In Year 1, displacement, velocity and acceleration were numbers with a sign along a straight line. In Year 2 (DfE content areas Q · Kinematics and R · Forces and Newton's laws) they become vectors in the plane, written with the unit vectors $\mathbf{i}$ and $\mathbf{j}$.
| Quantity | Vector | Magnitude |
|---|---|---|
| Position (from origin $O$) | $\mathbf{r} = x\mathbf{i} + y\mathbf{j}$ | distance from $O$ |
| Velocity | $\mathbf{v} = v_x\mathbf{i} + v_y\mathbf{j}$ | speed $= \sqrt{v_x^2 + v_y^2}$ |
| Acceleration | $\mathbf{a} = a_x\mathbf{i} + a_y\mathbf{j}$ | $|\mathbf{a}|$ |
| Force | $\mathbf{F} = m\mathbf{a}$ | $|\mathbf{F}| = m|\mathbf{a}|$ |
The Year 1 suvat equations work unchanged with vectors, provided the acceleration is constant.
($\mathbf{r}_0$ is the starting position; displacement $\mathbf{s} = \mathbf{r} - \mathbf{r}_0$)
A particle starts at the point with position vector $(2\mathbf{i} - 3\mathbf{j})$ m with velocity $(4\mathbf{i} + \mathbf{j})$ m s$^{-1}$ and constant acceleration $(-\mathbf{i} + 2\mathbf{j})$ m s$^{-2}$. Find its velocity, speed and position after $3$ s.
For the particle in Worked Example 1, find when it is moving parallel to $\mathbf{j}$.
A particle of mass $2$ kg moves with constant acceleration. Its velocity changes from $(3\mathbf{i} - 2\mathbf{j})$ m s$^{-1}$ to $(9\mathbf{i} + 6\mathbf{j})$ m s$^{-1}$ in $4$ s. Find the resultant force and its magnitude.
Two ships move with constant velocity. At time $t$ hours, ship $A$ is at $(2\mathbf{i} + 5\mathbf{j}) + t(3\mathbf{i} - \mathbf{j})$ km and ship $B$ at $(8\mathbf{i} + \mathbf{j}) + t(\mathbf{i} + \mathbf{j})$ km. Show the ships do not collide and find their least distance apart.
When the acceleration changes with time, use calculus as in Year 1 Variable Acceleration — but now apply it to each component separately.
$\mathbf{a} \;\xrightarrow{\;\text{integrate}\;}\; \mathbf{v} = \displaystyle\int \mathbf{a}\,\mathrm{d}t \;\xrightarrow{\;\text{integrate}\;}\; \mathbf{r} = \displaystyle\int \mathbf{v}\,\mathrm{d}t$
A particle has position $\mathbf{r} = (t^3 - 3t)\mathbf{i} + (2t^2 + 1)\mathbf{j}$ m at time $t$ s. Find $\mathbf{v}$ and $\mathbf{a}$, the speed at $t = 2$, and when the particle is moving parallel to $\mathbf{j}$.
A particle has acceleration $\mathbf{a} = 6t\,\mathbf{i} - 4\mathbf{j}$ m s$^{-2}$. At $t = 0$ it is at the point with position vector $\mathbf{i}$ m, moving with velocity $(2\mathbf{i} + 3\mathbf{j})$ m s$^{-1}$. Find $\mathbf{v}$ and $\mathbf{r}$ at time $t$, and its position at $t = 2$.
The path of the particle in Worked Example 6 for $0 \leqslant t \leqslant 2$. The velocity is always tangent to the path.
A particle of mass $0.5$ kg moves so that $\mathbf{r} = 3\cos(2t)\,\mathbf{i} + 3\sin(2t)\,\mathbf{j}$ m. Show that its speed is constant, and find the resultant force on it.
Forces $(2\mathbf{i} + 5\mathbf{j})$ N, $(4\mathbf{i} - 2\mathbf{j})$ N and $(-3\mathbf{i} + \mathbf{j})$ N act on a particle of mass $1.5$ kg, initially at rest at $O$. Find its position after $2$ s.
Constant $\mathbf{a}$
$\mathbf{v} = \mathbf{u} + \mathbf{a}t$, $\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \tfrac12\mathbf{a}t^2$.
Speed
$|\mathbf{v}| = \sqrt{v_x^2 + v_y^2}$.
Parallel to $\mathbf{j}$
$\mathbf{i}$-component of $\mathbf{v}$ is $0$.
At rest
Both components of $\mathbf{v}$ zero at the same $t$.
Differentiate
$\mathbf{r} \to \mathbf{v} \to \mathbf{a}$, each component.
Integrate
$\mathbf{a} \to \mathbf{v} \to \mathbf{r}$, with a vector constant.
Newton II
$\mathbf{F} = m\mathbf{a}$, sum the force vectors first.
Collide?
Same position at the same time.
