202104160754 Homework 1 (Q3)

A ship $A$ , which can sail at a constant speed $60\,\mathrm{km/hr}$ to meet a second ship $B$ which is $100\,\mathrm{km}$ away in the direction of $\mathrm{S60^\circ W}$ and is sailing due east at constant speed $30\,\mathrm{km/hr}$ . Find the sailing direction of $A$ and the time required to meet $B$ .

Solution.

(The solution below is based on the manuscript of 2014-2015 PHYS1250 Fundamental Physics Homework Solutions.)

Draw a diagram as follows:

Setup.

By the law of sines,

$\begin{aligned} \frac{V_A}{\sin 30^\circ} & = \frac{V_B}{\sin\theta} \\ \frac{60}{\sin 30^\circ} & = \frac{30}{\sin\theta} \\ \sin\theta & = 0.25 \\ \theta & = 14.5^\circ \end{aligned}$

Direction of $\mathbf{v}_A$ : $\mathrm{S45.5^\circ W}$

$\because 180^\circ -30^\circ -14.5^\circ - 90^\circ = 45.5^\circ$

Calculating $v_{AB}$ :

$\begin{aligned} |\mathbf{v}_{AB}| & = |\mathbf{v}_{A}|\cos\theta + |\mathbf{v}_{B}|\cos 30^\circ \\ & = 60 \cos 14.5^\circ + 30\cos 30^\circ \\ & = 84.1\,\mathrm{km/hr} \end{aligned}$

The time needed to meet ship $B$ is

$\begin{aligned} t & = \frac{100\,\mathrm{km}}{|\mathbf{v}_{AB}|} \\ & = \frac{100\,\mathrm{km}}{84.1\,\mathrm{km/hr}} \\ & = 1.19\,\mathrm{hr} \end{aligned}$

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