A train starts from rest and moves with uniform acceleration `alpha` for some time and acquires a velocity `v` it then moves with constant velocity for some time and then decelerates at rate `beta` and finally comes to rest at the next station. If L is distance between two stations then total time of travel is

Options

A. `(L)/(V)+(V)/(2)((1)/(alpha)+(1)/(beta))`
B. `(L)/(V)-(V)/(2)((1)/(alpha)+(1)/(beta))`
C. `(L)/(V)-(V)/(2)((1)/(alpha)-(1)/(beta))`
D. `(L)/(V)+(V)/(2)((1)/(alpha)-(1)/(beta))`

Explanation

To solve the problem step by step, we will break down the journey of the train into three parts: acceleration, constant velocity, and deceleration.

### Step 1: Calculate Time for Acceleration (T1)
The train starts from rest and accelerates uniformly with acceleration \( \alpha \) until it reaches a velocity \( v \).

Using the equation of motion:
\[ v = u + at \]
where \( u = 0 \) (initial velocity), \( a = \alpha \), and \( v \) is the final velocity.

Rearranging gives:
\[ t_1 = \frac{v - 0}{\alpha} = \frac{v}{\alpha} \]

### Step 2: Calculate Distance Covered During Acceleration (s1)
Using the equation of motion:
\[ s = ut + \frac{1}{2}at^2 \]
Substituting \( u = 0 \), we get:
\[ s_1 = 0 + \frac{1}{2} \alpha t_1^2 \]
Substituting \( t_1 = \frac{v}{\alpha} \):
\[ s_1 = \frac{1}{2} \alpha \left(\frac{v}{\alpha}\right)^2 = \frac{v^2}{2\alpha} \]

### Step 3: Calculate Time for Constant Velocity (T2)
During the constant velocity phase, the distance covered is \( s_2 \). The total distance \( L \) is given, and we need to find \( s_2 \):
\[ s_2 = L - s_1 - s_3 \]
where \( s_3 \) is the distance covered during deceleration.

### Step 4: Calculate Time for Deceleration (T3)
The train decelerates at a rate \( \beta \) until it comes to rest. Using the equation:
\[ 0 = v - \beta t_3 \]
Rearranging gives:
\[ t_3 = \frac{v}{\beta} \]

### Step 5: Calculate Distance Covered During Deceleration (s3)
Using the equation of motion:
\[ 0 = v^2 - 2\beta s_3 \]
Rearranging gives:
\[ s_3 = \frac{v^2}{2\beta} \]

### Step 6: Substitute s1 and s3 into s2
Now we can find \( s_2 \):
\[ s_2 = L - s_1 - s_3 = L - \frac{v^2}{2\alpha} - \frac{v^2}{2\beta} \]
Combining the terms gives:
\[ s_2 = L - \frac{v^2}{2}\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) \]

### Step 7: Calculate Time for Constant Velocity (T2)
Now we can find \( T_2 \):
\[ t_2 = \frac{s_2}{v} = \frac{L - \frac{v^2}{2}\left(\frac{1}{\alpha} + \frac{1}{\beta}\right)}{v} \]

### Step 8: Total Time of Travel
The total time \( T \) is the sum of all three times:
\[ T = t_1 + t_2 + t_3 \]
Substituting the values we found:
\[ T = \frac{v}{\alpha} + \left(\frac{L}{v} - \frac{v}{2}\left(\frac{1}{\alpha} + \frac{1}{\beta}\right)\right) + \frac{v}{\beta} \]
Simplifying gives:
\[ T = \frac{L}{v} + \frac{v}{2}\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) \]

### Final Answer
Thus, the total time of travel is:
\[ T = \frac{L}{v} + \frac{v}{2}\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) \]