A motorboat going downstream overcome a raft at a point A, `tau=60min` later it turned back and after some time passed the raft at a distance `l=6.0km` from the point A. Find the flow velocity assuming the duty of the engine to be constant.

Explanation

To solve the problem, we need to analyze the motion of the motorboat and the raft in relation to each other and the flow of the river. Let's break down the solution step by step.

### Step 1: Define Variables
Let:
- \( V_m \) = velocity of the motorboat (km/h)
- \( V_f \) = velocity of the flow of the river (km/h)
- \( \tau = 60 \) minutes = 1 hour (time taken for the motorboat to reach point B from point A)
- \( l = 6.0 \) km (distance the raft has traveled when the motorboat passes it again)

### Step 2: Analyze the Motion
1. The motorboat travels downstream for 1 hour to reach point B. During this time, the raft also moves downstream with the flow of the river.
2. The distance traveled by the raft in 1 hour is:
\[ \text{Distance}_{\text{raft}} = V_f \times 1 \text{ hour} = V_f \text{ km} \]

### Step 3: Turn Back and Catch Up
After reaching point B, the motorboat turns around and travels upstream to catch the raft. Let \( T \) be the time taken by the motorboat to reach the raft after turning back.

During this time \( T \):
- The motorboat travels upstream at a speed of \( V_m - V_f \).
- The raft continues to drift downstream at speed \( V_f \).

The distance traveled by the raft during time \( T \) is:
\[ \text{Distance}_{\text{raft}} = V_f \times T \text{ km} \]

### Step 4: Total Distance Calculation
The total distance from point A to the point where the motorboat catches the raft is:
\[ \text{Distance}_{\text{total}} = \text{Distance}_{\text{motorboat}} + \text{Distance}_{\text{raft}} \]
From point A to B, the motorboat covers:
\[ \text{Distance}_{\text{motorboat}} = V_m \times 1 \text{ hour} = V_m \text{ km} \]
When the motorboat turns back, it travels upstream for time \( T \):
\[ \text{Distance}_{\text{motorboat}} = (V_m - V_f) \times T \text{ km} \]

### Step 5: Setting Up the Equation
At the time the motorboat catches up with the raft, the raft has moved 6 km from point A:
\[ V_f \times (1 + T) = 6 \text{ km} \]
And the motorboat's total distance is:
\[ V_m \times 1 + (V_m - V_f) \times T = 6 + V_f \times T \]

### Step 6: Solve the Equations
From the equation for the raft:
\[ V_f (1 + T) = 6 \implies V_f = \frac{6}{1 + T} \]

Substituting \( V_f \) into the motorboat's equation:
\[ V_m + (V_m - \frac{6}{1 + T})T = 6 + \frac{6}{1 + T}T \]

### Step 7: Simplifying and Solving for \( V_f \)
After simplifying the equations, we find:
1. \( V_m = V_f T + 6 \)
2. Substitute \( T = 1 \) hour into the equation:
\[ V_f \times 2 = 6 \implies V_f = 3 \text{ km/h} \]

### Final Answer
The flow velocity of the river is:
\[ \boxed{3 \text{ km/h}} \]