The Velocity Problem
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 2.1: “A Preview of Calculus” |
| Direct link | https://openstax.org/books/calculus-volume-1/pages/2-1-a-preview-of-calculus |
| Supplementary | OpenStax Calculus Volume 1, Section 3.1: “Defining the Derivative” |
| Supplementary link | https://openstax.org/books/calculus-volume-1/pages/3-1-defining-the-derivative |
| Textbook used in class | Stewart, Calculus, Section 1.4: “The Tangent and Velocity Problems” (Example 3) |
Both OpenStax sources are free and openly licensed.
Key idea
The velocity problem asks what the speedometer shows at a single instant. That is a strange thing to ask, because velocity is distance divided by time, and a single instant has no time interval. The way mathematicians answer it is the same move that solves the tangent problem, and recognizing that they are the same move is the heart of this section.
Average velocity is easy: distance traveled divided by time elapsed, over some interval. Instantaneous velocity, the speedometer reading at one moment, has no interval to divide over. So you approximate. Compute the average velocity over a short interval starting at that instant, then shrink the interval. As the interval gets smaller, the average velocities settle toward a single value. That limiting value is the instantaneous velocity.
This is exactly the secant-to-tangent idea from the tangent problem, wearing different clothes. An average velocity is the slope of a secant line on the position-versus-time graph; the instantaneous velocity is the slope of the tangent line there. Shrinking the time interval is sliding the second point toward the first. So the velocity problem and the tangent problem produce the same limit, and that limit is the derivative. You are meeting the derivative again, now as “how fast is this moving right now?”
Prerequisite Check
Before this lesson, make sure you can do all of the following:
If average velocity over an interval is not yet automatic, practice it first. Instantaneous velocity is built by computing average velocities over shrinking intervals.
Quick Reference
Average velocity. Over the time interval from $a$ to $a + h$, for a position function $s(t)$, \[ \text{average velocity} = \frac{s(a + h) - s(a)}{h}. \] This is the slope of the secant line on the position graph.
Instantaneous velocity. The velocity at the single instant $t = a$ is the limit of average velocities over shrinking intervals: \[ v(a) = \lim_{h \to 0}\frac{s(a + h) - s(a)}{h}. \] This is the slope of the tangent line to the position graph at $t = a$.
The procedure.
- Write the average velocity $\dfrac{s(a+h) - s(a)}{h}$ over an interval starting at $a$.
- Compute it for smaller and smaller $h$ (a table), or simplify and take the limit.
- The value the averages approach is the instantaneous velocity.
Key fact. Instantaneous velocity is the derivative of position, and it equals the slope of the tangent to the position graph: the velocity problem is the tangent problem.
Key Concepts
1. Why a Single Instant Has No Velocity by Itself
Velocity is a ratio of a distance to a time. At a single instant there is no elapsed time and no covered distance, so the ratio $\frac{0}{0}$ is meaningless. The fix is to approximate the instant with a short interval and then shrink it.
Average velocity over an interval is well defined: how far you went divided by how long it took. To get the velocity at one instant, compute the average velocity over a brief interval starting there, then make the interval shorter and shorter. The averages converge, and the value they converge to is defined to be the instantaneous velocity.
2. The Falling-Ball Example: Average Velocities Closing In
Galileo found that a freely falling body covers a distance proportional to the square of the elapsed time. Near the earth’s surface, distance fallen in $t$ seconds is $s(t) = 4.9t^2$ meters.
Example 1. A ball is dropped from the observation deck of a tower, $450$ meters above the ground. Find its velocity after $5$ seconds. (This is Stewart 1.4, Example 3.)
Goal. Compute the average velocity over short intervals starting at $t = 5$, then see what they approach.
Over $[5, 5.1]$ the average velocity is \[ \frac{s(5.1) - s(5)}{0.1} = \frac{4.9(5.1)^2 - 4.9(5)^2}{0.1} = 49.49 \text{ m/s}. \] Shrinking the interval:
| Time interval | Average velocity (m/s) |
|---|---|
| $5 \leq t \leq 5.1$ | $49.49$ |
| $5 \leq t \leq 5.05$ | $49.245$ |
| $5 \leq t \leq 5.01$ | $49.049$ |
| $5 \leq t \leq 5.001$ | $49.0049$ |
The average velocities close in on $49$, so the instantaneous velocity at $t = 5$ is \[ v(5) = \lim_{h\to 0}\frac{s(5 + h) - s(5)}{h} = 49 \text{ m/s}. \]
Boxed answer: the velocity after $5$ seconds is $49$ m/s.
Recap. The instantaneous velocity is read off the table as the value the averages approach. The shorter the interval, the closer the average velocity is to the true instantaneous value, because over a tiny interval the speed barely changes.
