Let the width of the river be $d$, the speed of the ferry in still water be $v_f$, and the speed of the river current be $v_c$. We establish a coordinate system where the y-axis is directly across the river and the x-axis is downstream.

The velocity of the ferry relative to the ground, $\vec{v}_g$, is the vector sum of its velocity relative to the water, $\vec{v}_f$, and the velocity of the water (current) relative to the ground, $\vec{v}_c$.

$$\vec{v}_g = \vec{v}_f + \vec{v}_c$$

Let the ferry head at an angle $\theta$ with respect to the y-axis (the line directly across the river). The components of the velocities are:

$\vec{v}_f = (v_f \sin\theta) \hat{i} + (v_f \cos\theta) \hat{j}$ $\vec{v}_c = v_c \hat{i}$

The ferry's velocity relative to the ground is:

$\vec{v}_g = (v_f \sin\theta + v_c) \hat{i} + (v_f \cos\theta) \hat{j}$

The time $t$ to cross the river of width $d$ depends only on the y-component of the ground velocity, $v_{g,y}$:

$$t = \frac{d}{v_{g,y}} = \frac{d}{v_f \cos\theta}$$

[Q1] Direction for Shortest Time To minimize the crossing time $t$, the denominator $v_f \cos\theta$ must be maximized. Since $v_f$ is a constant speed, we must maximize $\cos\theta$. The maximum value of $\cos\theta$ is 1, which occurs when $\theta = 0^\circ$. Therefore, the ferry should head directly across the river, perpendicular to the current.

[Q2] Path Length for Shortest Time Crossing For the shortest time crossing, we set $\theta = 0^\circ$. The time taken is:

$$t_{min} = \frac{d}{v_f \cos(0^\circ)} = \frac{d}{v_f}$$

During this time, the ferry is carried downstream by the current. The distance it drifts downstream, $x_{drift}$, is determined by the x-component of its ground velocity, $v_{g,x} = v_f \sin(0^\circ) + v_c = v_c$.

$$x_{drift} = v_{g,x} \cdot t_{min} = v_c \frac{d}{v_f}$$

The total path length, $L$, is the magnitude of the total displacement vector, which has components $d$ (across) and $x_{drift}$ (downstream). Using the Pythagorean theorem:

$$L = \sqrt{d^2 + x_{drift}^2} = \sqrt{d^2 + \left(\frac{v_c d}{v_f}\right)^2}$$ $$L = d \sqrt{1 + \left(\frac{v_c}{v_f}\right)^2}$$

Substituting the given values:

$d = 600$ m, $v_f = 20$ m/s, $v_c = 17.32$ m/s. $$L = 600 \sqrt{1 + \left(\frac{17.32}{20}\right)^2} = 600 \sqrt{1 + (0.866)^2}$$ $$L = 600 \sqrt{1 + 0.749956} = 600 \sqrt{1.749956} \approx 600(1.32286)$$ $$L \approx 793.7 \text{ m}$$