| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Closest approach of two objects |
| Difficulty | Standard +0.3 This is a standard M3 closest approach problem requiring routine application of relative velocity concepts. Students find velocities from position vectors, establish relative position, then minimize distance using calculus (setting dr/dt = 0) or the dot product method. All steps are algorithmic with no novel insight required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10f Distance between points: using position vectors |
## Question 4(a) [4 marks]
M1: Find displacement of A: $\Delta \mathbf{r}_A = (-i + 3j) - (i + 2j) = (-2i + j)$ km in 0.5 hours
M1: Velocity of A: $\mathbf{v}_A = (-4i + 2j)$ kmh$^{-1}$
M1: Find displacement of B: $\Delta \mathbf{r}_B = (2i - j) - (-i + j) = (3i - 2j)$ km in 0.5 hours
M1: Velocity of B: $\mathbf{v}_B = (6i - 4j)$ kmh$^{-1}$
A1: Velocity of A relative to B: $\mathbf{v}_{A/B} = (-10i + 6j)$ kmh$^{-1}$
## Question 4(b) [3 marks]
M1: Position of A at $t = 0$: $(i + 2j)$ km
M1: Position of B at $t = 0$: $(-i + j)$ km
M1: $\mathbf{r} = [(i + 2j) - (-i + j)] + [(-10i + 6j)]t = (2 - 10t)i + (1 + 6t)j$
## Question 4(c) [5 marks]
M1: Distance squared: $d^2 = (2 - 10t)^2 + (1 + 6t)^2$
M1: $d^2 = 4 - 40t + 100t^2 + 1 + 12t + 36t^2 = 136t^2 - 28t + 5$
M1: $\frac{d(d^2)}{dt} = 272t - 28 = 0$
M1: $t = \frac{28}{272} = \frac{7}{68}$ (hours)
A1: $t = \frac{7}{68}$ hours (or equivalent)
## Question 4(d) [2 marks]
M1: $d^2 = 136 \left(\frac{7}{68}\right)^2 - 28 \left(\frac{7}{68}\right) + 5 = \frac{324}{68}$
A1: $d = \frac{9}{\sqrt{68}} = \frac{9\sqrt{17}}{34}$ km (or $0.346$ km to 3 s.f.)
---
# Mark Scheme - Question 54 Two boats, $A$ and $B$, are moving on straight courses with constant speeds. At noon, $A$ and $B$ have position vectors $( \mathbf { i } + 2 \mathbf { j } ) \mathrm { km }$ and $( - \mathbf { i } + \mathbf { j } ) \mathrm { km }$ respectively relative to a lighthouse. Thirty minutes later, the position vectors of $A$ and $B$ are ( $- \mathbf { i } + 3 \mathbf { j }$ ) km and $( 2 \mathbf { i } - \mathbf { j } ) \mathrm { km }$ respectively relative to the lighthouse.
\begin{enumerate}[label=(\alph*)]
\item Find the velocity of $A$ relative to $B$ in the form $( m \mathbf { i } + n \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }$, where $m$ and $n$ are integers.
\item The position vector of $A$ relative to $B$ at time $t$ hours after noon is $\mathbf { r } \mathrm { km }$.
Show that
$$\mathbf { r } = ( 2 - 10 t ) \mathbf { i } + ( 1 + 6 t ) \mathbf { j }$$
\item Determine the value of $t$ when $A$ and $B$ are closest together.
\item Find the shortest distance between $A$ and $B$.
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2014 Q4 [14]}}