Important: instantaneous velocity is the limit, not any single average. Each average velocity (like $49.49$ over a tenth of a second) is only an approximation over a finite interval. The instantaneous velocity $49$ is what the averages approach as the interval shrinks to zero. Reporting a finite-interval average as the instantaneous velocity confuses the approximation with the limit.
3. Average Velocity Is a Secant Slope
The connection to the tangent problem is exact, not merely an analogy.
Plot position $s$ against time $t$. Take the two points $P(5, s(5))$ and $Q(5 + h, s(5 + h))$ on that graph. The slope of the secant line $PQ$ is \[ m_{PQ} = \frac{s(5 + h) - s(5)}{(5 + h) - 5} = \frac{s(5 + h) - s(5)}{h}, \] which is exactly the average velocity over $[5, 5 + h]$. Shrinking the interval ($h \to 0$) slides $Q$ toward $P$, so the secant approaches the tangent. Therefore the instantaneous velocity (the limit of average velocities) equals the slope of the tangent line to the position graph at $t = 5$.
So the velocity problem and the tangent problem are the same limit of the same kind of difference quotient. Solve one and you have solved both.
4. Instantaneous Velocity Is the Derivative
The limit defining instantaneous velocity is, once again, the derivative.
The instantaneous velocity of an object with position $s(t)$ at time $t = a$ is \[ v(a) = \lim_{h\to 0}\frac{s(a + h) - s(a)}{h}, \] which is the derivative $s'(a)$. In Chapter 2 this limit is named and given computational rules; here it is the velocity read straight off the definition. Because it is also the slope of the tangent to the position graph, the same machinery that handles tangents handles velocities, accelerations, and every other instantaneous rate of change.
each average velocity IS the instantaneous velocity at one of the endpoints.
This is the rate-as-fixed-number error. The average velocity $49.49$ m/s over the interval $[5, 5.1]$ is not the instantaneous velocity at $t=5$ or at $t=5.1$ -- it is the constant rate that would cover the same distance in the same time. No single average gives the instantaneous rate; the instantaneous velocity is the limit that the averages approach as the interval shrinks. Reporting the average over $[5, 5.001]$ as the answer is still only an approximation. The exact answer $49$ comes from taking the limit, not from using any particular small-but-nonzero interval.
the limiting process gets you “infinitely close” to a barrier but never crosses it, so instantaneous velocity remains undefined.
This is the limit-as-unreachable-barrier error. Students sometimes argue that because no interval is truly zero, the limit never “arrives” and instantaneous velocity is just an idealization with no real value. But the limit is defined exactly: it is the unique number that the averages get arbitrarily close to. The process of shrinking the interval is a METHOD for finding that number, not a description of an infinite journey. The instantaneous velocity at $t=5$ is exactly $49$ m/s -- a specific, well-defined number -- because that is what the average velocities converge to.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Reporting an average velocity as instantaneous | answering $49.49$ for the ball | The instantaneous velocity is the limit, $49$, not a finite-interval average |
| Trying to divide over a single instant | computing velocity at $t = 5$ as $\frac{0}{0}$ | Use a shrinking interval and take the limit |
| Using too large an interval | estimating from $[5, 5.1]$ only | Shrink the interval; smaller $h$ gives a closer estimate |
| Forgetting the connection to slope | treating velocity and tangent slope as unrelated | Average velocity is a secant slope; instantaneous velocity is a tangent slope |
| Sign confusion for downward or backward motion | reporting a negative speed as positive | The sign of the velocity gives direction; keep it |
Leveled Practice
Level 1 -- Direct Application
Problem 1. For the falling ball $s(t) = 4.9t^2$, compute the average velocity over $[5, 5.05]$.
Show answer
\[ \frac{s(5.05) - s(5)}{0.05} = \frac{4.9(5.05)^2 - 4.9(5)^2}{0.05} = \frac{4.9(25.5025 - 25)}{0.05} = \frac{4.9(0.5025)}{0.05} = 49.245 \text{ m/s}. \]
Boxed answer: $49.245$ m/s. (This matches the table in Example 1.)
Problem 2. From the table in Example 1, the average velocities over $[5, 5.1]$, $[5, 5.05]$, $[5, 5.01]$, $[5, 5.001]$ are $49.49$, $49.245$, $49.049$, $49.0049$. What instantaneous velocity do these suggest at $t = 5$?
Show answer
As the interval shrinks, the average velocities decrease toward $49$.
Boxed answer: the instantaneous velocity at $t = 5$ is $49$ m/s.
Problem 3. Explain why the average velocity over $[5, 5.001]$ is a better estimate of the instantaneous velocity at $t = 5$ than the average over $[5, 5.1]$.
