Chapter 11.2, Problem 26E

To determine

To find: The true direction angle of the canoeist’s path and speed of the canoe.

Answer to Problem 26E

The true direction angle of the canoeist’s path and speed of the canoe are $106.325°$ and $2.947mph$ , respectively.

Explanation of Solution

Given information: River is flowing in east at speed $2mph$ . The canoeist crosses river at $4mph$ making angle of $135°$ .

Calculation:

  

Let r is the river vector,

  $\begin{array}{l}\overrightarrow{r}=\langle 2\cos 30°,2\sin 0°\rangle \\ \overrightarrow{r}=\langle 2,0\rangle \end{array}$

Similarly, c is the canoe vector,

  $\begin{array}{l}\overrightarrow{c}=\langle 4\cos 135°,4\sin 135°\rangle \\ \overrightarrow{c}=\langle -2\sqrt{2},2\sqrt{2}\rangle \end{array}$

Now, the sum of r and c vector will be the canoeist’s path vector p.

  $\begin{array}{l}\overrightarrow{p}=\overrightarrow{r}+\overrightarrow{c}\\ \overrightarrow{p}=\langle 2-2\sqrt{2},2\sqrt{2}\rangle \end{array}$

Finding the angle of the vector p to get the true directional angle,

  $\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{2\sqrt{2}}{2-2\sqrt{2}}\right)\\ \theta =-73.675°\end{array}$

The vector p is in the second quadrant, thus, add $180°$

  $\begin{array}{l}\theta =-73.675°+180°\\ \theta =106.325°\end{array}$

Calculating the magnitude of vector p to get how fast the canoe is travelling,

  $\begin{array}{l}\left|\overrightarrow{p}\right|=\sqrt{{\left(2-2\sqrt{2}\right)}^2+{\left(2\sqrt{2}\right)}^2}\\ \left|\overrightarrow{p}\right|=\sqrt{{\left(-0.828\right)}^2+{\left(2.8284\right)}^2}\\ \left|\overrightarrow{p}\right|=\sqrt{8.6832}\\ \left|\overrightarrow{p}\right|=2.947mph\end{array}$