The derivative gives velocity: $$v(t) = -9.8t$$

(Negative because downward is the negative direction.)

Common misconception

speed and velocity are the same thing.

This is the rate-as-fixed-number error merged with a sign confusion. Velocity is a signed quantity: positive means moving in the positive direction, negative means moving in the negative direction. Speed is the absolute value of velocity -- always non-negative. For a falling object with $v(t) = -9.8t$: at $t = 2$, the velocity is $-19.6$ m/s (downward) and the speed is $19.6$ m/s. Saying “the velocity is $19.6$ m/s” loses the directional information. When a problem asks “how fast,” it usually asks for speed; when it asks “what is the velocity,” the sign is essential.

Common misconception

the derivative of position at $t = a$ gives the average speed over the whole trip.

This is the height-vs-slope error in motion context. $v(a) = s'(a)$ is the instantaneous velocity at the single moment $t = a$. The average velocity over an interval $[t_1, t_2]$ is $\frac{s(t_2) - s(t_1)}{t_2 - t_1}$, a completely different quantity. For $s(t) = t^2$: $s'(2) = 4$ m/s (instantaneous at $t = 2$) while the average over $[0, 4]$ is $\frac{16 - 0}{4} = 4$ m/s here by coincidence. In general these differ: the average over $[0, 4]$ for $s(t) = t^3$ is $\frac{64}{4} = 16$, but $s'(2) = 12$.

Practice Problems

Level 1 Average vs. Instantaneous

A ball is thrown upward with position function $s(t) = 40t - 5t^2$ meters, where $t$ is in seconds.

(a) Find the average velocity over the interval $[1, 3]$.

(b) Is the instantaneous velocity at $t = 2$ the same as, greater than, or less than this average?

Thought Process

(a) Average velocity = $\frac{s(3) - s(1)}{3 - 1}$. Compute $s(3)$ and $s(1)$, then divide by 2.

(b) The instantaneous velocity at $t = 2$ is $s'(2)$. You can compute this, or reason about whether the ball is speeding up or slowing down in this interval.

Show Answer

(a) Compute positions:

$s(1) = 40(1) - 5(1) = 35$ m
$s(3) = 40(3) - 5(9) = 120 - 45 = 75$ m

Average velocity: $$v_{\text{avg}} = \frac{75 - 35}{3 - 1} = \frac{40}{2} = \boxed{20 \text{ m/s}}$$

(b) Compute $v(2) = s'(2)$: $$s'(t) = \lim_{h \to 0} \frac{[40(t+h) - 5(t+h)^2] - [40t - 5t^2]}{h}$$

$$= \lim_{h \to 0} \frac{40h - 5(2th + h^2)}{h} = \lim_{h \to 0}(40 - 10t - 5h) = 40 - 10t$$

So $v(2) = 40 - 10(2) = 20$ m/s.

The instantaneous velocity at $t = 2$ equals the average velocity over $[1, 3]$!

This is not a coincidence. For a quadratic position function, the instantaneous velocity at the midpoint of an interval equals the average velocity over that interval.

Level 2 Finding Velocity at a Specific Time

A particle moves along a line with position $s(t) = t^3 - 6t$ meters at time $t$ seconds.

Use the limit definition to find the instantaneous velocity at $t = 2$.

Thought Process

Use the velocity formula $v(2) = s'(2) = \lim_{h \to 0} \frac{s(2+h) - s(2)}{h}$.

Compute $s(2+h) = (2+h)^3 - 6(2+h)$
Compute $s(2) = 8 - 12 = -4$
Form the difference, simplify, cancel $h$, take the limit

Show Answer

$$v(2) = \lim_{h \to 0} \frac{s(2+h) - s(2)}{h}$$

Compute $s(2+h)$: $$(2+h)^3 - 6(2+h) = 8 + 12h + 6h^2 + h^3 - 12 - 6h = -4 + 6h + 6h^2 + h^3$$

And $s(2) = -4$.

Difference quotient: $$\frac{(-4 + 6h + 6h^2 + h^3) - (-4)}{h} = \frac{6h + 6h^2 + h^3}{h} = 6 + 6h + h^2$$

Take the limit: $$v(2) = \lim_{h \to 0}(6 + 6h + h^2) = \boxed{6 \text{ m/s}}$$

The particle is moving in the positive direction at 6 m/s when $t = 2$.

Level 3 When Does the Particle Stop?

A ball is thrown upward from ground level with position function $s(t) = 24t - 4t^2$ meters.

(a) Use the limit definition to find a formula for $v(t)$.

(b) At what time does the ball reach its maximum height?

Thought Process

(a) Find $s'(t)$ using the definition with general $t$, not a specific number.

(b) Maximum height occurs when the ball stops going up, i.e., when $v(t) = 0$. Solve $v(t) = 0$.

