Question

Ships A and B leave port together. For the next two hours, ship A travels at $20\mathrm{mph}$in a direction $30°$west of north while ship B travels $20°$east of north at $25\mathrm{mph}$.

a. What is the distance between the two ships two hours after they depart?

b. What is the speed of ship A as seen by ship B?

Short answer

a. $39\mathrm{mi}$

b.$20\mathrm{mph}$

Step-by-step solution

Given information

Ships A and B leave port together. For the next two hours, ship A travels at $20\mathrm{mph}$ in a direction$30°$ west of north while ship B travels $20°$ east of north at $25\mathrm{mph}$.

Part a) Step 1: Calculation

Distance of ship A is:

$$\begin{gathered} \overrightarrow{{\mathrm{r}}_{\mathrm{A}}}=\overrightarrow{{\mathrm{v}}_{\mathrm{A}}}\mathrm{t} \\ =20(-\sin 30°\hat{\mathrm{i}}+\cos 30°\hat{\mathrm{j}})\times 2 \\ =-20\hat{\mathrm{i}}+34.64\hat{\mathrm{j}}\mathrm{miles} \end{gathered}$$

Distance for ship B is:

$$\begin{gathered} \overrightarrow{{\mathrm{r}}_{\mathrm{B}}}=\overrightarrow{{\mathrm{v}}_{\mathrm{B}}}\mathrm{t} \\ =2\times (\sin 20°\hat{\mathrm{i}}+\cos 20°\hat{\mathrm{j}})\times 2 \\ =17.1\hat{\mathrm{i}}+46.98\hat{\mathrm{j}}\mathrm{miles} \end{gathered}$$

Now, distance between ship A and Ship B is:

$$\begin{gathered} \overrightarrow{\left|\mathrm{d}\right|}=\left|\overrightarrow{{\mathrm{r}}_{\mathrm{B}}}-\overrightarrow{{\mathrm{r}}_{\mathrm{A}}}\right| \\ \overrightarrow{\left|\mathrm{d}\right|}=\left|17.1\hat{\mathrm{i}}+46.98\hat{\mathrm{j}}+20\hat{\mathrm{i}}-34.64\hat{\mathrm{j}}\right| \\ \overrightarrow{\left|\mathrm{d}\right|}=\left|37.1\hat{\mathrm{i}}+12.34\hat{\mathrm{j}}\right| \\ \mathrm{d}=\sqrt{{(37.1)}^2+{(12.34)}^2} \\ \mathrm{d}=39.1\mathrm{miles} \end{gathered}$$

Part b) Step 1: Calculation

$$\begin{gathered} \overrightarrow{{\mathrm{v}}_{\mathrm{AB}}}=\overrightarrow{{\mathrm{v}}_{\mathrm{A}}}-\overrightarrow{{\mathrm{v}}_{\mathrm{B}}} \\ \overrightarrow{{\mathrm{v}}_{\mathrm{AB}}}=20(-\sin 30°\hat{\mathrm{i}}+\cos 30°\hat{\mathrm{j}})-25(\sin 20°\hat{\mathrm{i}}+\cos 20°\hat{\mathrm{j}}) \\ \overrightarrow{{\mathrm{v}}_{\mathrm{AB}}}=-18.55\hat{\mathrm{i}}-6.17\hat{\mathrm{j}} \\ \left|\overrightarrow{{\mathrm{v}}_{\mathrm{AB}}}\right|=\sqrt{{(-18.55)}^2+{(-6.17)}^2} \\ \left|\overrightarrow{{\mathrm{v}}_{\mathrm{AB}}}\right|=19.55\mathrm{mph} \\ \left|\overrightarrow{{\mathrm{v}}_{\mathrm{AB}}}\right|=20\mathrm{mph} \end{gathered}$$