91Ó°ÊÓ

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 $\stackrel{→}{V}$is,

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

The direction of vector is given by,

$\tan \mathrm{θ}=\frac{V_y}{V_x}$

Vector diagram:

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

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

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

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

The velocity of A with respect to B is,

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

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

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

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

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

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 $∆\underset{}{\overset{→}{x}}{}_{AB}=160nauticalmiles$. Thus, the time period after which the two ships are 160 nautical miles apart is,

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

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

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