Part 2: Average Speed and Speed at a Moment

In Part 1, speed at a moment was the number that makes the short-time line formula work. Now let us connect that idea to the more familiar idea of average speed.

Suppose your position is $x(t) = t^2$. This means you are speeding up as time goes on. Let us focus on the same moment as before, $t = 10$.

If you start at time $t$ and look again at time $t+h$, then the average speed over that interval is

$$\frac{x(t+h) - x(t)}{h}.$$

So the answer depends on which second time you choose. At time 10, the calculation looks like this.

• From time 10 to time 11 (h = 1), the average speed is (121 - 100) / 1 = 21.
• From time 10 to time 10.5 (h = 0.5), the average speed is (110.25 - 100) / 0.5 = 20.5.
• From time 10 to time 10.1 (h = 0.1), the average speed is (102.01 - 100) / 0.1 = 20.1.
• From time 10 to time 10.01 (h = 0.01), the average speed is (100.2001 - 100) / 0.01 = 20.01.

As the second point moves closer and closer to 10, the average speeds get closer and closer to 20. So it is natural to expect that the speed at time 10 is 20.

Now go back to the short-time formula from Part 1. We want the coefficient of $h$ in the expansion of $x(10+h)$. Since

$$x(10+h) = (10+h)^2,$$

we get

$$x(10+h) = (10+h)^2 = 100 + 20h + h^2.$$

Because $x(10) = 100$, this can be rewritten as

$$x(10+h) = x(10) + 20h + A(h^2).$$

So the coefficient of $h$ is 20, which means $v(10) = 20$. This is exactly the value the average speeds were approaching.

The same algebra works at any time. Starting from a general time $t$, we have

$$x(t+h) = (t+h)^2 = t^2 + 2th + h^2,$$

so

$$x(t+h) = x(t) + 2t\,h + A(h^2).$$

That means the speed at time $t$ is

$$v(t) = 2t.$$

This no longer depends on $h$. Average speed depends on two times, so it changes when you change the interval. Speed at a moment depends on just one time, and it is the number the average speeds are settling down to.