Kinematics with vectors in two dimensionsEdexcel A-Level Maths: Revision notes

In two dimensions, position, velocity and acceleration are vectors. Write them with perpendicular unit vectors $\mathbf{i}$ and $\mathbf{j}$, or as column vectors, so $(3\mathbf{i}-2\mathbf{j})$ and $\begin{pmatrix} 3 \\ -2 \end{pmatrix}$ mean the same thing.

• Position vector $\mathbf{r}$: location relative to a fixed origin $O$.
• Displacement $\mathbf{s}$: change in position, $\mathbf{r}_2-\mathbf{r}_1$.
• Velocity $\mathbf{v}$: rate of change of position (m s$^{-1}$). Speed is its magnitude, $|\mathbf{v}|=\sqrt{v_x^2+v_y^2}$.
• Acceleration $\mathbf{a}$: rate of change of velocity (m s$^{-2}$). The direction of motion is the direction of $\mathbf{v}$: the angle with $\mathbf{i}$ is $\tan^{-1}\left(\frac{v_y}{v_x}\right)$, taking care over the quadrant.

When $\mathbf{a}$ is constant, the straight-line formulae hold for vectors, applied to the whole vector (or to the $\mathbf{i}$ and $\mathbf{j}$ components separately): $\mathbf{v}=\mathbf{u}+\mathbf{a}t,\qquad \mathbf{s}=\mathbf{u}t+\tfrac12\mathbf{a}t^2,\qquad \mathbf{r}=\mathbf{r}_0+\mathbf{u}t+\tfrac12\mathbf{a}t^2.$ Worked example: $\mathbf{u}=(2\mathbf{i}-3\mathbf{j})$ and $\mathbf{a}=(4\mathbf{i}+2\mathbf{j})$. At $t=3$, $\mathbf{v}=(2+12)\mathbf{i}+(-3+6)\mathbf{j}=14\mathbf{i}+3\mathbf{j}$. At $t=2$ the displacement is $\mathbf{s}=(4\mathbf{i}-6\mathbf{j})+\frac12(4\mathbf{i}+2\mathbf{j})(4)=12\mathbf{i}-2\mathbf{j}$.

Many questions reduce to a condition on one component:

• 'Moving in the direction of $\mathbf{i}$' (or parallel to $\mathbf{i}$): the $\mathbf{j}$ component of $\mathbf{v}$ is zero.
• 'Moving in the direction of $\mathbf{j}$': the $\mathbf{i}$ component of $\mathbf{v}$ is zero.
• 'Instantaneously at rest': both components of $\mathbf{v}$ are zero at the same $t$.
• 'On the line through $O$ parallel to $\mathbf{i}$': the $\mathbf{j}$ component of $\mathbf{r}$ is zero. Example: a particle is at $(3\mathbf{i}-2\mathbf{j})$ with $\mathbf{u}=(5\mathbf{i}+2\mathbf{j})$, and at $t=4$ is at $(35\mathbf{i}+14\mathbf{j})$. Then $32\mathbf{i}+16\mathbf{j}=4\mathbf{u}+8\mathbf{a}$, so $\mathbf{a}=(1.5\mathbf{i}+\mathbf{j})$ and $\mathbf{v}(4)=11\mathbf{i}+6\mathbf{j}$, speed $\sqrt{157}\approx12.5$ m s$^{-1}$.

When position (or velocity) is a function of $t$, differentiate each component separately: $\mathbf{v}=\frac{d\mathbf{r}}{dt},\qquad \mathbf{a}=\frac{d\mathbf{v}}{dt}=\frac{d^2\mathbf{r}}{dt^2}.$ Example: $\mathbf{r}=(t^3-6t^2+9t)\mathbf{i}+(2t^2-8t)\mathbf{j}$ gives $\mathbf{v}=(3t^2-12t+9)\mathbf{i}+(4t-8)\mathbf{j}$ and $\mathbf{a}=(6t-12)\mathbf{i}+4\mathbf{j}$. At $t=1$, $\mathbf{v}=-4\mathbf{j}$ (moving parallel to $\mathbf{j}$ at $4$ m s$^{-1}$) and $\mathbf{a}=-6\mathbf{i}+4\mathbf{j}$. Substitute the value of $t$ only after differentiating.

Going the other way, integrate each component: $\mathbf{v}=\int\mathbf{a}\,dt,\qquad \mathbf{r}=\int\mathbf{v}\,dt.$ Each integration adds a constant vector $\mathbf{c}$ (one constant for each component), found from given conditions. Example: $\mathbf{a}=(2t\mathbf{i}-\mathbf{j})$ and $\mathbf{v}=(\mathbf{i}+3\mathbf{j})$ when $t=0$. Then $\mathbf{v}=(t^2+c_1)\mathbf{i}+(-t+c_2)\mathbf{j}$, and the condition gives $c_1=1$, $c_2=3$, so $\mathbf{v}=(t^2+1)\mathbf{i}+(3-t)\mathbf{j}$. Use calculus when the acceleration varies; use the constant-acceleration formulae only when it is constant.

1. Decide first: is the acceleration constant (use the vector formulae) or a function of $t$ (use calculus)?
2. Keep $\mathbf{i}$ and $\mathbf{j}$ components separate until the end.
3. Speed means magnitude $\sqrt{x^2+y^2}$; direction means an angle or a ratio of components.
4. Give units: m, m s$^{-1}$, m s$^{-2}$.
5. Give answers to 3 significant figures unless the question asks for an exact surd.