A train is moving forward with a horizontal acceleration a. A man sitting in the train drops a coin on the floor of the train . The acceleration of the coin w.r.t man is

Options

A. g downward
B. a downward
C. ` sqrt(a^(2) + g^(2)) at tan ^(-1)(g/a)` with horizontal in forward direction
D. ` sqrt ( a^(2) + g^(2) ) at tan ^(-1) (g/a)` with horizontal in backward direction

Answer

To solve the problem, we need to find the acceleration of the coin with respect to the man sitting in a train that is accelerating horizontally. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the scenario
The train is moving forward with a horizontal acceleration \( a \). When the man drops the coin, it will have two components of motion: the vertical motion due to gravity and the horizontal motion due to the acceleration of the train.

### Step 2: Identify the forces acting on the coin
1. **Vertical Motion**: The only force acting on the coin in the vertical direction is gravity, which accelerates the coin downwards with an acceleration \( g \).
2. **Horizontal Motion**: The coin, when dropped, does not have any horizontal force acting on it (assuming air resistance is negligible). However, since the train is accelerating, the man (and the train) will move forward with acceleration \( a \).

### Step 3: Set up the acceleration components
- The acceleration of the coin in the vertical direction is \( -g \) (downwards).
- The acceleration of the coin in the horizontal direction is \( 0 \) (since it is dropped and not given any horizontal velocity).

### Step 4: Calculate the acceleration of the coin with respect to the man
The acceleration of the coin with respect to the man can be expressed as:
\[ \vec{a}_{cm} = \vec{a}_{coin} - \vec{a}_{man} \]
Where:
- \( \vec{a}_{coin} = (0, -g) \) (coin's acceleration)
- \( \vec{a}_{man} = (a, 0) \) (man's acceleration in the train)

Thus, the relative acceleration becomes:
\[ \vec{a}_{cm} = (0, -g) - (a, 0) = (-a, -g) \]

### Step 5: Calculate the magnitude of the acceleration
The magnitude of the acceleration of the coin with respect to the man can be calculated using the Pythagorean theorem:
\[ |\vec{a}_{cm}| = \sqrt{(-a)^2 + (-g)^2} = \sqrt{a^2 + g^2} \]

### Step 6: Determine the direction of the acceleration
The direction of the acceleration can be found using the tangent function:
\[ \tan(\theta) = \frac{g}{a} \]
Thus, the angle \( \theta \) can be expressed as:
\[ \theta = \tan^{-1}\left(\frac{g}{a}\right) \]

### Final Result
The acceleration of the coin with respect to the man is:
\[ \text{Magnitude: } \sqrt{a^2 + g^2} \]
\[ \text{Direction: } \tan^{-1}\left(\frac{g}{a}\right) \text{ (backward relative to the direction of the train's acceleration)} \]