An engine can move at the speed of 20/3 m/s without any wagon attached and reduction in the speed of the train is directly proportional to the square root of the number of wagons attached to the engine. When there are 9 wagons attached its speed is 5 m/s. The greatest number of wagons with which the engine can move is :

Options

A. 144
B. 12
C. 143
D. 24

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Understand the problem
The engine's speed without any wagons is \( \frac{20}{3} \) m/s. The reduction in speed is directly proportional to the square root of the number of wagons attached.

### Step 2: Set up the equation for speed reduction
Let \( w \) be the number of wagons. The decrease in speed can be expressed as:
\[ \text{Decrease in speed} = k \sqrt{w} \]
where \( k \) is a constant.

### Step 3: Calculate the decrease in speed when 9 wagons are attached
When there are 9 wagons, the speed of the engine is 5 m/s. Thus, the decrease in speed when there are 9 wagons is:
\[ \text{Decrease in speed} = \frac{20}{3} - 5 \]
Calculating this gives:
\[ \frac{20}{3} - 5 = \frac{20}{3} - \frac{15}{3} = \frac{5}{3} \text{ m/s} \]

### Step 4: Substitute into the equation to find \( k \)
Now we can substitute into the equation:
\[ \frac{5}{3} = k \sqrt{9} \]
Since \( \sqrt{9} = 3 \), we have:
\[ \frac{5}{3} = k \cdot 3 \]
Solving for \( k \):
\[ k = \frac{5}{3} \cdot \frac{1}{3} = \frac{5}{9} \]

### Step 5: Set up the equation for maximum wagons
Now we can express the decrease in speed for any number of wagons \( w \):
\[ \text{Decrease in speed} = \frac{5}{9} \sqrt{w} \]
The speed of the engine becomes zero when the decrease in speed equals the original speed:
\[ \frac{20}{3} = \frac{5}{9} \sqrt{w} \]

### Step 6: Solve for \( w \)
To find \( w \), we rearrange the equation:
\[ \sqrt{w} = \frac{20}{3} \cdot \frac{9}{5} \]
Calculating the right side:
\[ \sqrt{w} = \frac{20 \cdot 9}{3 \cdot 5} = \frac{180}{15} = 12 \]
Now, squaring both sides gives:
\[ w = 12^2 = 144 \]

### Step 7: Determine the maximum number of wagons
The maximum number of wagons that can be attached without the engine stopping is one less than this value:
\[ \text{Maximum number of wagons} = 144 - 1 = 143 \]

### Final Answer
The greatest number of wagons with which the engine can move is **143**.

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