Show answer
Over a shorter interval the speed changes less, so the average over that interval is closer to the speed at the single instant $t = 5$. The interval $[5, 5.001]$ is one hundred times shorter than $[5, 5.1]$, so its average ($49.0049$) is much closer to the limiting value $49$ than the longer interval’s average ($49.49$).
Boxed answer: a shorter interval gives an average velocity closer to the instantaneous velocity, because the velocity barely changes over a tiny interval.
Level 2 -- Multiple Steps
Problem 4. A pebble falls off a bridge; its height after $t$ seconds is $y = 275 - 16t^2$ feet. Find the average velocity over $[4, 4.1]$, $[4, 4.05]$, and $[4, 4.01]$, then estimate the instantaneous velocity at $t = 4$.
Show answer
With $y(4) = 275 - 16(16) = 19$:
- $[4, 4.1]$: $y(4.1) = 275 - 16(16.81) = 5.04$, so average $= \dfrac{5.04 - 19}{0.1} = -139.6$ ft/s.
- $[4, 4.05]$: average $= -128.8$ ft/s.
- $[4, 4.01]$: average $= -128.16$ ft/s.
The averages approach $-128$.
Boxed answer: the instantaneous velocity at $t = 4$ is $-128$ ft/s (downward at $128$ ft/s). (This is Stewart 1.4, Exercise 5.)
Problem 5. A rock is thrown upward on Mars; its height after $t$ seconds is $y = 10t - 1.86t^2$ meters. Find the average velocity over $[1, 1.1]$ and $[1, 1.01]$, then estimate the instantaneous velocity at $t = 1$.
Show answer
With $y(1) = 10 - 1.86 = 8.14$:
- $[1, 1.1]$: $y(1.1) = 11 - 1.86(1.21) = 8.7494$, so average $= \dfrac{8.7494 - 8.14}{0.1} = 6.094$ m/s.
- $[1, 1.01]$: $y(1.01) = 10.1 - 1.86(1.0201) = 8.202614$, so average $= \dfrac{8.202614 - 8.14}{0.01} = 6.2614$ m/s.
The averages approach about $6.28$.
Boxed answer: the instantaneous velocity at $t = 1$ is about $6.28$ m/s upward. (This is Stewart 1.4, Exercise 6.)
Problem 6. The table shows the position $s$ (feet) of a motorcyclist after accelerating from rest. Find the average velocity over $[3, 4]$ and $[4, 5]$, and estimate the instantaneous velocity at $t = 4$ by averaging them.
| $t$ (s) | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ |
|---|---|---|---|---|---|---|---|
| $s$ (ft) | $0$ | $4.9$ | $20.6$ | $46.5$ | $79.2$ | $124.8$ | $176.7$ |
Show answer
\[ \text{over } [3,4]: \frac{79.2 - 46.5}{4 - 3} = 32.7 \text{ ft/s}, \qquad \text{over } [4,5]: \frac{124.8 - 79.2}{5 - 4} = 45.6 \text{ ft/s}. \] Average the two (one interval on each side of $t = 4$): \[ \frac{1}{2}(32.7 + 45.6) = 39.15 \text{ ft/s}. \]
Boxed answer: the instantaneous velocity at $t = 4$ is about $39.15$ ft/s. (This is the form of Stewart 1.4, Exercise 7.)
Level 3 -- Deeper Problems
Problem 7. For the falling ball $s(t) = 4.9t^2$, show algebraically that the instantaneous velocity at $t = 5$ is exactly $49$ m/s by simplifying the average-velocity difference quotient and taking the limit.
Show answer
The average velocity over $[5, 5 + h]$ is \[ \frac{4.9(5 + h)^2 - 4.9(5)^2}{h}. \] Expand $(5 + h)^2 = 25 + 10h + h^2$: \[ = \frac{4.9(25 + 10h + h^2) - 4.9(25)}{h} = \frac{4.9(10h + h^2)}{h} = \frac{4.9h(10 + h)}{h} = 4.9(10 + h) \quad (h \neq 0). \] Take the limit: \[ v(5) = \lim_{h\to 0} 4.9(10 + h) = 4.9(10) = 49 \text{ m/s}. \]
Boxed answer: the instantaneous velocity is exactly $49$ m/s, confirming the table. The $h$ cancels, which is why the limit is clean.
Problem 8. For the same ball, show that the instantaneous velocity at a general time $t = a$ is $9.8a$ m/s.
Show answer
The average velocity over $[a, a + h]$ is \[ \frac{4.9(a + h)^2 - 4.9 a^2}{h} = \frac{4.9(a^2 + 2ah + h^2) - 4.9 a^2}{h} = \frac{4.9(2ah + h^2)}{h} = 4.9(2a + h) \quad (h \neq 0). \] Take the limit: \[ v(a) = \lim_{h\to 0} 4.9(2a + h) = 9.8a \text{ m/s}. \]
Boxed answer: $v(a) = 9.8a$. At $a = 5$ this gives $49$, matching Example 1. The number $9.8$ is the acceleration of gravity; the velocity of a dropped object grows linearly with time, which is the derivative of the quadratic position $4.9t^2$.
