Two spaceships are observed from Earth to be approaching each other along a straight line. Ship A moves at 0.40c relative to the Earth observer, while ship B moves at 0.50c relative to the same observer. What speed does the captain of ship A report for the speed of ship B?

Options

A. 0.10c
B. 0.85c
C. 0.75c
D. 0.95c

Answer

To find the speed of spaceship B as reported by the captain of spaceship A, we need to use the principles of special relativity. The formula for the relativistic addition of velocities is given by:

\[ V_{AB} = \frac{V_{A} + V_{B}}{1 + \frac{V_{A} \cdot V_{B}}{c^2}} \]

Where:
- \( V_{AB} \) is the speed of ship B as observed from ship A.
- \( V_{A} \) is the speed of ship A relative to Earth.
- \( V_{B} \) is the speed of ship B relative to Earth.
- \( c \) is the speed of light.

### Step-by-step Solution:

1. **Identify the speeds of the spaceships relative to Earth**:
- Speed of ship A, \( V_{A} = 0.40c \)
- Speed of ship B, \( V_{B} = 0.50c \)

2. **Determine the relative speed of ship B as observed from ship A**:
- Since ship A is moving towards ship B, we need to consider the direction of the velocities. Thus, we will treat the speed of ship A as negative when calculating the relative speed:
\[ V_{A} = -0.40c \]
\[ V_{B} = 0.50c \]

3. **Substitute the values into the relativistic velocity addition formula**:
\[ V_{AB} = \frac{(-0.40c) + (0.50c)}{1 + \frac{(-0.40c)(0.50c)}{c^2}} \]

4. **Calculate the numerator**:
\[ V_{AB} = \frac{0.10c}{1 + \frac{-0.20c^2}{c^2}} = \frac{0.10c}{1 - 0.20} = \frac{0.10c}{0.80} \]

5. **Calculate the final speed**:
\[ V_{AB} = \frac{0.10c}{0.80} = 0.125c \]

### Final Answer:
The speed of spaceship B as reported by the captain of spaceship A is \( 0.125c \).

---