| Exam Board | WJEC |
|---|---|
| Module | Further Unit 3 (Further Unit 3) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Kinematics with position vectors |
| Difficulty | Standard +0.8 This is a multi-part further maths kinematics question requiring vector manipulation, collision conditions (equating position vectors and solving simultaneous equations), distance formula application, and solving a quadratic inequality. While the individual techniques are standard, the combination of steps, the need to work with 3D vectors throughout, and the conversion between units (km to m) makes this moderately challenging but still within typical further maths scope. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| i: \(60 + 168t = 62 + 160t\) | M1 | Comparison attempted for any component |
| \(t = 0 \cdot 25\) | A1 | |
| j: \(2 + 132t = pt\) | ||
| \(t = 0 \cdot 25 \Rightarrow p = 140\) | A1 | Convincing |
| k: \(4 = 3 + qt\) | ||
| \(t = 0 \cdot 25 \Rightarrow q = 4\) | A1 | Convincing |
| Answer | Marks | Guidance |
|---|---|---|
| (b) \(r_B - r_A = (2 - 8t)i + (-2 + 8t)j + (-1 + 4t)k\) | M1 | |
| \(AB^2 = (2 - 8t)^2 + (-2 + 8t)^2 + (-1 + 4t)^2\) | A1 | Correct method. Must lose i, j, k and be linear expressions |
| \((AB^2 = 144t^2 - 72t + 9)\) |
| Answer | Marks | Guidance |
|---|---|---|
| (c) \(AB^2 = 144t^2 - 72t + 9 = 0 \cdot 6^2\) | M1 | FT quadratic from (b) |
| \(AB^2 = 144t^2 - 72t + 8 \cdot 64 = 0\) | ||
| \((50t^2 - 25t + 3 = 0)\) | ||
| Solving quadratic | m1 | Attempt to solve resulting in at least one value of \(t\). |
| \(t = 0 \cdot 2, (0 \cdot 3)\)(hours) | A1 | |
| Alarms first activated at 9.12 (a.m.) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(AB^2 = (4t - 1)^2[(-2)^2 + 2^2 + 1] = 9(4t - 1)^2\) | (M1) | FT quadratic from (b) provided it is of the form \(a(4t - 1)^2\) |
| \(9(4t - 1)^2 = 0 \cdot 6^2\) or \(3(4t - 1) = 0 \cdot 6\) | (m1) | Attempt to solve resulting in at least one value of \(t\). |
| Solving quadratic | (A1) | |
| \(t = 0 \cdot 2, (0 \cdot 3)\) (hours) | (A1) | |
| Alarms first activated at 9.12 (a.m.) |
**(a)** Comparison of coefficients:
i: $60 + 168t = 62 + 160t$ | M1 | Comparison attempted for any component
$t = 0 \cdot 25$ | A1 |
j: $2 + 132t = pt$ | |
$t = 0 \cdot 25 \Rightarrow p = 140$ | A1 | Convincing
k: $4 = 3 + qt$ | |
$t = 0 \cdot 25 \Rightarrow q = 4$ | A1 | Convincing
**Subtotal: [4]**
**(b)** $r_B - r_A = (2 - 8t)i + (-2 + 8t)j + (-1 + 4t)k$ | M1 |
$AB^2 = (2 - 8t)^2 + (-2 + 8t)^2 + (-1 + 4t)^2$ | A1 | Correct method. Must lose i, j, k and be linear expressions
$(AB^2 = 144t^2 - 72t + 9)$ | |
**Subtotal: [2]**
**(c)** $AB^2 = 144t^2 - 72t + 9 = 0 \cdot 6^2$ | M1 | FT quadratic from (b)
$AB^2 = 144t^2 - 72t + 8 \cdot 64 = 0$ | |
$(50t^2 - 25t + 3 = 0)$ | |
Solving quadratic | m1 | Attempt to solve resulting in at least one value of $t$.
$t = 0 \cdot 2, (0 \cdot 3)$(hours) | A1 |
Alarms first activated at 9.12 (a.m.) | A1 |
**Subtotal: [4]**
**Alternative Solution**
Taking out a common factor of $(4t - 1)^2$ from the unsimplified form in (b):
$AB^2 = (4t - 1)^2[(-2)^2 + 2^2 + 1] = 9(4t - 1)^2$ | (M1) | FT quadratic from (b) provided it is of the form $a(4t - 1)^2$
$9(4t - 1)^2 = 0 \cdot 6^2$ or $3(4t - 1) = 0 \cdot 6$ | (m1) | Attempt to solve resulting in at least one value of $t$.
Solving quadratic | (A1) |
$t = 0 \cdot 2, (0 \cdot 3)$ (hours) | (A1) |
Alarms first activated at 9.12 (a.m.) | |
**Subtotal: [(4)]**
**Total for Question 3: 10**
---3. The position vectors $\mathbf { r } _ { A }$ and $\mathbf { r } _ { B }$, in kilometres, of two small aeroplanes $A$ and $B$ relative to a fixed point $O$ are given by
$$\begin{aligned}
& \mathbf { r } _ { A } = ( 60 \mathbf { i } + 2 \mathbf { j } + 4 \mathbf { k } ) + ( 168 \mathbf { i } + 132 \mathbf { j } ) t \\
& \mathbf { r } _ { B } = ( 62 \mathbf { i } + 3 \mathbf { k } ) + ( 160 \mathbf { i } + p \mathbf { j } + q \mathbf { k } ) t
\end{aligned}$$
where $t$ denotes the time in hours after 9:00 a.m. and $p , q$ are constants.\\
The aeroplanes $A$ and $B$ are on course to collide.
\begin{enumerate}[label=(\alph*)]
\item Show that $p = 140$ and $q = 4$.
\item Find an expression for the square of the distance between $A$ and $B$ at time $t$ hours after 9:00 a.m.
\item Both aeroplanes have systems that activate an alarm if they come within 600 m of each other. Determine the time when the alarms are first activated.
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 3 2019 Q3 [10]}}