Kinematics §

The polar form separates radial and tangential components. The term $-r\dot{\theta}^2$ is the centripetal acceleration (always inward), and $2\dot{r}\dot{\theta}$ is the Coriolis term (appears when the radial distance changes while the particle also rotates). In 3-D, spherical coordinates add a $\phi$-component; the full expression follows the same pattern: each coordinate’s second derivative plus cross-terms from the changing basis vectors.

For uniform circular motion ($r = \text{const}$, $\dot{r} = 0$), the formulas reduce to $\vec{v} = r\dot{\theta}\,\hat{\theta}$ and $\vec{a} = -r\dot{\theta}^2\,\hat{r}$, recovering the familiar centripetal-acceleration result. This connects directly to the Frenet–Serret decomposition below.

Uniform circular motion. A particle on a circle of radius $R$ at constant angular speed $\omega$ has $\dot{s} = R\omega$ (constant) and $\rho = R$. The tangential component vanishes ($\ddot{s} = 0$) and the entire acceleration is centripetal:

This is the special case where the Frenet normal $\hat{n}$ coincides with the inward radial direction $-\hat{r}$.

Projectile motion. For a projectile under gravity, the tangential component is $g\sin\alpha$ (where $\alpha$ is the angle between the velocity and the horizontal) and the normal component is $g\cos\alpha$. At the apex, $\alpha = 0$ and the full acceleration is normal (centripetal); at launch and landing, the tangential component is largest.

The Frenet–Serret frame is the natural language for any curved trajectory and is the bridge between the polar-coordinate formulas above and the general vector description.

These four equations are algebraically equivalent; any three determine the fourth. They apply to free fall ($a = -g$), uniformly accelerated vehicles, and any 1-D motion with constant acceleration. In 2-D or 3-D with constant $\vec{a}$, each component satisfies the same set independently.

Example: A ball thrown upward at $v_0 = 20\;\text{m/s}$. Time to peak: $t = v_0/g \approx 2.04\;\text{s}$. Maximum height: $h = v_0^2/(2g) \approx 20.4\;\text{m}$.

For constant acceleration the integrals evaluate to the closed-form equations in the previous section. For time-dependent acceleration (e.g., drag $\vec{a} = -k\vec{v}$), the integrals must be solved analytically or numerically. In 2-D and 3-D the integration is performed component-wise.

This inverse direction is what a student uses most often: the problem gives forces (hence acceleration) and initial conditions, and asks for the trajectory.

Ball dropped from a moving train. A passenger on a train moving at constant $\vec{u}$ drops a ball. In the train frame $S'$, the ball falls straight down: $x' = 0$, $y' = -\tfrac{1}{2}gt^2$. In the ground frame $S$, the ball follows a parabola: $x = ut$, $y = -\tfrac{1}{2}gt^2$. Both observers measure the same acceleration $\vec{a} = \vec{a}' = (0, -g)$ and the same time of flight. The trajectories differ, but the physics (acceleration, time) is invariant.

Acceleration is invariant under Galilean transformations, which is why Newton’s laws take the same form in all inertial frames. This invariance breaks down at relativistic speeds; see Interactions for the Lorentz transformation.

Decompose: $v_{0x} = v_0\cos\theta$, $v_{0y} = v_0\sin\theta$.

Time of flight. Using $y = y_0 + v_{0y}\,t + \tfrac{1}{2}a_y\,t^2$ with $y_0 = 0$, $a_y = -g$, and setting $y = 0$ at landing:

Range. Using $x = x_0 + v_{0x}\,t$ (no horizontal acceleration) with $t = T$:

Maximum height. Using $v_y^2 = v_{0y}^2 + 2a_y\,\Delta y$ with $v_y = 0$ at the peak:

Maximum range occurs at $\theta = 45°$ (since $\sin 2\theta$ is maximized there).