When two bodies move uniformly towards each other, the distance between them diminishes by `16 m` every `10 s`. If bodies move with velocities of the same magnitude and in the same direction as before the distance between then will decrease `3 m` every `5 s`. The velocity of each body is.

Explanation

To solve the problem step by step, we will analyze the two cases presented in the question.

### Step 1: Analyze the first case
In the first case, two bodies are moving towards each other. We denote the velocities of the two bodies as \( V_1 \) and \( V_2 \).

Given that the distance between them diminishes by 16 m every 10 s, we can express this as:
\[ \text{Relative speed} = \frac{\text{Distance}}{\text{Time}} = \frac{16 \, \text{m}}{10 \, \text{s}} = 1.6 \, \text{m/s} \]

Since they are moving towards each other, the relative speed is the sum of their speeds:
\[ V_1 + V_2 = 1.6 \quad \text{(Equation 1)} \]

### Step 2: Analyze the second case
In the second case, both bodies are moving in the same direction. The distance between them diminishes by 3 m every 5 s. We can express this as:
\[ \text{Relative speed} = \frac{\text{Distance}}{\text{Time}} = \frac{3 \, \text{m}}{5 \, \text{s}} = 0.6 \, \text{m/s} \]

Since they are moving in the same direction, the relative speed is the difference of their speeds:
\[ V_1 - V_2 = 0.6 \quad \text{(Equation 2)} \]

### Step 3: Solve the equations
Now we have two equations:
1. \( V_1 + V_2 = 1.6 \)
2. \( V_1 - V_2 = 0.6 \)

We can add these two equations to eliminate \( V_2 \):
\[ (V_1 + V_2) + (V_1 - V_2) = 1.6 + 0.6 \]
This simplifies to:
\[ 2V_1 = 2.2 \]
Now, divide both sides by 2:
\[ V_1 = 1.1 \, \text{m/s} \]

### Step 4: Find \( V_2 \)
Now that we have \( V_1 \), we can substitute it back into Equation 1 to find \( V_2 \):
\[ 1.1 + V_2 = 1.6 \]
Subtract \( 1.1 \) from both sides:
\[ V_2 = 1.6 - 1.1 = 0.5 \, \text{m/s} \]

### Conclusion
The velocities of the two bodies are:
- \( V_1 = 1.1 \, \text{m/s} \)
- \( V_2 = 0.5 \, \text{m/s} \)