Closest approach
Minimise $|\mathbf{r}_B - \mathbf{r}_A|^2$.
$\mathbf{i}$ and $\mathbf{j}$ are perpendicular unit vectors; distances in metres and times in seconds unless stated. Give answers to 3 significant figures where they are not exact.
A particle has initial velocity $(3\mathbf{i} + 4\mathbf{j})$ m s$^{-1}$ and constant acceleration $(2\mathbf{i} - \mathbf{j})$ m s$^{-2}$. Find its velocity and speed after $5$ s.
▶ Show solution
$\mathbf{v} = (3\mathbf{i} + 4\mathbf{j}) + 5(2\mathbf{i} - \mathbf{j}) = 13\mathbf{i} - \mathbf{j}$ m s$^{-1}$.
Speed $= \sqrt{169 + 1} = \sqrt{170} = 13.0$ m s$^{-1}$.
A particle leaves the origin with velocity $(5\mathbf{i} - 2\mathbf{j})$ m s$^{-1}$ and constant acceleration $(-\mathbf{i} + \mathbf{j})$ m s$^{-2}$. Show that it is on the $x$-axis when $t = 4$, and find its distance from $O$ then.
▶ Show solution
$\mathbf{r} = t(5\mathbf{i} - 2\mathbf{j}) + \tfrac12t^2(-\mathbf{i} + \mathbf{j})$.
At $t = 4$: $\mathbf{r} = (20 - 8)\mathbf{i} + (-8 + 8)\mathbf{j} = 12\mathbf{i}$.
The $\mathbf{j}$-component is $0$, so it is on the $x$-axis, $12$ m from $O$.
A constant force $(6\mathbf{i} - 8\mathbf{j})$ N acts on a particle of mass $4$ kg, initially at rest at $O$. Find the acceleration, the speed after $3$ s and the distance from $O$ then.
▶ Show solution
$\mathbf{a} = \tfrac14(6\mathbf{i} - 8\mathbf{j}) = 1.5\mathbf{i} - 2\mathbf{j}$ m s$^{-2}$ (magnitude $2.5$).
$\mathbf{v} = 3\mathbf{a} = 4.5\mathbf{i} - 6\mathbf{j}$; speed $= 7.5$ m s$^{-1}$.
$\mathbf{r} = \tfrac12(9)\mathbf{a} = 6.75\mathbf{i} - 9\mathbf{j}$; distance $= \sqrt{45.5625 + 81} = 11.25$ m.
(It moves in a straight line from rest, so distance $= \tfrac12 \times 2.5 \times 9$ ✓)
A particle has position $\mathbf{r} = (t^2 - 4t)\mathbf{i} + (t^3 - 3t)\mathbf{j}$. Find its speed when $t = 2$ and show that it is never at rest.
▶ Show solution
$\mathbf{v} = (2t - 4)\mathbf{i} + (3t^2 - 3)\mathbf{j}$.
At $t = 2$: $\mathbf{v} = 0\mathbf{i} + 9\mathbf{j}$, speed $= 9$ m s$^{-1}$.
$\mathbf{i}$-component is zero only at $t = 2$; $\mathbf{j}$-component is zero only at $t = \pm 1$. They are never zero together, so the particle is never at rest.
A particle has acceleration $(2\mathbf{i} - 6t\,\mathbf{j})$ m s$^{-2}$ and initial velocity $(\mathbf{i} + 4\mathbf{j})$ m s$^{-1}$. Find when it is moving parallel to $\mathbf{i}$ ($t > 0$), and its speed then.
▶ Show solution
$\mathbf{v} = (1 + 2t)\mathbf{i} + (4 - 3t^2)\mathbf{j}$.
Parallel to $\mathbf{i}$: $4 - 3t^2 = 0 \Rightarrow t = \dfrac{2}{\sqrt3} = 1.15$ s.
Speed $= 1 + 2 \times 1.155 = 3.31$ m s$^{-1}$.
Three forces $(2\mathbf{i} + 5\mathbf{j})$ N, $(4\mathbf{i} - 2\mathbf{j})$ N and $(p\mathbf{i} + q\mathbf{j})$ N act on a particle of mass $3$ kg, giving it acceleration $(3\mathbf{i} - \mathbf{j})$ m s$^{-2}$. Find $p$ and $q$.
▶ Show solution
$\mathbf{F} = m\mathbf{a} = 9\mathbf{i} - 3\mathbf{j}$.
$(6 + p)\mathbf{i} + (3 + q)\mathbf{j} = 9\mathbf{i} - 3\mathbf{j}$, so $p = 3$ and $q = -6$.
With $\mathbf{i}$ due east and $\mathbf{j}$ due north, a particle has velocity $\mathbf{v} = (3 + t)\mathbf{i} + (2t - 4)\mathbf{j}$ m s$^{-1}$. Find when it is moving north-east, and its speed then.
