Physics · Module 2
Kinematics 2D
Extending motion to a plane — decomposing vectors, coupling independent axes, and describing curved paths through space.
Prerequisites: Module 1 — Kinematics 1D · SUVAT equations · Trigonometry basics
§ 1 Foundation
Vectors in two dimensions
In 1D, direction was just a sign. In 2D, every kinematic quantity becomes a vector with two independent components — one along each axis. The $x$ and $y$ directions are completely decoupled: motion in $x$ has no effect on motion in $y$ and vice versa.
$\vec{r}$position vectorm
$\vec{v}$velocity vectorm s⁻¹
$\vec{a}$acceleration vectorm s⁻²
$\hat{i},\hat{j}$unit vectors (x, y)—
VECTOR NOTATION
$$\vec{r} = x\,\hat{i} + y\,\hat{j}$$
Position is a pair of coordinates. $x, y$ are scalars; the hat vectors carry direction.
$$|\vec{r}| = \sqrt{x^2 + y^2}$$
Magnitude (speed, distance) from Pythagoras. Direction: $\theta = \arctan(y/x)$.
The independence principle. Because $\hat{i} \perp \hat{j}$, the equations of motion for $x$ and $y$ are solved separately and independently. This is the entire foundation of 2D kinematics.
Vector addition follows the parallelogram law: lay two vectors tip-to-tail, the resultant goes from start to end. Subtraction reverses one vector first.
§ 2 Core Technique
Resolving into components
Every 2D problem begins with decomposition: breaking a vector at angle $\theta$ into its $x$ and $y$ parts using trigonometry. Once decomposed, each axis becomes an independent 1D problem solvable with SUVAT.
RESOLVING A VECTOR OF MAGNITUDE |v| AT ANGLE θ
$$v_x = |\vec{v}|\cos\theta$$
Horizontal component. Adjacent side of the right triangle. Maximum when $\theta = 0°$.
$$v_y = |\vec{v}|\sin\theta$$
Vertical component. Opposite side. Maximum when $\theta = 90°$.
X — Horizontal axis
SUVAT applied with $u_x, v_x, a_x$.
Under gravity only: $a_x = 0$, so $v_x$ is constant.
$x(t) = x_0 + u_x t$
Y — Vertical axis
SUVAT applied with $u_y, v_y, a_y$.
Under gravity: $a_y = -g \approx -9.8\ \text{m s}^{-2}$.
$y(t) = y_0 + u_y t - \tfrac{1}{2}g t^2$
Reconstruct the resultant after solving each axis: $|\vec{v}| = \sqrt{v_x^2 + v_y^2}$ and $\theta = \arctan(v_y / v_x)$. Always check the quadrant.
⟂
Independent axes
$x$ and $y$ share only one thing: time $t$. Otherwise they are entirely separate problems.
📐
SOH-CAH-TOA
$\sin$ for the component opposite to $\theta$, $\cos$ for the component adjacent to $\theta$.
⤡
Resultant speed
$|\vec{v}| = \sqrt{v_x^2+v_y^2}$. Speed is always non-negative; velocity carries sign information.
§ 3 Projectile Motion
Projectile motion in full
Projectile motion is the canonical 2D kinematics problem: constant horizontal velocity, uniformly accelerated vertical motion. The two axes are decoupled but share the same elapsed time $t$.
INITIAL CONDITIONS (launch speed v₀, angle θ)
$$u_x = v_0\cos\theta \qquad a_x = 0$$
No horizontal force in idealised projectile (ignores drag).
$$u_y = v_0\sin\theta \qquad a_y = -g$$
Only gravity acts vertically. Positive $y$ is upward.
Equations of motion
$$x(t) = v_0\cos\theta \cdot t$$
$$y(t) = v_0\sin\theta \cdot t - \tfrac{1}{2}g t^2$$
Trajectory — the parabola
$t = \dfrac{x}{v_0\cos\theta}$
Invert $x(t)$ to express time in terms of horizontal position
$y = v_0\sin\theta \cdot \dfrac{x}{v_0\cos\theta} - \tfrac{1}{2}g\left(\dfrac{x}{v_0\cos\theta}\right)^2$
Substitute $t$ into $y(t)$
$$\boxed{y = x\tan\theta - \frac{g}{2v_0^2\cos^2\theta}\,x^2}$$
A parabola opening downward. The coefficient of $x^2$ controls curvature.
Key results
| Quantity | Formula | When |
|---|
| Time of flight $T$ | $T = \dfrac{2v_0\sin\theta}{g}$ | Set $y=0$, trajectory symmetric |
| Max height $H$ | $H = \dfrac{v_0^2\sin^2\theta}{2g}$ | When $v_y = 0$ |
| Range $R$ | $R = \dfrac{v_0^2\sin 2\theta}{g}$ | Horizontal distance at $y=0$ |
| Max range $R_{\max}$ | $R_{\max} = \dfrac{v_0^2}{g}$ | $\theta = 45°$ — since $\sin 2\theta$ peaks at $90°$ |
| Speed at any $t$ | $|\vec{v}| = \sqrt{v_x^2 + (v_y - gt)^2}$ | Vector sum of components |
Complementary angles. Angles $\theta$ and $(90°-\theta)$ give the same range $R$ because $\sin 2\theta = \sin(180°-2\theta)$. A 30° launch and a 60° launch travel the same horizontal distance.
