Motion in a Plane: Quick Revision

1. Vectors — Definitions & Equality

Scalar: magnitude only (distance, mass, time). Combines by ordinary algebra.
Vector: magnitude and direction (displacement, velocity, force). Combines by the triangle/parallelogram law.
Two vectors are equal only if magnitude and direction both match.

Path length $\geq$ $|\text{displacement}|$ , equal only when the object never changes direction.

2. Vector Addition & Subtraction

\mathbf{A}+\mathbf{B}=\mathbf{B}+\mathbf{A}, \qquad (\mathbf{A}+\mathbf{B})+\mathbf{C}=\mathbf{A}+(\mathbf{B}+\mathbf{C})

$\mathbf{A}+(-\mathbf{A})=\mathbf{0}$ (no defined direction). Subtraction: $\mathbf{A}-\mathbf{B}=\mathbf{A}+(-\mathbf{B})$ .

Resultant of two velocities on one object: $\mathbf{v}=\mathbf{v}_1+\mathbf{v}_2$ — addition.
Relative velocity of 1 w.r.t. 2: $\mathbf{v}_{1/2}=\mathbf{v}_1-\mathbf{v}_2$ — subtraction. Easy to conflate with the above.

3. Resolving a Vector into Components

\mathbf{A} = A_x\hat{\mathbf{i}} + A_y\hat{\mathbf{j}}, \qquad A_x = A\cos\theta,\;\; A_y = A\sin\theta

A = \sqrt{A_x^2+A_y^2}, \qquad \theta = \tan^{-1}\!\left(\dfrac{A_y}{A_x}\right)

Sign of each component depends only on which quadrant $\theta$ falls in:

Quadrant I $(0^\circ\text{-}90^\circ)$ : $A_x>0,\; A_y>0$ .
Quadrant II $(90^\circ\text{-}180^\circ)$ : $A_x<0,\; A_y>0$ .
Quadrant III $(180^\circ\text{-}270^\circ)$ : $A_x<0,\; A_y<0$ .
Quadrant IV $(270^\circ\text{-}360^\circ)$ : $A_x>0,\; A_y<0$ .

4. Velocity & Acceleration in a Plane

\mathbf{v} = \dfrac{d\mathbf{r}}{dt} = v_x\hat{\mathbf{i}} + v_y\hat{\mathbf{j}}, \qquad v_x = \dfrac{dx}{dt},\;\; v_y = \dfrac{dy}{dt}

\mathbf{a} = \dfrac{d\mathbf{v}}{dt} = a_x\hat{\mathbf{i}} + a_y\hat{\mathbf{j}}, \qquad a_x = \dfrac{dv_x}{dt},\;\; a_y = \dfrac{dv_y}{dt}

Velocity is always tangent to the path. Unlike straight-line motion, the angle between $\mathbf{v}$ and $\mathbf{a}$ can be anywhere from $0^\circ$ to $180^\circ$ .

5. Motion with Constant Acceleration

\begin{gathered} \mathbf{v} = \mathbf{v}_0 + \mathbf{a}t \\[6px] \mathbf{r} = \mathbf{r}_0 + \mathbf{v}_0 t + \tfrac{1}{2}\mathbf{a}t^2 \end{gathered}

\begin{gathered} v_x = v_{0x} + a_x t, \qquad x = x_0 + v_{0x}t + \tfrac{1}{2}a_x t^2 \\[6px] v_y = v_{0y} + a_y t, \qquad y = y_0 + v_{0y}t + \tfrac{1}{2}a_y t^2 \end{gathered}

The two axes are independent: two ordinary 1D problems running at once.

6. Projectile Motion — Key Results

x(t) = (v_0\cos\theta_0)\,t, \qquad y(t) = (v_0\sin\theta_0)\,t - \tfrac{1}{2}gt^2

\begin{gathered} T_f = \dfrac{2v_0\sin\theta_0}{g}, \qquad h_m = \dfrac{v_0^2\sin^2\theta_0}{2g} \\[8px] R = \dfrac{v_0^2\sin 2\theta_0}{g} \end{gathered}

Maximum range: at $\theta_0=45^\circ$ (fixed launch speed).
Equal range: $\theta_0$ and $90^\circ-\theta_0$ give the same $R$ (same $\sin 2\theta_0$ ).
Caveat: both assume equal launch and landing height — see the concept page for the asymmetric case, where 45° is no longer optimal.

7. Uniform Circular Motion

Centripetal acceleration

a_c = \dfrac{v^2}{R} = \omega^2 R

$\omega = \Delta\theta/\Delta t$ , and since arc length $=R\,\Delta\theta$ , $v=\omega R$ .

\omega = 2\pi\nu, \qquad v = 2\pi R\nu, \qquad a_c = 4\pi^2\nu^2 R

$\mathbf{a}_c$ always points toward the centre; its magnitude is constant but its direction keeps changing, so it is not a constant vector — the constant-acceleration equations of §5 do not apply to circular motion.

Full derivations and worked examples: detailed notes →