Determining Swimmer's Effective Velocity Perpendicular to River Flow

This problem involves understanding relative velocities, specifically when a swimmer moves in a river with a current. We need to find the swimmer's speed directly across the river, considering both their swimming ability and the river's flow, under the specific condition that they land exactly opposite their starting point.

Understanding Relative Velocity in River Crossing

When dealing with movement in fluids like water, we often consider relative velocities. The swimmer's speed in still water is their velocity relative to the water. The river's current is its velocity relative to the bank. The swimmer's actual path and speed relative to the bank (their effective velocity) is the vector sum of their velocity relative to the water and the velocity of the water relative to the bank.

Mathematically, this is represented as:

$ \vec{v}_{effective} = \vec{v}_{swimmer\_relative\_to\_water} + \vec{v}_{river\_relative\_to\_bank} $

The key condition given is that the swimmer aims to land directly opposite their starting point. This means that the swimmer's resultant velocity relative to the bank must be purely directed across the river, perpendicular to the flow. Consequently, the component of the swimmer's effective velocity parallel to the river flow must be zero.

Calculating the Swimmer's Velocity Components

Let's define the velocities involved:

• $v_s$: The swimmer's speed in still water, which is given as $4$ km/h. This represents the magnitude of the swimmer's velocity relative to the water, so $|\vec{v}_{swimmer\_relative\_to\_water}| = 4$ km/h.
• $v_r$: The speed of the river current, given as $2$ km/h. This is the magnitude of the river's velocity relative to the bank, so $|\vec{v}_{river\_relative\_to\_bank}| = 2$ km/h.

To analyze the motion, we can set up a coordinate system. Let the direction directly across the river be the y-axis, and the direction along the river flow be the x-axis.

The river's velocity is entirely along the x-axis (the direction of flow). So, in vector form:

$ \vec{v}_{river\_relative\_to\_bank} = (v_r, 0) = (2, 0) \text{ km/h} $

The swimmer's velocity relative to the water, $\vec{v}_{swimmer\_relative\_to\_water}$, has a magnitude of $v_s = 4$ km/h. Let its components along the x and y axes be $(v_{sx}, v_{sy})$. According to the Pythagorean theorem in vector space:

$ v_{sx}^2 + v_{sy}^2 = v_s^2 = 4^2 = 16 $

The effective velocity of the swimmer relative to the bank, $\vec{v}_{effective}$, is the vector sum:

$ \vec{v}_{effective} = \vec{v}_{swimmer\_relative\_to\_water} + \vec{v}_{river\_relative\_to\_bank} $

Substituting the components, we get:

$ \vec{v}_{effective} = (v_{sx}, v_{sy}) + (2, 0) = (v_{sx} + 2, v_{sy}) $

The condition that the swimmer lands directly opposite their starting point implies that their effective velocity relative to the bank has no component parallel to the river flow (the x-component must be zero). Let the components of the effective velocity be $(v_{ex}, v_{ey})$:

$ v_{ex} = v_{sx} + 2 = 0 $

Solving this equation for $v_{sx}$, we find the component of the swimmer's velocity (relative to water) needed in the direction against the current:

$ v_{sx} = -2 \text{ km/h} $

This negative sign indicates that the swimmer must swim partially upstream relative to the water to counteract the river's current and achieve a resultant path straight across.

Finding the Magnitude of Perpendicular Velocity

The question asks for the magnitude of the swimmer's effective velocity perpendicular to the river flow. This is the y-component of the effective velocity, $v_{ey}$, which is equal to $v_{sy}$ from our component analysis.

We can find the value of $v_{sy}$ using the Pythagorean relationship for the swimmer's velocity relative to the water:

$ v_{sx}^2 + v_{sy}^2 = v_s^2 $

Substitute the values we know ($v_{sx} = -2$ km/h and $v_s = 4$ km/h):

$ (-2)^2 + v_{sy}^2 = 4^2 $

$ 4 + v_{sy}^2 = 16 $

Now, we solve for $v_{sy}^2$:

$ v_{sy}^2 = 16 - 4 $

$ v_{sy}^2 = 12 $

To find $v_{sy}$, we take the square root. Since $v_{sy}$ represents the speed component across the river, we consider the positive value:

$ v_{sy} = \sqrt{12} $

Simplify the square root:

$ v_{sy} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3} \text{ km/h} $

Thus, the magnitude of the swimmer's effective velocity perpendicular to the river flow is $2\sqrt{3}$ km/h.