An engine can pull four coaches at a maximum speed of `20ms^-1`. The mass of the engine is twice the mass of every coach. Assuming resistive forces to be proportional to the weight, approximate maximum speeds of the engine, when it pulls `12` and `6` coaches, are

Options

A. (a) `8.5ms^-1` and `15ms^-1`, respectively
B. (b) `6.5ms^-1` and `8ms^-1`, respectively
C. (c) `8.5ms^-1` and `13ms^-1` respectively
D. (d) `10.5ms^-1` and `15ms^-1`, respectively

Explanation

To solve the problem step by step, we will analyze the given information and apply the principles of physics related to work, power, and energy.

### Given Information:
- An engine can pull 4 coaches at a maximum speed of \(20 \, \text{m/s}\).
- The mass of the engine is twice the mass of each coach.
- We need to find the maximum speeds when the engine pulls 12 and 6 coaches.

### Step 1: Define Variables
Let:
- The mass of one coach = \(m\)
- The mass of the engine = \(2m\)
- Total mass when pulling 4 coaches = \(2m + 4m = 6m\)

### Step 2: Calculate Power
The power \(P\) when pulling 4 coaches at maximum speed can be expressed as:
\[ P = \text{Force} \times \text{Velocity} \]
The resistive force \(F\) is proportional to the weight, which can be expressed as:
\[ F = k \cdot \text{Total mass} \cdot g = k \cdot 6m \cdot g \]
Thus, the power can be written as:
\[ P = (k \cdot 6m \cdot g) \cdot 20 \]

### Step 3: Calculate Maximum Speed for 12 Coaches
When the engine pulls 12 coaches:
- Total mass = \(2m + 12m = 14m\)
- The power remains the same, so:
\[ P = (k \cdot 14m \cdot g) \cdot v_{12} \]
Setting the two expressions for power equal gives:
\[ k \cdot 6m \cdot g \cdot 20 = k \cdot 14m \cdot g \cdot v_{12} \]
Cancelling \(k\), \(m\), and \(g\) from both sides:
\[ 6 \cdot 20 = 14 \cdot v_{12} \]
Solving for \(v_{12}\):
\[ v_{12} = \frac{6 \cdot 20}{14} = \frac{120}{14} \approx 8.57 \, \text{m/s} \]

### Step 4: Calculate Maximum Speed for 6 Coaches
When the engine pulls 6 coaches:
- Total mass = \(2m + 6m = 8m\)
- The power is still the same:
\[ P = (k \cdot 8m \cdot g) \cdot v_{6} \]
Setting the two expressions for power equal gives:
\[ k \cdot 6m \cdot g \cdot 20 = k \cdot 8m \cdot g \cdot v_{6} \]
Cancelling \(k\), \(m\), and \(g\) from both sides:
\[ 6 \cdot 20 = 8 \cdot v_{6} \]
Solving for \(v_{6}\):
\[ v_{6} = \frac{6 \cdot 20}{8} = \frac{120}{8} = 15 \, \text{m/s} \]

### Final Answers:
- The maximum speed when pulling 12 coaches is approximately \(8.57 \, \text{m/s}\).
- The maximum speed when pulling 6 coaches is \(15 \, \text{m/s}\).