To solve this problem, we need to break down the swimmer's motion into components and understand how the current affects their trajectory. Here's the approach:

Step 1: Understanding the components

• The swimmer is trying to swim from point A to point B.
• d1 = 200 mm is the distance from point A to C (the starting point to the point on the opposite bank directly across the river).
• d2 = 150 mm is the distance from point C to point B (downstream, influenced by the river's current).
• vr = 5 km/h is the velocity of the river's current, which is moving horizontally (perpendicular to the direction the swimmer wants to go).

The swimmer's velocity relative to the water is at an angle $\theta = 45^\circ$ to the line from A to C. This means the swimmer's velocity relative to the water has two components:

• A component that aims directly across the river (along the line from A to C).
• A component that tries to counteract the effect of the current and prevent the swimmer from drifting downstream.

Step 2: Breakdown of the swimmer’s velocity components

Let $v_s$ be the swimmer’s speed relative to the water.

1. The swimmer's component along the river’s flow (horizontal direction, parallel to the current): $v_{s_x} = v_s \sin(\theta)$ Since $\theta = 45^\circ$, we know: $v_{s_x} = v_s \sin(45^\circ) = \frac{v_s}{\sqrt{2}}$

2. The swimmer's component perpendicular to the river’s flow (vertical direction, across the river): $v_{s_y} = v_s \cos(\theta)$ Since $\theta = 45^\circ$, we know: $v_{s_y} = v_s \cos(45^\circ) = \frac{v_s}{\sqrt{2}}$

Step 3: Balancing the current

To swim directly from A to B, the swimmer must counteract the current. This means the horizontal component of the swimmer's velocity ($v_{s_x}$) must cancel out the effect of the river’s current velocity ($v_r$).

Thus, we set the horizontal velocity component equal to the velocity of the current: $v_{s_x} = v_r$ Substituting $v_{s_x} = \frac{v_s}{\sqrt{2}}$ and $v_r = 5 \, \text{km/h}$, we get: $\frac{v_s}{\sqrt{2}} = 5 \, \text{km/h}$

Step 4: Solving for $v_s$

Solving for $v_s$: $v_s = 5 \sqrt{2} \, \text{km/h}$ $v_s \approx 7.07 \, \text{km/h}$

Conclusion:

The swimmer needs to swim at a speed of approximately 7.07 km/h relative to the water to swim directly from point A to point B.

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Do you need further clarification on any part of this solution? Or would you like to explore more examples?

Additional Questions:

1. How does the angle $\theta$ affect the swimmer’s required speed?
2. If the swimmer swims slower than 7.07 km/h, how would their path change?
3. What if the current was stronger, say $10 \, \text{km/h}$?
4. How would the swimmer’s trajectory look if they swam at an angle of $30^\circ$ instead of $45^\circ$?
5. Can the swimmer reach point B directly if they swim faster than the required speed? How?

Tip: Always break down a vector problem into components along the directions of motion to simplify the analysis.