A train moves from rest with acceleration `alpha` and in time `t_(1)` covers a distance x. It then decelerates rest at constant retardation `beta` for distance y in time `t_(2)`. Then

Options

A. `(x)/(y) = (beta)/(alpha)`
B. `(beta)/(alpha)=(t_(1))/(t_(2))`
C. x=y
D. `(x)/(y)=(betat_(1))/(alphat_(2))`

Explanation

To solve the problem step by step, we will analyze the motion of the train in two phases: acceleration and deceleration.

### Step 1: Analyze the acceleration phase
The train starts from rest and accelerates with an acceleration \( \alpha \) for a time \( t_1 \) covering a distance \( x \).

Using the equation of motion:
\[ x = ut + \frac{1}{2} a t^2 \]
where \( u = 0 \) (initial velocity), \( a = \alpha \), and \( t = t_1 \):
\[ x = 0 + \frac{1}{2} \alpha t_1^2 \]
Thus, we have:
\[ x = \frac{1}{2} \alpha t_1^2 \quad \text{(1)} \]

### Step 2: Find the final velocity after acceleration
The final velocity \( v_0 \) at the end of the acceleration phase can be calculated using:
\[ v = u + at \]
Substituting \( u = 0 \), \( a = \alpha \), and \( t = t_1 \):
\[ v_0 = 0 + \alpha t_1 = \alpha t_1 \quad \text{(2)} \]

### Step 3: Analyze the deceleration phase
In the second phase, the train decelerates with a constant retardation \( \beta \) and comes to rest after covering a distance \( y \) in time \( t_2 \).

Using the equation of motion again:
\[ y = v_0 t_2 - \frac{1}{2} \beta t_2^2 \]
Substituting \( v_0 = \alpha t_1 \):
\[ y = (\alpha t_1) t_2 - \frac{1}{2} \beta t_2^2 \]
Thus, we have:
\[ y = \alpha t_1 t_2 - \frac{1}{2} \beta t_2^2 \quad \text{(3)} \]

### Step 4: Relate the distances and times
From equations (1) and (3), we can find a relationship between \( x \), \( y \), \( t_1 \), and \( t_2 \).

Using the relationship between the slopes of the velocity-time graph:
\[ \frac{\beta}{\alpha} = \frac{t_1}{t_2} \quad \text{(4)} \]

### Step 5: Find the relationship between \( x \) and \( y \)
From equation (1):
\[ t_1^2 = \frac{2x}{\alpha} \]
From equation (4):
\[ t_1 = \frac{\beta}{\alpha} t_2 \]
Substituting \( t_1 \) into the expression for \( y \):
\[ y = \alpha \left(\frac{\beta}{\alpha} t_2\right) t_2 - \frac{1}{2} \beta t_2^2 \]
This simplifies to:
\[ y = \beta t_2^2 - \frac{1}{2} \beta t_2^2 = \frac{1}{2} \beta t_2^2 \]
Now we can relate \( x \) and \( y \):
From (1):
\[ x = \frac{1}{2} \alpha t_1^2 = \frac{1}{2} \alpha \left(\frac{\beta}{\alpha} t_2\right)^2 = \frac{1}{2} \frac{\beta^2}{\alpha} t_2^2 \]
Thus:
\[ \frac{x}{y} = \frac{\frac{1}{2} \frac{\beta^2}{\alpha} t_2^2}{\frac{1}{2} \beta t_2^2} = \frac{\beta}{\alpha} \]

### Final Result
The final relationship we derived is:
\[ \frac{x}{y} = \frac{\beta}{\alpha} \]