The greatest acceleration or deceleration that a train may have is a. The minimum time in which the train may reach form one station to the other seprated by a distance d is-

Options

A. `4sqrt(d/a)`
B. `sqrt((2d)/a)`
C. `1/2 sqrt(d/a)`
D. `2sqrt(d/a)`

Answer

To solve the problem of finding the minimum time \( T \) for a train to travel a distance \( D \) with a maximum acceleration \( A \), we can break down the solution into the following steps:

### Step 1: Understand the Motion
The train starts from rest, accelerates to a maximum velocity, and then decelerates back to rest. For minimum time, the train will accelerate for half the time and decelerate for the other half.

### Step 2: Define Variables
- Let \( A \) be the maximum acceleration (and deceleration).
- Let \( D \) be the total distance between the two stations.
- Let \( T \) be the total time taken to travel the distance \( D \).
- The time spent accelerating is \( \frac{T}{2} \) and the time spent decelerating is also \( \frac{T}{2} \).

### Step 3: Calculate Maximum Velocity
Using the formula for velocity under constant acceleration:
\[ V = A \cdot t \]
During the acceleration phase (time \( \frac{T}{2} \)):
\[ V = A \cdot \frac{T}{2} \]

### Step 4: Calculate Distance During Acceleration
The distance covered during the acceleration phase can be calculated using the formula for distance under constant acceleration:
\[ d = \frac{1}{2} A t^2 \]
For the acceleration phase:
\[ D_1 = \frac{1}{2} A \left(\frac{T}{2}\right)^2 = \frac{1}{2} A \cdot \frac{T^2}{4} = \frac{AT^2}{8} \]

### Step 5: Calculate Distance During Deceleration
The distance covered during the deceleration phase is the same as the acceleration phase since the train comes to rest at the end:
\[ D_2 = \frac{1}{2} A \left(\frac{T}{2}\right)^2 = \frac{AT^2}{8} \]

### Step 6: Total Distance
The total distance \( D \) is the sum of the distances during acceleration and deceleration:
\[ D = D_1 + D_2 = \frac{AT^2}{8} + \frac{AT^2}{8} = \frac{AT^2}{4} \]

### Step 7: Solve for Time \( T \)
Now, we can set the total distance equal to \( D \):
\[ D = \frac{AT^2}{4} \]
Rearranging gives:
\[ T^2 = \frac{4D}{A} \]
Taking the square root of both sides:
\[ T = \sqrt{\frac{4D}{A}} = 2\sqrt{\frac{D}{A}} \]

### Final Answer
Thus, the minimum time \( T \) for the train to travel the distance \( D \) is:
\[ T = 2\sqrt{\frac{D}{A}} \]