Question
At $1: 00 \mathrm{pm}$ a ship is $1 \mathrm{~km}$ east and $4 \mathrm{~km}$ north of a harbour. A coordinate system is defined with the harbour at the origin. The position vector of the ship at $1: 00 \mathrm{pm}$ is given by $\left(\begin{array}{l}1 \\ 4\end{array}\right)$.
The ship has a constant velocity of $\left(\begin{array}{c}1.2 \\ -0.6\end{array}\right)$ kilometres per hour $\left(\mathrm{kmh}^{-1}\right)$.
a. Write down an expression for the position vector $r$ of the ship, $t$ hours after $1: 00 \mathrm{pm}$.
b. Find the time at which the bearing of the ship from the harbour is $045^{\circ}$.
▶️Answer/Explanation
a. $(\boldsymbol{r}=)\left(\begin{array}{l}1 \\ 4\end{array}\right)+t\left(\begin{array}{c}1.2 \\ -0.6\end{array}\right) \quad \boldsymbol{A 1}$
Note: Do not condone the use of $\lambda$ or any other variable apart from $t$.
[1 mark]
b. when the bearing from the port is $045^{\circ}$, the distance east from the port is equal to the distance north from the port
(M1)
$$
1+1.2 t=4-0.6 t
$$
1. $8 t=3$
$t=\frac{5}{3} \quad$ (1.6666.., 1 hour 40 minutes)
(A1)
time is $2: 40 \mathrm{pm} \quad(14: 40)$
A1
[4 marks]