91Ó°ÊÓ

The vector quantity displacement refers to how far an object is. In this problem, the original displacement and the new displacement are given in the vector form as$\stackrel{\mathbf{→}}{{\mathbf{d}}_{\mathbf{0}}}\stackrel{\mathbf{→}}{{\mathbf{d}}_{\mathbf{0}}}$and$\stackrel{\mathbf{→}}{{\mathbf{d}}_{\mathrm{n}}}\stackrel{\mathbf{→}}{{\mathbf{d}}_{\mathrm{n}}}$. By comparing these distances, the east and north components of the new displacement can be found out. These components are used to find displacement and angle.

The new displacement is written as

role="math" localid="1656303977083" $$\begin{gathered} \stackrel{→}{{\mathrm{d}}_{\mathrm{n}}}=\mathrm{A}\hat{\mathrm{i}}+\mathrm{B}\hat{\mathrm{j}}\left(1\right) \\ \left|\stackrel{→}{{\mathrm{d}}_{\mathrm{n}}}\right|=\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2}\left(2\right) \end{gathered}$$

The direction angle can be given as

$$\begin{gathered} {\tan }^{-1}\left(\frac{\mathrm{B}}{\mathrm{A}}\right) \\ \mathrm{Given}, \\ \mathrm{North}\mathrm{distance}=120\mathrm{km}\mathrm{and} \\ \mathrm{East}\mathrm{distance}=100\mathrm{km} \end{gathered}$$

$$\begin{gathered} \mathrm{Considering}\mathrm{east}\mathrm{is}\stackrel{→}{\mathrm{i}}\mathrm{and}\mathrm{north}\mathrm{is}\stackrel{→}{\mathrm{j}}, \\ \mathrm{original}\mathrm{displacement} \\ \stackrel{→}{{\mathrm{d}}_0}=(120\mathrm{km})\stackrel{→}{\mathrm{j}} \\ \mathrm{displacement}\mathrm{by}\mathrm{storm}\stackrel{→}{{\mathrm{d}}_{\mathrm{s}}}\mathrm{written}\mathrm{as} \\ \stackrel{→}{{\mathrm{d}}_{\mathrm{s}}}=\left(100\mathrm{km}\right)\stackrel{→}{\mathrm{i}} \\ \mathrm{and}\mathrm{new}\mathrm{displacement}\stackrel{→}{{\mathrm{d}}_{\mathrm{n}}}\mathrm{is}\mathrm{given}\mathrm{by}. \\ \stackrel{→}{{\mathrm{d}}_{\mathrm{n}}}=\mathrm{A}\hat{\mathrm{i}}+\mathrm{B}\hat{\mathrm{j}} \\ \mathrm{The}\mathrm{original}\mathrm{displacement}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as} \\ \stackrel{→}{{\mathrm{d}}_0}=\stackrel{→}{{\mathrm{d}}_{\mathrm{s}}}+\stackrel{→}{{\mathrm{d}}_{\mathrm{n}}} \\ \mathrm{From}\mathrm{the}\mathrm{above}\mathrm{given}\mathrm{information}, \\ \mathrm{A}=100\mathrm{km}\mathrm{and}\mathrm{B}=120\mathrm{km} \\ \mathrm{Substituting}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{in}\mathrm{equation}\left(\mathrm{ii}\right)\mathrm{the}\mathrm{magnitude}\mathrm{of}\mathrm{the}\mathrm{new} \\ \mathrm{displacement}\mathrm{is}, \\ \left|\stackrel{→}{{\mathrm{d}}_{\mathrm{n}}}\right|=\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2}=\sqrt{{\left(100\right)}^2+{\left(120\right)}^2}=156\mathrm{km} \end{gathered}$$

Using equation (iii) the direction is calculate as

$$\begin{gathered} {\tan }^{-1}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)={\tan }^{-1}\left(\frac{120}{100}\right)=50.2^{°} \\ \mathrm{Now}\mathrm{from}\mathrm{the}\mathrm{figure},\mathrm{we}\mathrm{can}\mathrm{conclude}\mathrm{that}\mathrm{the}\mathrm{angle}\mathrm{found}\mathrm{above}\mathrm{is}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{the} \\ \mathrm{negative}\mathrm{x}-\mathrm{axis}.\mathrm{The}\mathrm{ship}\mathrm{must}\mathrm{sail}129.8^{°}\mathrm{counter}\mathrm{clockwise}\mathrm{from}\mathrm{east}\mathrm{or}50.2^{°}\mathrm{north} \\ \mathrm{from}\mathrm{west}\mathrm{or}39.8^{°}\mathrm{west}\mathrm{of}\mathrm{north}. \\ \mathrm{Thus}\mathrm{the}\mathrm{direction}\mathrm{is}129.8^{°} \end{gathered}$$