▶ Show solution
North-east needs equal, positive components: $3 + t = 2t - 4 \Rightarrow t = 7$.
$\mathbf{v} = 10\mathbf{i} + 10\mathbf{j}$ (both positive ✓). Speed $= 10\sqrt2 = 14.1$ m s$^{-1}$.
A particle starts from $O$ with velocity $\mathbf{v} = 6\mathrm{e}^{-2t}\,\mathbf{i} + (4 - 2t)\,\mathbf{j}$ m s$^{-1}$, $t \geqslant 0$. (a) Find the magnitude of its acceleration when $t = 0$. (b) Find $\mathbf{r}$ at time $t$. (c) Describe its position when $t = 4$.
▶ Show solution
(a) $\mathbf{a} = -12\mathrm{e}^{-2t}\,\mathbf{i} - 2\mathbf{j}$; at $t = 0$, $\mathbf{a} = -12\mathbf{i} - 2\mathbf{j}$, magnitude $\sqrt{148} = 12.2$ m s$^{-2}$.
(b) $\mathbf{r} = -3\mathrm{e}^{-2t}\,\mathbf{i} + (4t - t^2)\mathbf{j} + \mathbf{c}$; $\mathbf{r} = \mathbf{0}$ at $t = 0$ gives $\mathbf{c} = 3\mathbf{i}$.
$\mathbf{r} = 3\left(1 - \mathrm{e}^{-2t}\right)\mathbf{i} + (4t - t^2)\mathbf{j}$.
(c) At $t = 4$: $\mathbf{r} = 3(1 - \mathrm{e}^{-8})\mathbf{i} + 0\mathbf{j} \approx 3.00\mathbf{i}$ — back on the $x$-axis, almost exactly $3$ m from $O$.
Particles $A$ and $B$ have positions $\mathbf{r}_A = (\mathbf{i} + 2\mathbf{j}) + t(2\mathbf{i} + \mathbf{j})$ and $\mathbf{r}_B = (7\mathbf{i} - 4\mathbf{j}) + t(-\mathbf{i} + 3\mathbf{j})$. Show that they do not collide and find their least distance apart.
▶ Show solution
$\mathbf{i}$: $1 + 2t = 7 - t \Rightarrow t = 2$. $\mathbf{j}$: $2 + t = -4 + 3t \Rightarrow t = 3$. Different times, so no collision.
$\mathbf{r}_B - \mathbf{r}_A = (6 - 3t)\mathbf{i} + (2t - 6)\mathbf{j}$.
$d^2 = (6 - 3t)^2 + (2t - 6)^2 = 13t^2 - 60t + 72$.
$\dfrac{\mathrm{d}(d^2)}{\mathrm{d}t} = 26t - 60 = 0 \Rightarrow t = \dfrac{30}{13}$.
$d^2 = 72 - \dfrac{900}{13} = \dfrac{36}{13}$, so $d = \dfrac{6}{\sqrt{13}} = 1.66$ m.
A particle $P$ of mass $2$ kg moves so that its velocity at time $t$ s is $\mathbf{v} = (3t^2 - 12)\mathbf{i} + (4t - 2)\mathbf{j}$ m s$^{-1}$. At $t = 0$, $P$ is at the point with position vector $(2\mathbf{i} - \mathbf{j})$ m.
(a) Find the magnitude of the resultant force on $P$ when $t = 1$. (b) Find the value of $t$ ($t \geqslant 0$) when $P$ is moving parallel to $\mathbf{j}$. (c) Find the position vector of $P$ at that time. (d) Find the distance of $P$ from $O$ at that time.
▶ Show solution
(a) $\mathbf{a} = 6t\,\mathbf{i} + 4\mathbf{j}$; at $t = 1$, $\mathbf{a} = 6\mathbf{i} + 4\mathbf{j}$.
$\mathbf{F} = 2\mathbf{a} = 12\mathbf{i} + 8\mathbf{j}$, $|\mathbf{F}| = \sqrt{208} = 14.4$ N.
(b) $3t^2 - 12 = 0 \Rightarrow t = 2$. (Then $\mathbf{v} = 6\mathbf{j} \neq \mathbf{0}$ ✓)
(c) $\mathbf{r} = (t^3 - 12t)\mathbf{i} + (2t^2 - 2t)\mathbf{j} + \mathbf{c}$; at $t = 0$, $\mathbf{c} = 2\mathbf{i} - \mathbf{j}$.
$\mathbf{r} = (t^3 - 12t + 2)\mathbf{i} + (2t^2 - 2t - 1)\mathbf{j}$.
At $t = 2$: $\mathbf{r} = (8 - 24 + 2)\mathbf{i} + (8 - 4 - 1)\mathbf{j} = -14\mathbf{i} + 3\mathbf{j}$ m.
(d) $|\mathbf{r}| = \sqrt{196 + 9} = \sqrt{205} = 14.3$ m.