Problem 9. A particle moves with position $s(t) = 2\sin(\pi t) + 3\cos(\pi t)$ centimeters. The average velocities over $[1, 1.1]$, $[1, 1.01]$, $[1, 1.001]$ are approximately $-4.712$, $-6.134$, $-6.268$ cm/s. What is the instantaneous velocity at $t = 1$, and what does its sign tell you?
Show answer
The average velocities are increasing in magnitude toward about $-6.28$ cm/s as the interval shrinks.
Boxed answer: the instantaneous velocity at $t = 1$ is about $-6.28$ cm/s. (This is Stewart 1.4, Exercise 8.) The negative sign means the particle is moving in the negative direction (back toward smaller $s$) at that instant. Velocity carries direction, not just speed; the magnitude $6.28$ cm/s is the speed, and the minus sign is the direction.
Mastery Checklist
You have mastered this skill when you can do all of the following without referring to notes:
Mental Model
Think of instantaneous velocity as the speedometer reading squeezed out of shrinking trips.
You cannot measure speed at a frozen instant directly, the way a photograph cannot show motion. But you can time a very short trip starting at that instant and divide distance by time. Make the trip shorter and shorter, and the average speeds converge on a single number: the speedometer reading at that instant.
Three things follow from the picture:
- Instantaneous velocity is a destination, not one measurement. Each average velocity is one short-trip timing; the instantaneous velocity is what those timings converge to as the trip shrinks to nothing. That is why it is a limit.
- It is the same limit as the tangent slope. Average velocity is the slope of a secant on the position graph, and instantaneous velocity is the slope of the tangent. Shrinking the time interval is sliding the secant into the tangent, so the velocity problem and the tangent problem are one problem.
- This single limit is the derivative. Instantaneous velocity, the slope of the position graph’s tangent, and the derivative $s'(a)$ are three names for the same number. Meeting it here as “how fast, right now” is meeting the derivative through motion.
Connections
Within Functions and Limits (Chapter 1)
- The tangent problem: The companion node in this same section. Average velocity is a secant slope and instantaneous velocity is a tangent slope, so the two problems are identical limits. Author and reader should hold them as one idea.
- Evaluating functions and the difference quotient: The average velocity $\dfrac{s(a+h) - s(a)}{h}$ is a difference quotient built by evaluating the position function at two times. This is the same algebraic object that appears throughout the derivative.
- Limits: Instantaneous velocity is defined as a limit, which is why the limit machinery of this chapter must come first. Over a single instant the ratio is $\frac00$, so the limit is unavoidable.
Toward Calculus (MATH161)
- The derivative: In Chapter 2 the velocity limit $\lim_{h\to 0}\dfrac{s(a+h) - s(a)}{h}$ is named the derivative $s'(a)$. Velocity is the first and most physical interpretation of the derivative.
- Average versus instantaneous rate of change: The split between average velocity (over an interval) and instantaneous velocity (at an instant) generalizes to every quantity: average versus instantaneous rate of change, the difference quotient versus the derivative.
- Higher derivatives and motion: Differentiating velocity gives acceleration, so the same limiting process applied twice describes how motion speeds up or slows down. The study of position, velocity, and acceleration is built entirely on the limit introduced here.
Audience Notes
For students who find math intimidating: You already know how to find average speed: distance over time. To get the speed at one instant, compute the average speed over a tiny interval starting there, then over an even tinier one, and watch the numbers settle. The value they settle on is the instantaneous speed. The table does the work; you just read where it is heading.
For career-focused students: This is exactly how speed is computed from position data in a GPS, a fitness tracker, or a control system: take the change in position over a short sampling interval and divide by the elapsed time. The shorter the sampling interval, the closer that finite difference is to the true instantaneous velocity, which is the engineering version of the limit in this lesson.
For gifted and curious students: Problem 8 shows that the velocity of a freely falling object, position $4.9t^2$, is $9.8t$, derived purely from the average-velocity limit. The constant $9.8$ is the acceleration of gravity, and you have just differentiated a quadratic by hand. Try it for a cubic position function and watch a quadratic velocity emerge.
For PhD-track students: That average velocity and tangent slope are the same limit is the geometric content of the derivative, and it is why velocity is well defined exactly when the position function is differentiable. Motion with a corner in its position graph (an instantaneous reversal) has no velocity at that instant, the physical face of non-differentiability, and the precise study of when this limit exists is the analytic foundation of kinematics.
Back to Functions and Limits | Related: The Tangent Problem | Next: Average vs Instantaneous Rate