The principle of relative velocity states that the velocity of the ferry relative to the ground ($\vec{v}_{fg}$) is the vector sum of its velocity relative to the water ($\vec{v}_{fw}$) and the velocity of the water relative to the ground ($\vec{v}_{wg}$).

$$\vec{v}_{fg} = \vec{v}_{fw} + \vec{v}_{wg}$$

Let the y-axis be perpendicular to the river banks (direction of crossing) and the x-axis be parallel to the banks (direction of current). The vectors are:

• Resultant velocity: $\vec{v}_{fg} = v_{fg} \hat{j}$ (since the path is perpendicular to the banks)
• Current velocity: $\vec{v}_{wg} = v_{wg} \hat{i}$
• Ferry velocity relative to water: $\vec{v}_{fw} = v_{fw,x} \hat{i} + v_{fw,y} \hat{j}$

Substituting into the vector equation and separating components:

• x-component: $0 = v_{fw,x} + v_{wg} \implies v_{fw,x} = -v_{wg}$
• y-component: $v_{fg} = v_{fw,y}$

The magnitude of the ferry's velocity relative to water, $v_{fw}$, is given by the Pythagorean theorem on its components: $v_{fw}^2 = v_{fw,x}^2 + v_{fw,y}^2$. Substituting the relations above yields a right-triangle relationship between the speed magnitudes:

$$v_{fw}^2 = v_{wg}^2 + v_{fg}^2$$

[Q1] What is the speed of the river current? First, we find the ferry's resultant speed across the river, $v_{fg}$, using the river width $w$ and crossing time $t$.

$$v_{fg} = \frac{w}{t}$$

Next, we solve the Pythagorean relation for the speed of the current, $v_{wg}$.

$$v_{wg} = \sqrt{v_{fw}^2 - v_{fg}^2}$$

Substituting the expression for $v_{fg}$:

$$v_{wg} = \sqrt{v_{fw}^2 - \left(\frac{w}{t}\right)^2}$$

Now, we substitute the given values: $w = 600$ m, $t = 1.0$ min $= 60$ s, and $v_{fw} = 20$ m/s.

$$v_{fg} = \frac{600 \, \text{m}}{60 \, \text{s}} = 10 \, \text{m/s}$$ $$v_{wg} = \sqrt{(20 \, \text{m/s})^2 - (10 \, \text{m/s})^2} = \sqrt{300} \, \text{m/s} \approx 17.3 \, \text{m/s}$$

[Q2] What is the angle between the boat's heading and the direction of the current? Let $\theta$ be the angle between the heading vector $\vec{v}_{fw}$ and the current vector $\vec{v}_{wg}$. From the vector component analysis, the heading vector is $\vec{v}_{fw} = -v_{wg} \hat{i} + v_{fg} \hat{j}$ and the current vector is $\vec{v}_{wg} = v_{wg} \hat{i}$.

The angle can be found from the geometry of the vector triangle. Let $\phi$ be the angle the ferry heads upstream relative to the direct perpendicular path. In the right triangle of speeds, this angle is between the hypotenuse $v_{fw}$ and the leg $v_{fg}$.

$$\cos(\phi) = \frac{v_{fg}}{v_{fw}}$$

The direction of the current is perpendicular ($90^\circ$) to the resultant path. Since the ferry heads upstream, the total angle $\theta$ between the heading and the current is:

$$\theta = 90^\circ + \phi = 90^\circ + \arccos\left(\frac{v_{fg}}{v_{fw}}\right)$$

Substituting the values:

$$\theta = 90^\circ + \arccos\left(\frac{10 \, \text{m/s}}{20 \, \text{m/s}}\right) = 90^\circ + \arccos(0.5)$$ $$\theta = 90^\circ + 60^\circ = 150^\circ$$