Back to physics. In one dimension velocity and acceleration were numbers along a line. In space they are vectors: $\mathbf{v}(t) = \mathbf{r}'(t)$, speed $|\mathbf{v}(t)|$, and $\mathbf{a}(t) = \mathbf{r}''(t)$. The new question is what acceleration does: part of it speeds you up or slows you down, and part of it turns you.
Key idea
Acceleration splits into a tangential part and a normal part:
$$\mathbf{a} = a_T\,\mathbf{T} + a_N\,\mathbf{N}, \qquad a_T = \frac{d|\mathbf{v}|}{dt} = \frac{\mathbf{v} \cdot \mathbf{a}}{|\mathbf{v}|}, \qquad a_N = \kappa|\mathbf{v}|^2 = \frac{|\mathbf{v} \times \mathbf{a}|}{|\mathbf{v}|}.$$
$a_T$ changes the speed; $a_N \ge 0$ changes the direction.
Why it works
Write the velocity as speed times direction: $\mathbf{v} = |\mathbf{v}|\,\mathbf{T}$. Differentiate with the product rule:
$$\mathbf{a} = \frac{d|\mathbf{v}|}{dt}\,\mathbf{T} + |\mathbf{v}|\,\mathbf{T}'.$$
From the curvature lesson, $\mathbf{T}' = |\mathbf{T}'|\,\mathbf{N} = \kappa|\mathbf{v}|\,\mathbf{N}$. So $\mathbf{a} = \frac{d|\mathbf{v}|}{dt}\mathbf{T} + \kappa|\mathbf{v}|^2\mathbf{N}$. That is why a car taking a curve twice as fast needs four times the sideways force.
Worked examples
Worked example: Velocity, speed and both components
For $\mathbf{r}(t) = \langle t^2, 2t, \ln t \rangle$, find $\mathbf{v}$, the speed, $a_T$ and $a_N$ at $t = 1$.
- $\mathbf{v} = \left\langle 2t, 2, \frac{1}{t} \right\rangle$ and $\mathbf{a} = \left\langle 2, 0, -\frac{1}{t^2} \right\rangle$; at $t = 1$: $\mathbf{v} = \langle 2, 2, 1 \rangle$, $\mathbf{a} = \langle 2, 0, -1 \rangle$.
- Speed $|\mathbf{v}| = 3$.
- $a_T = \frac{\mathbf{v} \cdot \mathbf{a}}{|\mathbf{v}|} = \frac{4 + 0 - 1}{3} = 1$: the speed grows by 1 unit per second per second.
- $\mathbf{v} \times \mathbf{a} = \langle -2, 4, -4 \rangle$ with length ${}6$, so $a_N = \frac{6}{3} = 2$. Check: $a_T^2 + a_N^2 = 1 + 4 = 5 = |\mathbf{a}|^2$.
Answer: $\mathbf{v} = \langle 2, 2, 1 \rangle$, speed ${}3$, $a_T = 1$, $a_N = 2$
Worked example: A projectile
A ball leaves the ground with $\mathbf{v}(0) = \langle 20, 10, 64 \rangle$ ft/s; gravity gives $\mathbf{a} = \langle 0, 0, -32 \rangle$. Where does it land, and how high does it go?
- Integrate: $\mathbf{v}(t) = \langle 20, 10, 64 - 32t \rangle$ and, starting at the origin, $\mathbf{r}(t) = \langle 20t, 10t, 64t - 16t^2 \rangle$.
- Landing: ${}64t - 16t^2 = 16t(4 - t) = 0$ at $t = 4$ s, at $(80, 40, 0)$.
- Top: vertical velocity ${}64 - 32t = 0$ at $t = 2$, height ${}128 - 64 = 64$ ft.
Answer: lands at $(80, 40, 0)$; maximum height ${}64$ ft
Common mistake
Thinking constant speed means no acceleration. A car going around a circle at a steady 30 mph has $a_T = 0$, but its direction keeps changing, so $a_N = \kappa|\mathbf{v}|^2 \ne 0$. Acceleration is the change in the velocity vector. Also: $a_N$ is never negative (it is a length along $\mathbf{N}$), while a negative $a_T$ simply means slowing down. A useful check is $a_T^2 + a_N^2 = |\mathbf{a}|^2$.