A train is moving forward at a velocity of 2.0m/s. At the instant the train begins to accelerate at `0.80 m//s^(2)` a passenger drops a coin which takes 0.50s to fall to the floor. Relative to a spot on the floor directly under the coin at release, it lands. 

Options

A. 1.1 m towards the rear of the train
B. 1.0 m towards the rear of the train
C. 0.10 m towards the rear of the train
D. 0.90 m towards towards the front of the train

Answer

To solve the problem step by step, we need to analyze the motion of the coin and the train. Here's the detailed solution:

### Step 1: Understand the initial conditions
- The train is moving forward with a velocity \( v_t = 2.0 \, \text{m/s} \).
- The train starts to accelerate at \( a_t = 0.80 \, \text{m/s}^2 \).
- A coin is dropped from the train, and it takes \( t = 0.50 \, \text{s} \) to fall to the floor.

### Step 2: Determine the motion of the coin
When the coin is dropped, it has the same horizontal velocity as the train, which is \( 2.0 \, \text{m/s} \). Therefore, the initial horizontal velocity of the coin \( v_c = 2.0 \, \text{m/s} \).

### Step 3: Calculate the distance traveled by the coin in the horizontal direction
Since the coin is dropped and not thrown, it will continue to move horizontally with the same velocity as the train for the duration of its fall. The horizontal distance \( d_c \) traveled by the coin while it falls can be calculated using the formula:
\[ d_c = v_c \cdot t \]
Substituting the known values:
\[ d_c = 2.0 \, \text{m/s} \cdot 0.50 \, \text{s} = 1.0 \, \text{m} \]

### Step 4: Calculate the distance the train moves during the fall
During the time the coin is falling, the train is also accelerating. We can calculate the distance \( d_t \) that the train travels during this time using the formula for distance under constant acceleration:
\[ d_t = v_t \cdot t + \frac{1}{2} a_t \cdot t^2 \]
Substituting the known values:
\[ d_t = 2.0 \, \text{m/s} \cdot 0.50 \, \text{s} + \frac{1}{2} \cdot 0.80 \, \text{m/s}^2 \cdot (0.50 \, \text{s})^2 \]
Calculating each term:
\[ d_t = 1.0 \, \text{m} + \frac{1}{2} \cdot 0.80 \cdot 0.25 = 1.0 \, \text{m} + 0.1 \, \text{m} = 1.1 \, \text{m} \]

### Step 5: Determine the relative position of the coin when it lands
The coin moves horizontally \( 1.0 \, \text{m} \) while the train moves \( 1.1 \, \text{m} \). Therefore, the relative position of the coin with respect to the train when it lands is:
\[ \text{Relative position} = d_t - d_c = 1.1 \, \text{m} - 1.0 \, \text{m} = 0.1 \, \text{m} \]

### Conclusion
The coin lands \( 0.1 \, \text{m} \) towards the rear of the train from the point directly below where it was released.