1. Analyzing the Two Boats

Let $v_b = 1.0$ m/s be the speed of the boats relative to the water, and $v_w$ be the river current velocity.

Boat 1 (Minimum Path): Appears moving perpendicular to the river current to a ground observer. This means its resultant velocity is directly across the river.

The velocity component along the river must be zero:

$$ v_b \sin \theta = v_w $$

The resultant velocity across the river (y-direction) is:

$$ v_{y1} = \sqrt{v_b^2 – v_w^2} $$

Boat 2 (Minimum Time): Appears moving perpendicular to the shoreline to a boy on a raft. Since the raft moves with the river, the relative velocity is just the boat’s velocity relative to water. This implies the boat heads straight across.

The velocity component across the river is simply:

$$ v_{y2} = v_b $$

2. Using Displacement Data

In a certain time interval $t$, the distances from the shoreline are $\Delta y_1 = 4.0$ m and $\Delta y_2 = 5.0$ m.

$$ \Delta y_1 = v_{y1} t \quad \text{and} \quad \Delta y_2 = v_{y2} t $$

Taking the ratio:

$$ \frac{\Delta y_1}{\Delta y_2} = \frac{v_{y1}}{v_{y2}} = \frac{\sqrt{v_b^2 – v_w^2}}{v_b} $$

Substituting the values:

$$ \frac{4}{5} = \sqrt{1 – \left(\frac{v_w}{v_b}\right)^2} $$

Squaring both sides:

$$ \frac{16}{25} = 1 – \left(\frac{v_w}{v_b}\right)^2 $$

$$ \left(\frac{v_w}{v_b}\right)^2 = 1 – \frac{16}{25} = \frac{9}{25} $$

$$ \frac{v_w}{v_b} = \frac{3}{5} = 0.6 $$

Since $v_b = 1.0$ m/s, the river current velocity is:

$$ v_w = 0.6 \, \text{m/s} $$