For `1//m` of the distance between two stations a train is uniformly accelerated and for `1//n` of the distance it is uniformly retarded. It starts from rest at one station and comes to rest at the other. Then the ratio of its greatest velocity to its average velocity is

Options

A. `1+1//m:1//n`
B. `1+1//m+1//n:1`
C. `1+1//n:1//m`
D. None of these

Answer

To solve the problem, we need to find the ratio of the greatest velocity (V_max) to the average velocity (V_avg) of a train that accelerates uniformly for a distance of \( \frac{s}{m} \) and then decelerates uniformly for a distance of \( \frac{s}{n} \), starting from rest and coming to rest at the other station.

### Step-by-Step Solution:

1. **Define the total distance**:
Let the total distance between the two stations be \( s \). The train accelerates over a distance of \( \frac{s}{m} \) and decelerates over a distance of \( \frac{s}{n} \).

2. **Calculate maximum velocity during acceleration**:
Using the equation of motion \( v^2 = u^2 + 2as \), where \( u = 0 \) (initial velocity), we can find the maximum velocity \( V_{max} \) at the end of the acceleration phase.
\[ V_{max}^2 = 0 + 2a \left(\frac{s}{m}\right) \implies V_{max} = \sqrt{2a \frac{s}{m}} \]
Here, \( a \) is the acceleration.

3. **Calculate acceleration**:
From the equation \( V_{max}^2 = 2a \frac{s}{m} \), we can express acceleration \( a \) as:
\[ a = \frac{V_{max}^2 m}{2s} \]

4. **Calculate deceleration**:
For the deceleration phase, we use the same equation of motion:
\[ 0 = V_{max}^2 + 2(-r)\left(\frac{s}{n}\right) \implies V_{max}^2 = 2r \left(\frac{s}{n}\right) \implies r = \frac{V_{max}^2 n}{2s} \]
Here, \( r \) is the deceleration.

5. **Calculate time taken during acceleration (t1)**:
The time taken to accelerate can be calculated using:
\[ t_1 = \frac{V_{max}}{a} = \frac{V_{max}}{\frac{V_{max}^2 m}{2s}} = \frac{2s}{V_{max} m} \]

6. **Calculate time taken during deceleration (t2)**:
Similarly, the time taken to decelerate is:
\[ t_2 = \frac{V_{max}}{r} = \frac{V_{max}}{\frac{V_{max}^2 n}{2s}} = \frac{2s}{V_{max} n} \]

7. **Calculate time taken during constant velocity (t3)**:
The distance covered during constant velocity is:
\[ t_3 = \frac{s - \left(\frac{s}{m} + \frac{s}{n}\right)}{V_{max}} = \frac{s\left(1 - \frac{1}{m} - \frac{1}{n}\right)}{V_{max}} \]

8. **Total time (T)**:
The total time taken for the journey is:
\[ T = t_1 + t_2 + t_3 = \frac{2s}{V_{max} m} + \frac{2s}{V_{max} n} + \frac{s\left(1 - \frac{1}{m} - \frac{1}{n}\right)}{V_{max}} \]

9. **Average velocity (V_avg)**:
The average velocity is defined as:
\[ V_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{s}{T} \]

10. **Calculate the ratio \( \frac{V_{max}}{V_{avg}} \)**:
Finally, we can find the ratio of maximum velocity to average velocity:
\[ \frac{V_{max}}{V_{avg}} = 1 + \frac{1}{m} + \frac{1}{n} \]

### Final Answer:
The ratio of the greatest velocity to the average velocity is:
\[ \frac{V_{max}}{V_{avg}} = 1 + \frac{1}{m} + \frac{1}{n} \]