Question

A boat is headed at 115° and is traveling downstream at 28 mph. The stream is flowing S38°E at 11 mph. In what direction is the boat traveling? At what rate can it travel in still water?

Answer 1

A boat is headed at $115{}^{\circ}$ and is traveling downstream at 28 mph. The stream is flowing S38°E at 11 mph. In what direction is the boat traveling? At what rate can it travel in still water?

**Solution:**

$$S38{}^{\circ}E = 180{}^{\circ} - 38{}^{\circ} = 142{}^{\circ}$$

$\overrightarrow{v_1}$ – vector of the boat velocity in still water;

$\overrightarrow{v_2}$ – vector of the stream velocity;

$$|\overrightarrow{v_2}| = 11 \text{ mph}$$

$\vec{v}$ – the resultant velocity of the boat,

$$|\vec{v}| = 28 \text{ mph}$$

$\alpha$ – the angle between the headed direction of boat and the direction of stream,

$\beta$ – the angle between the headed direction of boat and the direction the boat is traveling.

$$\alpha = 142{}^{\circ} - 115{}^{\circ} = 27{}^{\circ}$$ $$\alpha = \beta + \gamma$$

Use the sine law:

$$\frac{|\vec{v}|}{\sin(180{}^{\circ} - \alpha)} = \frac{|\overrightarrow{v_1}|}{\sin\beta} = \frac{|\overrightarrow{v_2}|}{\sin\gamma}$$ $$\gamma = \arcsin\left(\frac{\overrightarrow{v_2} \cdot \sin(180{}^{\circ} - \alpha)}{\vec{v}}\right) = \arcsin\left(\frac{11 \cdot \sin153{}^{\circ}}{28}\right) = 10,27{}^{\circ}$$

The direction the boat is traveling is the direction the boat is headed plus $\gamma$

$$115{}^{\circ} + 10,27{}^{\circ} = 125,27{}^{\circ}$$ $$\beta = 27{}^{\circ} - 10,27{}^{\circ} = 16,73{}^{\circ}$$ $$|\overrightarrow{v_1}| = \frac{\sin\beta \cdot |\overrightarrow{v_2}|}{\sin\gamma} = 17,76 \text{ mph}$$

**Answer:**

The boat is traveling $125,27{}^{\circ}$. It can travel at $17,76 \text{ mph}$ in still water.