Chapter 9.1, Problem 64E

Velocity of a Boat The boater in Exercise 63 wants to arrive at a point on the north shore of the river directly opposite the starting point. In what direction should the boat be headed?

To determine

The direction of the boat to head so that boat arrives at opposite point on north shore.

Answer to Problem 64E

The boat should be headed in a direction such that angle of $120°$ is made with the shore.

Explanation of Solution

Given:

The speed of river flow is $10\text{mi/h}$ due east and the speed of motorboat is $20\text{mi/h}$.

Calculation:

Take a coordinate system with the origin as the starting point of motorboat. Vector you represent velocity vector of heading of motorboat, vector v represents velocity of river along positive $x$ direction and vector p represents the resultant true course of motoboat.

Figure (1)

In Figure (1), the velocity v of river is along positive $x$ direction and magnitue of velocity of river is $10\text{mi/h}$, therefore,

$\begin{array}{c}v=10i+0j\\ =10i\end{array}$

The velocity vector of river flow is $10i$.

The magnitude of velocity of motorboat is $20\text{mi/h}$.

 If $\theta$ is the angle made by vector u with the positive $x$ direction then, horizontal component $u_1$ of velocity u of motorboat is,

$u_1=\left|u\right|\cos \theta$

Substitute $20$ for $\left|u\right|$ in above formula.

$\begin{array}{c}{u}_1=\left|u\right|\cos \theta \\ =20\cos \theta \end{array}$

Vertical component of velocity u of swimmer is,

$u_2=\left|u\right|\sin \theta$

Substitute $20$ for $\left|u\right|$ in above formula.

$\begin{array}{c}{u}_2=\left|u\right|\sin \theta \\ =20\sin \theta \end{array}$

Velocity vector u of swimmer is given by $u=u_{1}i+u_{2}j$.

Substitute $20\cos \theta$ for $u_1$ and $20\sin \theta$ for $u_2$ in above relation.

$\begin{array}{c}u=u_{1}i+u_{2}j\\ =\left(20\cos \theta \right)i+\left(20\sin \theta \right)j\end{array}$

If p represents the resultant true course of motorboat then,

$p=u+v$

Substitute $\{\left(20\cos \theta \right)i+\left(20\sin \theta \right)j\}$ for u and $10i$ for v in above relation.

$\begin{array}{c}p=\left(20\cos \theta \right)i+\left(20\sin \theta \right)j+10i\\ =\left(20\cos \theta +10\right)i+\left(20\sin \theta \right)j\end{array}$

Since the motorboat has to reach at opposite point of starting point, therefore horizontal component of vector p should be zero.

$20\cos \theta +10=0$

Solve for $\cos \theta$.

$\begin{array}{c}\cos \theta =\frac{-10}{20}\\ =-\frac{1}{2}\end{array}$

Therefore, $\theta$ is given by,

$\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{-1}{2}\right)\\ =120°\end{array}$

Thus, the boat should be headed in a direction such that angle of $120°$ is made with the shore.