§ 4 Relative Motion
Relative velocity in 2D
Velocity is always measured relative to a reference frame. When the observer is also moving, velocities add as vectors. This is the classical (Galilean) velocity addition rule, valid for speeds much less than $c$.
GALILEAN VELOCITY ADDITION
$$\vec{v}_{A/C} = \vec{v}_{A/B} + \vec{v}_{B/C}$$
Velocity of A relative to C = velocity of A relative to B, plus velocity of B relative to C.
The inner subscripts cancel: $A/B + B/C = A/C$.
Classic examples
✈️
Wind & aircraft
$\vec{v}_\text{ground} = \vec{v}_\text{airspeed} + \vec{v}_\text{wind}$. To fly north in a crosswind, aim slightly into the wind.
🚢
River crossing
Aim perpendicular to reach the opposite bank. Aim upstream to land directly across. The fastest crossing isn't the shortest path.
🚗
Overtaking cars
A car at 100 km/h passes one at 90 km/h at a relative speed of only 10 km/h — negligible from the ground, slow from inside the car.
River crossing problem
A boat aims perpendicular to a river of width $d$ with current $\vec{v}_c$ (along river) and boat speed $\vec{v}_b$ (across river). The resultant velocity has both components:
$$|\vec{v}_\text{res}| = \sqrt{v_b^2 + v_c^2}$$
Resultant speed over ground.
$$\text{drift} = v_c \cdot \frac{d}{v_b}$$
How far downstream the boat lands.
To land directly opposite, aim at angle $\phi = \arcsin(v_c / v_b)$ upstream. This requires $v_b > v_c$ — if the current is faster than the boat, you cannot go straight across.
Key trick: Write out subscripts methodically. $\vec{v}_{B/E}$ = velocity of boat relative to earth. If you know $\vec{v}_{B/W}$ (boat rel. water) and $\vec{v}_{W/E}$ (water rel. earth), add them component-by-component.
§ 5 Circular Motion
Uniform circular motion
An object moving at constant speed around a circle has constant magnitude of velocity but changing direction — which means it is accelerating. This centripetal acceleration always points toward the centre.
Counterintuitive: constant speed does not mean zero acceleration. Acceleration measures the rate of change of the velocity vector, which has both magnitude and direction. Changing direction alone produces acceleration.
$r$radius of circlem
$v$speed (constant)m s⁻¹
$T$period (one orbit)s
$\omega$angular velocityrad s⁻¹
$a_c$centripetal accel.m s⁻²
Angular quantities
$$\omega = \frac{2\pi}{T} = \frac{v}{r}$$
Angular velocity: angle swept per second. $2\pi$ radians per orbit.
$$v = \omega r$$
Tangential speed: faster if you're further from the centre or spinning faster.
Centripetal acceleration
As time $\delta t \to 0$, the velocity vector rotates by $\delta\theta = \omega\,\delta t$
Definition of angular velocity
$|\delta\vec{v}| = v\,\delta\theta = v\omega\,\delta t$
Arc length on the velocity-space circle (radius $v$, angle $\delta\theta$)
$$a_c = \frac{|\delta\vec{v}|}{\delta t} = v\omega = \frac{v^2}{r} = \omega^2 r$$
Direction: always toward the centre (centripetal = "centre-seeking")
CENTRIPETAL ACCELERATION
$$a_c = \frac{v^2}{r}$$
$$a_c = \omega^2 r$$
Position and velocity as functions of time
$$x(t) = r\cos(\omega t + \phi_0)$$
$$y(t) = r\sin(\omega t + \phi_0)$$
Position traces a circle of radius $r$. $\phi_0$ is the starting angle.
$$v_x = -r\omega\sin(\omega t)$$
$$v_y = r\omega\cos(\omega t)$$
Velocity is always tangent to the circle — perpendicular to the radius vector.
| Quantity | Linear | Angular |
|---|
| Displacement | $s$ (m) | $\theta$ (rad) |
| Velocity | $v$ (m s⁻¹) | $\omega$ (rad s⁻¹) |
| Acceleration | $a$ (m s⁻²) | $\alpha$ (rad s⁻²) |
| Bridge equation | $v = \omega r \quad s = \theta r \quad a_t = \alpha r$ |
Two accelerations in non-uniform circular motion. If speed is also changing, there is a tangential acceleration $a_t = \alpha r$ (along the path) in addition to centripetal $a_c = v^2/r$ (toward centre). The total acceleration is $|\vec{a}| = \sqrt{a_c^2 + a_t^2}$.
§ 6 Interactive
Live simulations
Three 2D scenarios — each computed frame-by-frame from the equations above. Drag the sliders to explore.
Projectile motion
2D · gravity only
River crossing — relative velocity
2D · reference frames
Uniform circular motion
2D · centripetal