To ships 10 km apart on a line running south to north . The one father north is steaming west at 20 km/h. The other is steaming north at 20 km/h. The distance of closest approach is . . . . to reach it the time taken is . . . .

Answer

To solve the problem of finding the distance of closest approach and the time taken for two ships moving in different directions, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Initial Setup**:
- We have two ships: Ship A is 10 km south of Ship B.
- Ship A is moving west at a speed of 20 km/h.
- Ship B is moving north at a speed of 20 km/h.

2. **Setting Up the Coordinate System**:
- Let Ship B be at point (0, 10) (10 km north).
- Let Ship A be at point (0, 0) (0 km, at the origin).
- Ship A moves west, so its position at time \( t \) is given by \( (-20t, 0) \).
- Ship B moves north, so its position at time \( t \) is given by \( (0, 10 + 20t) \).

3. **Finding the Distance Between the Ships**:
- The distance \( d \) between the two ships at time \( t \) can be expressed using the distance formula:
\[ d = \sqrt{((-20t) - 0)^2 + (0 - (10 + 20t))^2} \]
- Simplifying this gives:
\[ d = \sqrt{(20t)^2 + (-(10 + 20t))^2} \]
\[ = \sqrt{400t^2 + (10 + 20t)^2} \]
\[ = \sqrt{400t^2 + (100 + 400t + 400t^2)} \]
\[ = \sqrt{800t^2 + 400t + 100} \]

4. **Finding the Minimum Distance**:
- To find the minimum distance, we need to minimize the function \( d^2 = 800t^2 + 400t + 100 \).
- We can find the minimum by taking the derivative and setting it to zero:
\[ \frac{d(d^2)}{dt} = 1600t + 400 = 0 \]
\[ 1600t = -400 \implies t = -\frac{1}{4} \text{ (not valid as time cannot be negative)} \]
- We need to check the critical points or use the second derivative test to find the minimum distance.

5. **Using Geometry for Closest Approach**:
- The closest distance can also be found geometrically. The ships are moving at right angles to each other.
- The closest approach occurs when the line connecting the two ships is perpendicular to their velocities.
- The distance of closest approach can be calculated using the right triangle formed by their paths:
\[ \text{Distance of closest approach} = \sqrt{(10)^2 + (20)^2} = \sqrt{100 + 400} = \sqrt{500} = 10\sqrt{5} \text{ km} \]

6. **Calculating the Time Taken**:
- The relative velocity of the two ships can be calculated as:
\[ v_{rel} = \sqrt{(20)^2 + (20)^2} = 20\sqrt{2} \text{ km/h} \]
- The time taken to reach the closest approach can be calculated using:
\[ t = \frac{\text{Distance}}{\text{Relative Velocity}} = \frac{10\sqrt{5}}{20\sqrt{2}} = \frac{\sqrt{5}}{4\sqrt{2}} \text{ hours} \]

### Final Answers:
- **Distance of Closest Approach**: \( 10\sqrt{5} \) km
- **Time Taken**: \( \frac{\sqrt{5}}{4\sqrt{2}} \) hours