Chapter 9, Problem 4T

A river is flowing due east at 8 mi/h. A man heads his motorboat in the direction N 30° E in the river. The speed of the motorboat relative to the water is 12 mi/h.

1. (a) Express the true velocity of the motorboat as a vector.
2. (b) Find the true speed and direction of the motorboat.

To determine

The vector which express the true velocity of the motorboat.

Answer to Problem 4T

The vector which express the true velocity of the motorboat is $14i+6\sqrt{3}j$.

Explanation of Solution

Given:

The river is flowing at the speed of 8 mi/h towards east.

The speed of motorboat is 12 mi/h towards $N30°\text{E}$.

Formula used: The vector with magnitude $\left|v\right|$ and direction $\theta$ is given below,

$v=\left|v\right|\cos \theta i+\left|v\right|\sin \theta j$ (1)

The addition of the vectors $u=a_{1}i+a_{2}j$ and $v=b_{1}i+b_{2}j$ is given below,

$u+v=\left(a_1+b_1\right)i+\left(a_2+b_2\right)j$ (2)

Calculation:

The river is flowing at the speed of 8 mi/h towards east. The speed of motorboat is 12 mi/h towards $N30°\text{E}$.

In the below figure vector u represent the river and vector v represent the motorboat.

Figure 1

From Figure 1, it is noticed that the direction of the vector v is $60°$.

The river is flowing at the speed of 8 mi/h towards east, therefore the length of the vector u is 8 and the direction of the vector u is 0.

Substitute 8 for $\left|v\right|$ and 0 for $\theta$ in equation (1), to find the vector u,

$\begin{array}{c}u=\left(8\right)\cos \left(0°\right)i+\left(8\right)\sin \left(0°\right)j\\ =8\left(1\right)i+8\left(0\right)j\left(\because \sin 0°=0,\cos 0°=1\right)\\ =8i+0j\\ =8i\end{array}$

The speed of motorboat is 12 mi/h, therefore the length of the vector v is 12 and the direction of the vector v is $60°$.

Substitute 12 for $\left|v\right|$ and $60°$ for $\theta$ in equation (1), to find the vector v,

$\begin{array}{c}v=\left(12\right)\cos 60°i+\left(12\right)\sin 60°j\\ =12\left(\frac{1}{2}\right)i+12\left(\frac{\sqrt{3}}{2}\right)j\left(\because \sin 60°=\frac{\sqrt{3}}{2},\cos 60°=\frac{1}{2}\right)\\ =6i+6\sqrt{3}j\end{array}$

The true velocity of the motorboat is $u+v$.

Substitute 8 for $a_1$, 0 for $a_2$, 6 for $b_1$ and $6\sqrt{3}$ for $b_2$ in equation (3), to find the true velocity $u+v$,

$\begin{array}{c}u+v=\left(8+6\right)i+\left(0+6\sqrt{3}\right)j\\ =14i+6\sqrt{3}j\end{array}$

Thus, the vector which express the true velocity of the motorboat is $14i+6\sqrt{3}j$.

To determine

To evaluate: The true speed and direction of the motorboat.

Answer to Problem 4T

The true speed of the motorboat is $17.44\text{mi/h}$ and the direction of the motorboat is $\text{N}53.41°\text{E}$.

Explanation of Solution

Given:

The vector which expresses the true velocity of the motorboat is given below,

$w=14i+6\sqrt{3}j$

Formula used: The magnitude and length of a vector $v=\langle a_1,a_2\rangle$ is given below,

$\left|v\right|=\sqrt{a_1^2+a_2^2}$ (3)

The formula to calculate the direction of a vector $v=\langle a_1,a_2\rangle$ is given below,

$\tan \theta =\frac{a_2}{a_1}$ (4)

Where,

• $\theta$ is the direction of the vector v.

Calculation:

Section1:

Substitute 14 for $a_1$ and $6\sqrt{3}$ for $a_2$ in equation (3), to find the true speed of the motorboat,

$\begin{array}{c}\left|w\right|=\sqrt{{\left(14\right)}^2+{\left(6\sqrt{3}\right)}^2}\\ =\sqrt{196+{\left(6\right)}^2{\left(\sqrt{3}\right)}^2}\\ =\sqrt{196+36\left(3\right)}\end{array}$

Simplify the above expression,

$\begin{array}{c}\left|w\right|=\sqrt{196+108}\\ =\sqrt{304}\\ =17.4356\\ \approx 17.44\end{array}$

Thus, the true speed of the motorboat is $17.44\text{mi/h}$.

Section 2:

Substitute 14 for $a_1$ and $6\sqrt{3}$ for $a_2$ in equation (4), to find the true direction of the motorboat,

$\begin{array}{c}\tan \theta =\frac{6\sqrt{3}}{14}\\ \theta ={\tan }^{-1}\left(\frac{10.39}{14}\right)\\ ={\tan }^{-1}\left(0.7423\right)\\ =36.59\end{array}$

The direction from x-axis is 36.59.

The direction of the motorboat from y-axis is given below,

$\begin{array}{c}\theta =90-\theta \\ =90-36.59\\ =53.41\end{array}$

Thus, the true direction of the motorboat is $\text{N}53.41°\text{E}$.