In two or three dimensions, position is a vector that changes with time, $\vec{r}(t) = \langle x(t), y(t), z(t)\rangle$: a set of Parametric Equations. Velocity and acceleration are its derivatives, and since Adding Vectors by Components works component by component, so does calculus.
Key idea
$$\vec{v} = \dfrac{d\vec{r}}{dt} = \left\langle \dfrac{dx}{dt}, \dfrac{dy}{dt}, \dfrac{dz}{dt} \right\rangle, \qquad \vec{a} = \dfrac{d\vec{v}}{dt}.$$ The velocity is always tangent to the path, and the speed is $|\vec{v}|$. The acceleration points wherever the velocity is changing, which on a curved path is toward the inside of the curve, not along the path.
Why it works
A derivative of a vector is the limit of $\Delta\vec{r}/\Delta t$, and subtracting two vectors subtracts their components. So each component is differentiated on its own. The velocity is tangent because $\Delta\vec{r}$ is a chord of the path, and chords over shorter and shorter times line up with the tangent. The acceleration is the limit of $\Delta\vec{v}/\Delta t$; when the direction of $\vec{v}$ turns, $\Delta\vec{v}$ points toward the side the path is bending, even if the speed stays the same.
Worked example: Velocity and speed
A drone has $\vec{r}(t) = \langle 2t^2 - 3t,\ t^3 - 4t \rangle$ (m, s). Find its velocity and speed at $t = 2\ \text{s}$.
- $\vec{v}(t) = \langle 4t - 3,\ 3t^2 - 4 \rangle$.
- At $t = 2$: $\vec{v} = \langle 5, 8 \rangle\ \text{m/s}$.
- Speed $= \sqrt{5^2 + 8^2} = \sqrt{89} \approx 9.43\ \text{m/s}$.
Answer: $\vec{v} = \langle 5, 8 \rangle\ \text{m/s}$; speed $\approx 9.43\ \text{m/s}$.
Worked example: Is the acceleration along the motion?
For the same drone, find $\vec{a}(2)$ and the angle between $\vec{v}$ and $\vec{a}$.
- $\vec{a}(t) = \langle 4,\ 6t \rangle$, so $\vec{a}(2) = \langle 4, 12 \rangle\ \text{m/s}^2$.
- $$\cos\theta = \dfrac{\vec{v}\cdot\vec{a}}{|\vec{v}||\vec{a}|} = \dfrac{20 + 96}{9.434 \times 12.65} = 0.972$$
- $\theta \approx 13.6^\circ$. Less than 90°, so the drone speeds up; not zero, so its path bends.
Answer: $\vec{a} = \langle 4, 12 \rangle\ \text{m/s}^2$, about $13.6^\circ$ from $\vec{v}$.
Common mistake
Assuming the acceleration points along the velocity. That's only true on a straight path. A car rounding a bend at steady speed has an acceleration pointing sideways, into the bend. Check with components: if $\vec{a}$ were along $\vec{v}$, their cross product (or $v_xa_y - v_ya_x$ in two dimensions) would be zero. For the drone above, $5(12) - 8(4) = 28 \neq 0$.
The component of $\vec{a}$ along $\vec{v}$ changes the speed; the component perpendicular to $\vec{v}$ bends the path. You'll split these exactly in Tangential and Centripetal Acceleration.