A dog weighing `5kg` is standing on a flat boat so that it is `10` metres from the shore. It walks `4m` on the boat towards the shore and then halts. The boat weighs `20kg` and one can assume that there is no friction between it and water. The dog from the shore at the end of this time is

Options

A. `3.4 m`
B. `6.8 m`
C. `12.6m`
D. `10 m`

Explanation

To solve the problem step-by-step, we need to analyze the situation involving the dog and the boat.

### Step 1: Understand the Initial Conditions
- The dog weighs \(5 \, \text{kg}\) and is initially \(10 \, \text{m}\) from the shore.
- The boat weighs \(20 \, \text{kg}\).
- The dog walks \(4 \, \text{m}\) towards the shore.

### Step 2: Define Variables
- Let \(x\) be the distance the boat moves away from the shore when the dog walks \(4 \, \text{m}\) towards it.
- After the dog walks \(4 \, \text{m}\), the effective distance the dog has moved towards the shore is \(4 - x\).

### Step 3: Apply the Center of Mass Principle
Since there are no external forces acting on the system (dog + boat), the center of mass of the system will remain stationary.

The position of the center of mass before the dog walks can be calculated as:
\[ \text{Initial Center of Mass} = \frac{m_d \cdot d_d + m_b \cdot d_b}{m_d + m_b} \]
Where:
- \(m_d = 5 \, \text{kg}\) (mass of the dog)
- \(d_d = 10 \, \text{m}\) (initial distance of the dog from the shore)
- \(m_b = 20 \, \text{kg}\) (mass of the boat)
- \(d_b = 10 \, \text{m}\) (initial distance of the boat from the shore)

Calculating the initial center of mass:
\[ \text{Initial Center of Mass} = \frac{5 \cdot 10 + 20 \cdot 10}{5 + 20} = \frac{50 + 200}{25} = \frac{250}{25} = 10 \, \text{m} \]

### Step 4: Calculate the Final Positions
After the dog walks \(4 \, \text{m}\) towards the shore, the new positions will be:
- The dog is now at \(10 - (4 - x)\) meters from the shore.
- The boat moves \(x\) meters away from the shore, so its new position is \(10 + x\) meters from the shore.

### Step 5: Set Up the Equation for Center of Mass
Since the center of mass does not change, we have:
\[ \text{Final Center of Mass} = \frac{m_d \cdot (10 - (4 - x)) + m_b \cdot (10 + x)}{m_d + m_b} \]
Setting the initial and final center of mass equal gives:
\[ 10 = \frac{5 \cdot (6 + x) + 20 \cdot (10 + x)}{25} \]

### Step 6: Solve for \(x\)
Expanding and simplifying:
\[ 10 = \frac{30 + 5x + 200 + 20x}{25} \]
\[ 10 = \frac{230 + 25x}{25} \]
Multiplying both sides by \(25\):
\[ 250 = 230 + 25x \]
\[ 25x = 20 \implies x = \frac{20}{25} = 0.8 \, \text{m} \]

### Step 7: Find the Final Position of the Dog
Now, substituting \(x\) back to find the final position of the dog:
\[ \text{Final position of the dog} = 10 - (4 - 0.8) = 10 - 3.2 = 6.8 \, \text{m} \]

### Conclusion
The dog is \(6.8 \, \text{m}\) from the shore at the end of this time.