A passenger reaches the platform and finds that the second last boggy of the train is passing him. The second last boggy takes `3 s` to pass the passenger, and the last boggy takes `2 s` to pass him. Find the time by which the passnger late for the departure of the trains? Assume that the train accelerates at constant rate and all the boggies are of equal length.

Explanation

To solve the problem, we need to analyze the motion of the train and the time taken by the bogies to pass the passenger. Let's break it down step by step:

### Step 1: Define Variables
Let:
- \( L \) = length of each bogey
- \( N \) = total number of bogies in the train
- \( T \) = time taken for the third last bogey to pass the passenger
- \( a \) = acceleration of the train

### Step 2: Time Taken by Each Bogey
From the problem:
- The second last bogey takes \( 3 \) seconds to pass the passenger.
- The last bogey takes \( 2 \) seconds to pass the passenger.

### Step 3: Distance Calculation
The distance traveled by the train while the second last bogey passes the passenger can be expressed as:
\[ \text{Distance} = \text{Speed} \times \text{Time} \]
For the second last bogey, the distance it covers while passing the passenger is:
\[ L = (v + a(T + 3)) \cdot 3 \]
Where \( v \) is the initial speed of the train when the second last bogey starts passing the passenger.

For the last bogey, the distance is:
\[ L = (v + a(T + 2)) \cdot 2 \]

### Step 4: Set Up Equations
From the above two equations, we can set up the following:
1. \( L = 3(v + a(T + 3)) \) (1)
2. \( L = 2(v + a(T + 2)) \) (2)

### Step 5: Equate the Two Equations
Since both equations equal \( L \), we can set them equal to each other:
\[ 3(v + a(T + 3)) = 2(v + a(T + 2)) \]

### Step 6: Simplify the Equation
Expanding both sides:
\[ 3v + 3a(T + 3) = 2v + 2a(T + 2) \]
Rearranging gives:
\[ 3v - 2v + 3aT + 9a = 2aT + 4a \]
This simplifies to:
\[ v + aT + 5a = 0 \]
Thus, we have:
\[ v = -aT - 5a \]

### Step 7: Substitute Back
Now, substitute \( v \) back into either equation (1) or (2) to find \( T \). Let's use equation (1):
\[ L = 3(-aT - 5a + a(T + 3)) \]
This simplifies to:
\[ L = 3(-aT - 5a + aT + 3a) \]
\[ L = 3(-2a) \]
This gives us:
\[ L = -6a \]

### Step 8: Calculate the Time Late
The passenger is late by the time it takes for the third last bogey to pass the passenger:
\[ T + 3 = 3 + 3 = 6 \text{ seconds} \]
Thus, the passenger is late by \( 3.5 \) seconds.

### Final Answer
The passenger is late for the departure of the train by **3.5 seconds**.

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