Problem 11.120 Shore-based radar indicates that a ferry leaves its slip with a velocity v 18 km/h 70°, while instruments aboard the ferry indicate a speed of 18.4 km/h and a heading of 30° west of south relative to the river. Determine the velocity of the river.

To determine the velocity of the river, we will need to use vector addition. The ferry's velocity according to the shore-based radar represents the resultant velocity that combines the ferry's own velocity relative to the water and the river's velocity.

Let's assume: - \( \vec{V}_f \) is the velocity of the ferry relative to the river. - \( \vec{V}_r \) is the velocity of the river. - \( \vec{V}_{rf} \) is the resultant velocity of the ferry relative to the shore.

According to the shore-based radar: \( \vec{V}_{rf} = 18 \text{ km/h at } 70 ^\circ \text{ to the east of north.} \)

According to the instruments aboard the ferry: \( \vec{V}_f = 18.4 \text{ km/h at } 30 ^\circ \text{ to the west of south or equivalently, } 180^\circ - 30^\circ = 150 ^\circ \text{ to the east of north.} \)

The relationship between these vectors is given by the vector sum: \( \vec{V}_f + \vec{V}_r = \vec{V}_{rf} \)

To solve for \( \vec{V}_r \), we can rearrange the equation: \( \vec{V}_r = \vec{V}_{rf} - \vec{V}_f \)

Now let's break down the vectors into their northward and eastward components.

For \( \vec{V}_{rf} \): Northward component: \( V_{rf_N} = 18 \cos(70^\circ) \) Eastward component: \( V_{rf_E} = 18 \sin(70^\circ) \)

For \( \vec{V}_f \): Northward component: \( V_{f_N} = 18.4 \cos(150^\circ) \) Eastward component: \( V_{f_E} = 18.4 \sin(150^\circ) \)

Now we can compute these components using trigonometric functions: \( V_{rf_N} = 18 \times \cos(70^\circ) \) \( V_{rf_E} = 18 \times \sin(70^\circ) \)

\( V_{f_N} = 18.4 \times \cos(150^\circ) \) \( V_{f_E} = 18.4 \times \sin(150^\circ) \)

Since we are looking for \( \vec{V}_r \), the components of the river's velocity will be: \( V_{r_N} = V_{rf_N} - V_{f_N} \) \( V_{r_E} = V_{rf_E} - V_{f_E} \)

After calculating the northward and eastward components of \( \vec{V}_r \), we can then find the magnitude \( V_r \) and direction \( \theta \) of the river's velocity:

Magnitude: \( V_r = \sqrt{V_{r_N}^2 + V_{r_E}^2} \)

Direction: \( \theta = \arctan(\frac{V_{r_E}}{V_{r_N}}) \)

Note that the direction will be relative to the north (if \( V_{r_N} \) is positive) or south (if \( V_{r_N} \) is negative), and either east or west depending on the sign of \( V_{r_E} \).

Remember to keep track of the signs of the components to accurately determine the direction, and if necessary, adjust the angle based on which quadrant your resultant vector falls into.

Extra: Vectors in physics represent quantities that have both magnitude and direction. Examples include forces, velocities, and accelerations. To add vectors, you can use the parallelogram rule where you place the vectors tip to tail and draw the resultant from the free tail to the free tip of the combined vectors.

Ferry problems like this are a classic application of relative motion in physics. The actual motion of the ferry as seen from the shore (the resultant velocity) is the vector addition of the water's motion (in this case, the river's current) and the ferry's motion through the water.

To do vector addition analytically, you typically break down vectors into their components along the x-axis (east-west direction) and y-axis (north-south direction). By convention, east and north are considered positive directions, while west and south are negative. Calculations involving sines, cosines, and arctangents as you might recall from trigonometry classes are necessary to resolve these vectors into their components and to calculate the resultant vector from component sums or differences.