Chapter 9.1, Problem 56E

Velocity Suppose that in Exercise 55 the current is flowing at 1.2 mi/h due south. In what direction should the swimmer head in order to arrive at a landing point due east of his starting point?

To determine

The direction of the swimmer to land at a point due east of his starting point.

Answer to Problem 56E

The direction of the swimmer to land at a point due east of his starting point is $\text{N}\left(\text{53}\text{.14}\right)°\text{E}$.

Explanation of Solution

Given:

The water current is flowing at $1.2\text{mi/h}$ due south and the swimmer swims at a speed of

$2\text{mi/h}$.

Calculation:

Consider a coordinate system with the origin as the starting point of swimmer. Vector u represents velocity vector of heading of swimmer, vector v represents velocity of river along negative $y$ direction and vector p represents the resultant true course of swimmer.

Figure (1)

In figure (1), the velocity of river v is along negative y direction and magnitue of velocity of river is $1.2\text{mi/h}$, therefore,

$v=-1.2j$.

The magnitude of velocity of swimmer is $2\text{mi/h}$.

If $\theta$ is the angle made by vector u with the positive $x$ direction then, horizontal component $u_1$ of velocity of swimmer u is $u_1=\left|u\right|\cos \theta$.

Substitute 2 for $\left|u\right|$ in above formula.

$\begin{array}{c}{u}_1=\left|u\right|\cos \theta \\ =2\cos \theta \end{array}$

Vertical component of velocity u of swimmer is $u_2=\left|u\right|\sin \theta$.

Substitute 2 for $\left|u\right|$ in above formula.

$\begin{array}{c}{u}_2=\left|u\right|\sin \theta \\ =2\sin \theta \end{array}$

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

Substitute $2\cos \theta$ for $u_1$ and $2\sin \theta$ for $u_2$ in above relation.

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

If p represents the resultant true course of swimmer then, $p=u+v$.

Substitute $\{\left(2\cos \theta \right)i+\left(2\sin \theta \right)j\}$ for u and $-1.2j$ for v in above relation.

$\begin{array}{c}p=\left(2\cos \theta \right)i+\left(2\sin \theta \right)j+\left(-1.2\right)j\\ =\left(2\cos \theta \right)i+\left(2\sin \theta -1.2\right)j\end{array}$

Since the swimmer has to land at a point due east of starting point, therefore vertical component of vector p should be 0.

$\begin{array}{c}2\sin \theta -1.2=0\\ 2\sin \theta =1.2\\ \sin \theta =\frac{1.2}{2}\\ \sin \theta =0.6\end{array}$

Therefore $\theta$ is given by,

$\begin{array}{c}\theta ={\sin }^{-1}\left(0.6\right)\\ =36.86\end{array}$

Thus, direction is given by,

$\text{N}\left(90-36.86\right)°\text{E=N}\left(53.14\right)°\text{E}$.

Therefore, the swimmer should head in the direction $\text{N}\left(53.14\right)°\text{E}$.