91Ó°ÊÓ

1. The ship A is located 4.0km north and 2.5 km east of ship B.
2. The velocity of A,$V_A=22\mathrm{km}/\mathrm{h}$ in south direction.
3. The velocity of B,${\mathrm{V}}_{\mathrm{B}}=40\mathrm{km}/\mathrm{h}$ in a direction

When two frames of reference Aand Bare moving relative to each other at a constant velocity, the velocity of a particle Pas measured by an observer in frame Ais different than the velocity measured in frame B. The velocities measured in two reference frames are related by,

${\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{P}\mathbf{A}}\mathbf{=}{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{P}\mathbf{B}}\mathbf{+}{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{B}\mathbf{A}}$

Here${\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{BA}}$ is the velocity of B with respect to A.

Using the relative motion concept, we can find the magnitude and the direction of the resultant vector. Using these magnitudes of the velocity vectors, we can find the required time and least separation between the ships.

Formulae:

The magnitude of the vector $\stackrel{â‡\P Ä}{V}$is given by, $V=\sqrt{V_x^2+V_y^2}$ (i)

The velocity vector in terms of unit-vector notation, $\stackrel{â‡\P Ä}{v}=v\cos \mathrm{θ}\hat{i}+v\sin \mathrm{θ}\hat{j}$ (ii)

Vector diagram:

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

$\stackrel{→}{V_A}=\left(-22\right)\hat{j}$

The velocity of B can be written in the vector notation using equation (ii) as:

$$\begin{gathered} \stackrel{→}{V_B}=40\cos 37°\hat{i}+40\sin 37°\hat{j} \\ =31.95\hat{i}+24.07\hat{j} \end{gathered}$$

The velocity of A with respect to B is given as:

$$\begin{gathered} \stackrel{→}{V_{AB}}=\stackrel{→}{V_A}-\stackrel{→}{V_B} \\ =\left(-22\right)\hat{j}-\left(31.95\hat{i}+24.07\hat{j}\right) \\ =-32\hat{i}-46\hat{j} \end{gathered}$$

Now, the magnitude of the velocity vector can be given using equation (i) as follows:

$$\begin{gathered} V_{AB}=\sqrt{{\left(-32\right)}^2+{\left(-46\right)}^2} \\ =56.04\mathrm{km}/\mathrm{hr} \\ ≈56\mathrm{km}/\mathrm{hr} \end{gathered}$$

Hence, the value of the velocity is 56 km/hr.

The position of A with respect to B is given as follows:

$$\begin{gathered} {\stackrel{→}{r}}_{AB}-{\stackrel{→}{r}}_{0AB}=∫\stackrel{→}{V_{AB}}dt \\ =∫\left(-32\hat{i}-46\hat{j}\right)dt \\ =∫\left(-32t\hat{i}-46t\hat{j}\right) \\ {\stackrel{→}{r}}_{AB}=\left(2.5-32t\right)\hat{i}+\left(4-46t\right)\hat{j} \end{gathered}$$

Hence, the expression for the position is$\left(2.5-32t\right)\hat{i}+\left(4-46t\right)\hat{j}$ .

The separation between two boats will be minimum if$\frac{d{r}_{AB}}{dt}=0$

Thus, solving the expression for the position, we can get the time value as follows:

$$\begin{gathered} \frac{d}{dt}\left[\sqrt{{\left(2.5-32t\right)}^2{\left(4-46t\right)}^2}\right]=0 \\ \frac{6286t-528}{\sqrt[2]{{\left(2.5-32t\right)}^2{\left(4-46t\right)}^2}}=0 \\ 6286t=528 \\ t=0.08399 \\ ≈0.084\mathrm{h} \end{gathered}$$

Hence, the value of the time is $0.084\mathrm{h}$.

The minimum separation between two boats at t=0.084 h is given as:

$$\begin{gathered} r_{AB}=\sqrt{{\left(2.5-32\left(0.084\right)\right)}^2+{\left(4-46\left(0.084\right)\right)}^2} \\ =0.225 \\ ~0.2\mathrm{km} \end{gathered}$$

Hence, the value of the separation is 0.2 km.