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Date	May 2022	Marks available	4	Reference code	22M.1.AHL.TZ1.13
Level	Additional Higher Level	Paper	Paper 1	Time zone	Time zone 1
Command term	Find	Question number	13	Adapted from	N/A

Question

At $1 : 00 pm$ a ship is $1 km$ east and $4 km$ north of a harbour. A coordinate system is defined with the harbour at the origin. The position vector of the ship at $1 : 00 pm$ is given by $(\begin{matrix} 1 \\ 4 \end{matrix})$ .

The ship has a constant velocity of $(\begin{matrix} 1.2 \\ - 0.6 \end{matrix})$ kilometres per hour ( ${km h}^{- 1}$ ).

Write down an expression for the position vector $r$ of the ship, $t$ hours after $1 : 00 pm$ .

[1]

Find the time at which the bearing of the ship from the harbour is $045 ˚$ .

[4]

Markscheme

$(r =) (\begin{matrix} 1 \\ 4 \end{matrix}) + t (\begin{matrix} 1.2 \\ - 0.6 \end{matrix})$ A1

Note: Do not condone the use of $λ$ or any other variable apart from $t$ .

[1 mark]

when the bearing from the port is $045 ˚$ , the distance east from the port is equal to the distance north from the port (M1)

$1 + 1.2 t = 4 - 0.6 t$ (A1)

$1.8 t = 3$

$t = \frac{5}{3}$ ( $1.6666 \dots,$ $1$ hour $40$ minutes) (A1)

time is $2 : 40 pm (14 : 40)$ A1

[4 marks]

Examiners report

Most candidates were able to answer part (a) correctly but there were some very poor examples of vector notation. The question asked for an expression of $r$ in terms of $t$ and although a failure to write $r$ was condoned the use of $λ$ or some other variable was penalized. In part (b) few candidates recognized that the eastern and northern distances would be equal with a bearing of 045°. Those who correctly obtained a value of $t = \frac{5}{3}$ often did not use this to find the time as required.

[N/A]

Syllabus sections

Topic 3—Geometry and trigonometry » AHL 3.11—Vector equation of a line in 2d and 3d

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