Show Answer

(a) Find $v(t) = s'(t)$:

$$v(t) = \lim_{h \to 0} \frac{[24(t+h) - 4(t+h)^2] - [24t - 4t^2]}{h}$$

$$= \lim_{h \to 0} \frac{24h - 4(2th + h^2)}{h} = \lim_{h \to 0} \frac{24h - 8th - 4h^2}{h}$$

$$= \lim_{h \to 0}(24 - 8t - 4h) = \boxed{24 - 8t}$$

(b) Set $v(t) = 0$: $$24 - 8t = 0 \implies t = \boxed{3 \text{ seconds}}$$

(c) Maximum height: $$s(3) = 24(3) - 4(9) = 72 - 36 = \boxed{36 \text{ meters}}$$

Level 4 Impact Velocity

A stone is dropped from a bridge 80 meters above a river. Its height above the water is $h(t) = 80 - 5t^2$ meters after $t$ seconds.

(a) How long does it take the stone to hit the water?

(b) What is the velocity of the stone when it hits the water?

Thought Process

(a) The stone hits the water when $h(t) = 0$. Solve $80 - 5t^2 = 0$.

(b) First find $v(t) = h'(t)$ using the limit definition. Then evaluate at the time from part (a).

Show Answer

(a) Solve $h(t) = 0$: $$80 - 5t^2 = 0 \implies t^2 = 16 \implies t = \boxed{4 \text{ seconds}}$$

(We take the positive root since $t \geq 0$.)

(b) Find $v(t) = h'(t)$:

$$v(t) = \lim_{h \to 0} \frac{[80 - 5(t+h)^2] - [80 - 5t^2]}{h}$$

$$= \lim_{h \to 0} \frac{-5(t+h)^2 + 5t^2}{h} = \lim_{h \to 0} \frac{-5(2th + h^2)}{h}$$

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

At $t = 4$: $$v(4) = -10(4) = \boxed{-40 \text{ m/s}}$$

The negative sign indicates downward motion.

(c) Speed is the absolute value: $$\vert v(4)\vert = \vert -40\vert = \boxed{40 \text{ m/s}}$$

Level 5 Comparing Two Objects

Object A moves with position $s_A(t) = t^2 + 2t$ and Object B moves with position $s_B(t) = 3t + 1$ (both in meters, with $t$ in seconds).

(a) At what time(s) do the objects have the same position?

(b) At what time(s) do the objects have the same velocity?

Thought Process

(a) Same position: Solve $s_A(t) = s_B(t)$.

(b) Same velocity: First find $v_A(t)$ and $v_B(t)$ (derivatives), then solve $v_A(t) = v_B(t)$.

Show Answer

(a) Same position: $$t^2 + 2t = 3t + 1$$ $$t^2 - t - 1 = 0$$ $$t = \frac{1 \pm \sqrt{5}}{2}$$

Since $t \geq 0$, only $t = \frac{1 + \sqrt{5}}{2} \approx 1.618$ seconds works.

(b) Find velocities using the limit definition:

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

For $v_B(t)$: $$v_B(t) = \lim_{h \to 0} \frac{3(t+h) + 1 - 3t - 1}{h} = \lim_{h \to 0} 3 = 3$$

Same velocity: $$2t + 2 = 3 \implies t = \boxed{0.5 \text{ seconds}}$$

(c) Objects meet at $t = \frac{1+\sqrt{5}}{2} \approx 1.618$ s.

At this time:

$v_A(1.618) = 2(1.618) + 2 \approx 5.24$ m/s
$v_B(1.618) = 3$ m/s

So Object A is moving faster at the moment it passes Object B (or B passes A).

Since $v_A > v_B$ at the meeting point, Object A comes from behind and passes Object B.

Conceptual Questions (CCI-Style)

Level 2 Interpreting Velocity Sign

A ball is thrown upward. At the instant when the ball reaches its maximum height, what is true?

(A) The velocity is positive and decreasing

(B) The velocity is zero

(D) The velocity is undefined

Thought Process

At the maximum height:

The ball has stopped going up
The ball hasn’t started coming down yet
This is the “turnaround point”

What must the velocity be at this instant?

Show Answer

(B) The velocity is zero.

At the maximum height, the ball is momentarily at rest - it has stopped rising but hasn’t yet started falling. This corresponds to $v(t) = 0$.

Note: The velocity is definitely defined (it’s zero). The ball will immediately start moving downward with negative velocity.

Mastery Checklist

I can distinguish between average velocity and instantaneous velocity
I can compute instantaneous velocity using the limit definition
I can interpret the sign of velocity (positive = forward, negative = backward)
I can find when velocity equals zero (object momentarily at rest)
I can distinguish between velocity (signed) and speed (unsigned)
I can solve problems about objects in free fall

Mental Model

The Speedometer Reading: Think of instantaneous velocity as what a perfect speedometer would show at each instant. The speedometer doesn’t care how fast you were going a second ago or will be going a second from now - it shows your velocity right now.

Mathematically, we can’t measure an instant directly, but we can see what average velocities approach as we measure over shorter and shorter intervals.

Connections

Looking back:

Instantaneous velocity is precisely the derivative of position
The limit process “zooms in” to capture velocity at an instant

Looking ahead:

Taking the derivative of velocity gives acceleration
These ideas generalize to any rate of change, not just motion

Real-world connections:

GPS devices calculate velocity by taking position differences over small time intervals
Radar guns use the Doppler effect but report what is effectively an instantaneous velocity

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Last updated: 2026-01-22