A boat is capable of a maximum speed of 6 km/hr in calm water. The boat needs to cross a river that is 0.5 km wide to a point directly across from the starting point. If the speed of the water in the river is 4 km/hr, what is the minimum time required to cross and end up at a point directly across from the starting point?

To solve this problem, we need to find the minimum time required for the boat to cross the river and end up directly across from the starting point. Let's analyze the situation step by step.

### Problem Breakdown
1. **Boat's Speed in Calm Water**: \( v_b = 6 \) km/hr.
2. **River's Width**: \( d = 0.5 \) km.
3. **River's Current Speed**: \( v_r = 4 \) km/hr.
4. **Objective**: The boat must end up at a point directly across from the starting point.

To achieve this, the boat must have a component of its velocity that counters the river's current, ensuring it moves straight across.

### Solution Strategy

1. **Boat's Velocity Relative to the River**:
   The boat moves with a speed of \( 6 \) km/hr in still water. However, due to the river's current of \( 4 \) km/hr, we need to find the angle at which the boat should be directed to counteract the current and achieve a straight crossing.

   Let \( \theta \) be the angle the boat makes with the river flow direction, aiming upstream.

   The boat's velocity components:
   - **Across the river (perpendicular component)**: \( v_b \cos \theta \).
   - **Against the river's current (upstream component)**: \( v_b \sin \theta \).

   The boat must counteract the river's speed, so:
   \[
   v_b \sin \theta = v_r
   \]

2. **Finding the Angle \( \theta \):**
   \[
   6 \sin \theta = 4 \implies \sin \theta = \frac{4}{6} = \frac{2}{3}
   \]
   \[
   \theta = \arcsin \left( \frac{2}{3} \right)
   \]

3. **Velocity Across the River:**
   The component of the boat's velocity across the river (perpendicular to the current) is:
   \[
   v_{\text{across}} = v_b \cos \theta
   \]

   Using the Pythagorean identity:
   \[
   \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left( \frac{2}{3} \right)^2} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}
   \]

   Thus:
   \[
   v_{\text{across}} = 6 \cdot \frac{\sqrt{5}}{3} = 2\sqrt{5} \, \text{km/hr}
   \]

4. **Minimum Time to Cross the River:**
   The time to cross the river is given by:
   \[
   \text{Time} = \frac{\text{Distance}}{\text{Velocity across the river}} = \frac{d}{v_{\text{across}}}
   \]

   Substituting the values:
   \[
   \text{Time} = \frac{0.5}{2\sqrt{5}} \, \text{hours}
   \]

   Simplify:
   \[
   \text{Time} = \frac{1}{4\sqrt{5}} \, \text{hours} \approx \frac{1}{4 \times 2.236} \approx \frac{1}{8.944} \approx 0.112 \, \text{hours}
   \]

   Converting to minutes:
   \[
   \text{Time} \approx 0.112 \times 60 \approx 6.72 \, \text{minutes}
   \]

### Conclusion
The minimum time required for the boat to cross the river and end up directly across from the starting point is approximately **6.72 minutes**.

Would you like a more detailed explanation or have any other questions?

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Here are some related questions you might find interesting:
1. How would the crossing time change if the river's speed increased?
2. What if the boat could travel faster in calm water? How would that affect the crossing time?
3. How would the boat's path change if it didn't aim directly to counteract the river's current?
4. How can we calculate the boat's path if it aims at a different angle?
5. What is the impact of wind resistance on the boat's time to cross the river?

**Tip:** When dealing with problems involving relative motion, it's crucial to resolve velocity components along and perpendicular to the direction of motion.

Find out how to calculate the minimum time for a boat to cross a 0.5 km wide river, considering a 6 km/hr boat speed and a 4 km/hr current. Using relative velocity, trigonometric identities, and time-distance relationships, this solution explains the step-by-step process. Suitable for students in Grades 9-12.

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