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.
| Property | Value |
|---|---|
| Concept | Tangent and Velocity Problems |
| Chapter | 1.4 |
| Difficulty | Intermediate |
| Time | ~20 minutes |
| 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 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:
position s(t)
│
│ ·
│ ·
│ · · curve s(t)
│ · ╱
│ · ╱ tangent line
│ · ╱
│· ╱
└─────────────────── time t
↑
t = a
The slope of the tangent = instantaneous velocity at t = aThe slope of the position-time graph at any point equals the instantaneous velocity at that time.
| 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.
| 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.
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.
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.)
A particle moves along a line with position $s(t) = t^2 + 3t$ meters. Find the instantaneous velocity at $t = 2$ seconds.
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.
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 $v(t) = 0$?)
(c) What is the maximum height?
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 $t = 1.5$.
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.
Looking back:
Looking ahead:
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.
| Previous | Up | Next |
|---|---|---|
| Tangent Slope via Limits | Skills Index | Derivative Definition |
Last updated: 2026-01-22