Frame `S_(1)` moves in the positive x direction at 0.6c with respect to frame `S_(2)`. A particle moves in the positive x direction at 0.4c as measured by an observer in `S_(1)`. The speed of the particle as measured by an observer in `S_(2)` is

Options

A. `c//5`
B. `5c//19`
C. `8c//25`
D. `25c//31`

Answer

To find the speed of the particle as measured by an observer in frame \( S_2 \), we can use the relativistic velocity addition formula. The formula states:

\[ v_{pS_2} = \frac{v_{pS_1} + v_{S_1S_2}}{1 + \frac{v_{pS_1} \cdot v_{S_1S_2}}{c^2}} \]

Where:
- \( v_{pS_2} \) is the speed of the particle as measured in frame \( S_2 \).
- \( v_{pS_1} \) is the speed of the particle as measured in frame \( S_1 \) (which is \( 0.4c \)).
- \( v_{S_1S_2} \) is the speed of frame \( S_1 \) with respect to frame \( S_2 \) (which is \( 0.6c \)).
- \( c \) is the speed of light.

### Step 1: Identify the speeds
From the problem:
- \( v_{pS_1} = 0.4c \)
- \( v_{S_1S_2} = 0.6c \)

### Step 2: Substitute into the formula
Substituting the values into the relativistic velocity addition formula:

\[ v_{pS_2} = \frac{0.4c + 0.6c}{1 + \frac{(0.4c)(0.6c)}{c^2}} \]

### Step 3: Simplify the numerator
The numerator simplifies to:

\[ 0.4c + 0.6c = 1.0c \]

### Step 4: Simplify the denominator
Now, simplify the denominator:

\[ 1 + \frac{(0.4)(0.6)c^2}{c^2} = 1 + 0.24 = 1.24 \]

### Step 5: Combine the results
Now we can combine the results:

\[ v_{pS_2} = \frac{1.0c}{1.24} \]

### Step 6: Calculate the final speed
Calculating the final speed:

\[ v_{pS_2} = \frac{100c}{124} = \frac{25c}{31} \]

### Conclusion
Thus, the speed of the particle as measured by an observer in frame \( S_2 \) is:

\[ v_{pS_2} = \frac{25c}{31} \]