Pre-U Pre-U 9795/2 2012 June — Question 8 8 marks

Exam BoardPre-U
ModulePre-U 9795/2 (Pre-U Further Mathematics Paper 2)
Year2012
SessionJune
Marks8
TopicVectors Introduction & 2D
TypeInterception: find bearing/direction to intercept (exact intercept)
DifficultyChallenging +1.2 This is a relative velocity/interception problem requiring vector addition and trigonometry. Students must set up position vectors, account for the moving target, and solve for an unknown direction using the constraint that speeds are fixed. While it involves multiple steps and careful geometric reasoning, it follows a standard mechanics template for interception problems that Further Maths students practice regularly. The calculation is straightforward once the vector equation is established.
Spec1.10h Vectors in kinematics: uniform acceleration in vector form3.02e Two-dimensional constant acceleration: with vectors
8 \includegraphics[max width=\textwidth, alt={}, center]{d8ca5464-435f-45e0-8e19-1830415a7c60-4_757_729_260_708} An aircraft carrier, \(A\), is heading due north at \(40 \mathrm {~km} \mathrm {~h} ^ { - 1 }\). A destroyer, \(D\), which is 8 km south-west of \(A\), is ordered to take up a position 3 km east of \(A\) as quickly as possible. The speed of \(D\) is \(60 \mathrm {~km} \mathrm {~h} ^ { - 1 }\) (see diagram). Find the bearing, \(\theta\), of the course that \(D\) should take, giving your answer to the nearest degree.
Question 8
\(x^2 = 3^2 + 8^2 - 2 \times 3 \times 8\cos 135° \Rightarrow x = 10.341\) M1A1
\(\frac{\sin\theta}{3} = \frac{\sin 135°}{8} \Rightarrow \theta = 11.837°\) (\(\theta\) as in distance diagram.) M1A1
\(180° - (45° + 11.837°) = 123.163°\) B1
\(\frac{\sin\alpha}{40} = \frac{\sin 123.163°}{60} \Rightarrow \alpha = 33.923°\) (\(\alpha\) as in velocities diagram.) M1A1
Course is: \((45° + 11.837°) - 33.923° = 22.914°\)
Course is \(023°\) (Bearing notation not required.) A1 [8]
Alternative solution:
Attempt at either equation M1
\(60\cos\theta t = 4\sqrt{2} + 40t\) A1
\(60\sin\theta t = 4\sqrt{2} + 3\) A1
Attempt to eliminate \(t\). M1
Obtain e.g. \((4\sqrt{2}+3)\cos\theta - 4\sqrt{2}\sin\theta = \frac{2}{3}(4\sqrt{2}+3)\) A1
or \(8.656...\cos\theta - 5.656...\sin\theta = 5.7712...\)
Rearrange to obtain \(\cos(\theta + 33.16) = 0.5580...\) A1
Solve: \(\theta + 33.16 = 56.076... \Rightarrow \theta = 22.9\)
Course is \(023°\) (Bearing notation not required.) M1A1 [8] [8]
**Question 8**
$x^2 = 3^2 + 8^2 - 2 \times 3 \times 8\cos 135° \Rightarrow x = 10.341$ **M1A1**

$\frac{\sin\theta}{3} = \frac{\sin 135°}{8} \Rightarrow \theta = 11.837°$ ($\theta$ as in distance diagram.) **M1A1**

$180° - (45° + 11.837°) = 123.163°$ **B1**

$\frac{\sin\alpha}{40} = \frac{\sin 123.163°}{60} \Rightarrow \alpha = 33.923°$ ($\alpha$ as in velocities diagram.) **M1A1**

Course is: $(45° + 11.837°) - 33.923° = 22.914°$

Course is $023°$ (Bearing notation not required.) **A1** [8]

**Alternative solution:**
Attempt at either equation **M1**

$60\cos\theta t = 4\sqrt{2} + 40t$ **A1**

$60\sin\theta t = 4\sqrt{2} + 3$ **A1**

Attempt to eliminate $t$. **M1**

Obtain e.g. $(4\sqrt{2}+3)\cos\theta - 4\sqrt{2}\sin\theta = \frac{2}{3}(4\sqrt{2}+3)$ **A1**

or $8.656...\cos\theta - 5.656...\sin\theta = 5.7712...$

Rearrange to obtain $\cos(\theta + 33.16) = 0.5580...$ **A1**

Solve: $\theta + 33.16 = 56.076... \Rightarrow \theta = 22.9$

Course is $023°$ (Bearing notation not required.) **M1A1** [8] **[8]**
8\\
\includegraphics[max width=\textwidth, alt={}, center]{d8ca5464-435f-45e0-8e19-1830415a7c60-4_757_729_260_708}

An aircraft carrier, $A$, is heading due north at $40 \mathrm {~km} \mathrm {~h} ^ { - 1 }$. A destroyer, $D$, which is 8 km south-west of $A$, is ordered to take up a position 3 km east of $A$ as quickly as possible. The speed of $D$ is $60 \mathrm {~km} \mathrm {~h} ^ { - 1 }$ (see diagram). Find the bearing, $\theta$, of the course that $D$ should take, giving your answer to the nearest degree.

\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2012 Q8 [8]}}
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