This problem involves analyzing the motion of a boat crossing a river, taking into account the river's width $y_1$, the boat's speed $v_B$, and the current's velocity $v_x$ in two cases. Let’s break down each part:

1. Given Data:
   • River width $y_1 = 100 \, \text{m}$
   • Boat's speed $v_B = 1.00 \, \text{m/s}$
   • Constant flow speed of the river $v_F = 0.80 \, \text{m/s}$
   • Coefficient $c = 0.33 \times 10^{-1} \, \text{m}^{-1} \cdot \text{s}^{-1}$

Part (a)

Question: What is the displacement $x_1$ of the boat if the current $v_x$ is constant?

Since $v_x = v_F$ is constant, the boat’s horizontal displacement $x_1$ due to the river's current can be calculated as it crosses the river.

Solution for (a):

1. Time taken $t$ to cross the river vertically (width $y_1$): $t = \frac{y_1}{v_B} = \frac{100 \, \text{m}}{1.00 \, \text{m/s}} = 100 \, \text{s}$

2. The horizontal displacement $x_1$: $x_1 = v_x \cdot t = v_F \cdot t = 0.80 \, \text{m/s} \times 100 \, \text{s} = 80 \, \text{m}$

Thus, $x_1 = 80 \, \text{m}$.

Part (b)

Question: What is the displacement $x_1$ if the current $v_x$ depends on the distance from the bank, modeled as $v_x = c y (y_1 - y)$?

In this scenario, $v_x$ is variable, so we would integrate to find $x_1$.

Solution for (b): We would integrate the current velocity over the time taken to cross the river. Given $v_x = c y (y_1 - y)$, set up the integration with respect to $y$ from $0$ to $y_1$: $x_1 = \int_0^{y_1} v_x \, dt = \int_0^{y_1} c y (y_1 - y) \, \frac{dy}{v_B}$

Performing this integration will yield the displacement $x_1$ under the variable current condition.

Part (c)

Question: Under what angle $\alpha$ should the boat be steered relative to the bank normal to arrive directly opposite the starting point in case (a)?

To counteract the drift, the boat should be directed at an angle $\alpha$ so that its resultant velocity vector points directly across the river.

Solution for (c): Using trigonometry: $\tan(\alpha) = \frac{v_x}{v_B}$ $\alpha = \arctan\left(\frac{v_F}{v_B}\right) = \arctan\left(\frac{0.80}{1.00}\right)$ $\alpha = \arctan(0.80) \approx 38.66^\circ$

Thus, the boat should be steered at approximately $38.66^\circ$ against the current.

Would you like further details on these solutions or have any questions?

Here are some additional questions related to this problem:

1. How would the time to cross the river change if $v_B$ increased?
2. What effect would an increase in $y_1$ have on the boat's displacement in part (a)?
3. How would the answer in part (b) change if $c$ were different?
4. How does the steering angle $\alpha$ vary with changes in $v_F$?
5. What would happen to the boat's trajectory if the current direction changed?

Tip: When dealing with variable speeds, always consider integration over the distance or time to capture the cumulative effect.