A bomber plane moves due east at `100 km h^-1` over a town `T` at a certain instant of time. Six minutes later, an interceptor plane sets off flying due north - east from the station `S` which is `40 km` south of `T`. If both maintain their courses, find the velocity with which the interceptor plane must fly in order to just overtake the bomber.

Explanation

To solve the problem, we need to determine the velocity of the interceptor plane so that it can overtake the bomber plane. Let's break down the solution step by step.

### Step 1: Understand the positions of the planes
- The bomber plane is flying due east at a speed of 100 km/h.
- The interceptor plane starts flying northeast from a station that is 40 km south of town T, 6 minutes after the bomber has already taken off.

### Step 2: Calculate the distance traveled by the bomber before the interceptor takes off
- The bomber flies for 6 minutes before the interceptor starts.
- Convert 6 minutes to hours:
\[ 6 \text{ minutes} = \frac{6}{60} \text{ hours} = 0.1 \text{ hours} \]
- Distance traveled by the bomber in this time:
\[ \text{Distance} = \text{Speed} \times \text{Time} = 100 \text{ km/h} \times 0.1 \text{ h} = 10 \text{ km} \]

### Step 3: Determine the position of the bomber when the interceptor starts
- At the time the interceptor starts, the bomber is 10 km east of town T.

### Step 4: Calculate the time taken by the bomber to reach point A
- Let point A be the point where the interceptor overtakes the bomber. The horizontal distance from T to A is:
\[ TA = 10 \text{ km} + x \text{ km} \quad (\text{where } x \text{ is the additional distance the bomber travels}) \]
- The time taken by the bomber to reach point A:
\[ T_b = \frac{TA}{100} \text{ hours} \]

### Step 5: Calculate the time taken by the interceptor to reach point A
- The interceptor starts flying 6 minutes later, so the time taken by the interceptor to reach point A is:
\[ T_i = T_b - 0.1 \text{ hours} \]

### Step 6: Set up the relationship between distances
- The interceptor is flying northeast, which means it has to cover a distance of:
\[ SA = \sqrt{(10 + x)^2 + 40^2} \]
- The time taken by the interceptor to reach point A is:
\[ T_i = \frac{SA}{V} \text{ where } V \text{ is the speed of the interceptor} \]

### Step 7: Equate the times
- We know that:
\[ T_i = T_b - 0.1 \]
- Substitute the expressions for \( T_i \) and \( T_b \):
\[ \frac{\sqrt{(10 + x)^2 + 1600}}{V} = \frac{10 + x}{100} - 0.1 \]

### Step 8: Solve for V
- Rearranging gives:
\[ V = \frac{\sqrt{(10 + x)^2 + 1600}}{\frac{10 + x}{100} - 0.1} \]
- Substitute \( x = 40 \sqrt{2} \) (the distance from S to A) and solve for V.

### Step 9: Final calculation
- After substituting and simplifying, we find:
\[ V = \frac{400 \sqrt{2}}{3} \text{ km/h} \]

Thus, the velocity with which the interceptor plane must fly in order to just overtake the bomber is \( \frac{400 \sqrt{2}}{3} \) km/h.