Two ships, A and B, leave port at the same time. Ship A travels northwest at $\mathbf{24}\mathbf{}\mathit{k}\mathit{n}\mathit{o}\mathit{t}\mathit{s}$, and ship B travels at $\mathbf{28}\mathbf{}\mathit{k}\mathit{n}\mathit{o}\mathit{t}\mathit{s}$in a direction ${\mathbf{40}}^{\mathbf{°}}$west of south. ($\left(1knot=1nauticalmileperhour\right)$).What are the (a) magnitude and (b) direction of the velocity of ship A relative to B? (c) After what time will the ships be 160 nautical miles apart? (d) What will be the bearing of B (the direction of B’s position) relative to A at that time?

1. The velocity of A $\left(V_A\right)=24knots$in a northwest direction.

2. The velocity of B $\left(V_B\right)=28knots$in a direction ${40}^{°}$west of south.

The operation of combining two or more vectors into a vector sum is known as vector addition. Using the vector diagram and relative motion concept, we can find the magnitude and the direction of the resultant vector.

Formula:

Magnitude of the vector $\overrightarrow{V}$is,

$V=\sqrt{V_x^2+V_y^2}$

The direction of vector is given by,

$\tan \theta =\frac{V_y}{V_x}$

Vector diagram:

The velocity of A can be written in the vector notation as,

$$\begin{gathered} \overrightarrow{V}\theta =V_A\cos \theta \hat{j}+{}_A\sin \\ =\left(-24knots\right)\sin \left({45}^{°}\right)\widehat{i+}\left(24knots\right)\cos \left({45}^{°}\right)\hat{j} \\ =\left(-16.97knots\right)\hat{i}+\left(16.97knots\right)\hat{j} \end{gathered}$$

The velocity of B can be written in the vector notation as,

$$\begin{gathered} \overrightarrow{V}\theta =V_B\cos \theta \hat{j}+{}_B\sin \\ =\left(-28knots\right)\sin \left({40}^{°}\right)\hat{i}+\left(-28knots\right)\cos \left({40}^{°}\right)\hat{j} \\ =\left(-18knots\right)\hat{i}+\left(-21.45knots\right)\hat{j} \end{gathered}$$

The velocity of A with respect to B is,

$$\begin{gathered} {\overrightarrow{V}}_{AB}={\overrightarrow{V}}_A-{\overrightarrow{V}}_B \\ =\left[\left(-16.97knots\right)\hat{i}+\left(16.97knots\right)\hat{j}\right]-\left[\left(-18knots\right)\hat{i}+\left(-21knots\right)\hat{j}\right] \\ =\left(-1.03knots\right)\hat{i}+\left(38.4knots\right)\hat{j} \end{gathered}$$

The magnitude of velocity of ship A relative to ship B is

$$\begin{gathered} V_{AB}=\sqrt{{V_{ABx}}^2+{V_{ABy}}^2} \\ =\sqrt{{\left(1.03knots\right)}^2+{\left(38.4knots\right)}^2} \\ =38.41knots \\ \approx 38knots \end{gathered}$$

Therefore, the magnitude of velocity of ship A relative to ship B is $38knots$.

The direction of velocity of ship A relative to ship B is given by the angle mad by $V_{AB}$with north direction

$$\begin{gathered} \tan \theta =\frac{V_{ABx}}{V_{ABy}} \\ =\frac{1.03knots}{38.4knots} \\ \theta ={\tan }^{-1}\left(\frac{1.03knots}{38.4knots}\right) \\ =1.{53}^{°} \\ \approx 1.5^{°} \end{gathered}$$

Therefore, the direction of velocity of ship A relative to ship B $1.5^{°}$east of north.

The distance between two ships is 160 nautical miles. Therefore, the position vector $∆\overrightarrow{\underset{}{x}}{}_{AB}=160nauticalmiles$. Thus, the time period after which the two ships are 160 nautical miles apart is,

$$\begin{gathered} t=\frac{∆{\displaystyle \overrightarrow{\underset{}{X}}}{}_{AB}}{∆{\overrightarrow{V}}_{AB}} \\ =\frac{160nauticalmiles}{38.4knots} \\ =4.2h \end{gathered}$$

Therefore, after $4.2h$two ships are 160 nautical miles apart from each other.

The velocity ${\overrightarrow{V}}_{AB}$and $∆{\overrightarrow{V}}_{AB}$are in the same direction. Also, ${\overrightarrow{V}}_{AB}$does not change with time. To view the situation relative to A, we have to reverse the direction of ${\overrightarrow{V}}_{AB}$. So, we have ${\overrightarrow{V}}_{AB}=-{\overrightarrow{V}}_{BA}$and $∆{\overrightarrow{V}}_{AB}=-∆{\overrightarrow{V}}_{BA}$. Thus, we can conclude that B is at a bearing of $1.5^{°}$west of south relative to A during the journey.