A wave is travelling along positive x- direction with velocity `2m//s`.
Further, `y(x)` equation of the wave pulse at `t=0` is
`y=(10)/(2+(2x+4)^(2))`
(a) From the given information make complete `y(x, t)` equation.
(b) Find `y(x)` equation at `t = 1s`

Explanation

To solve the given problem step by step, we will break it down into two parts as specified in the question.

### Part (a): Finding the complete y(x, t) equation

1. **Identify the form of the wave equation**:
The general form of a wave traveling in the positive x-direction is given by:
\[ y(x, t) = f(kx - \omega t) \]
where \( k \) is the wave number and \( \omega \) is the angular frequency.

2. **Given information**:
The wave pulse at \( t = 0 \) is given by:
\[ y(x, 0) = \frac{10}{2 + (2x + 4)^2} \]

3. **Rewrite the equation for \( t = 0 \)**:
Since at \( t = 0 \), we can express the wave function as:
\[ y(x, 0) = \frac{10}{2 + (2x - 0 + 4)^2} \]
This suggests that the function \( f \) can be represented as:
\[ f(kx - \omega t) = \frac{10}{2 + (2x + 4)^2} \]

4. **Determine the wave parameters**:
The wave is traveling with a velocity \( v = 2 \, \text{m/s} \). We know that:
\[ v = \frac{\omega}{k} \]
Therefore, we can express \( \omega \) in terms of \( k \):
\[ \omega = 2k \]

5. **Find the wave number \( k \)**:
The wave number \( k \) can be determined from the spatial part of the equation. From the equation \( (2x + 4) \), we can see that:
\[ k = 2 \, \text{(from the term \( 2x \))} \]

6. **Calculate \( \omega \)**:
Using the relationship \( \omega = 2k \):
\[ \omega = 2 \times 2 = 4 \]

7. **Construct the complete wave equation**:
Substitute \( \omega \) and \( k \) into the wave equation:
\[ y(x, t) = \frac{10}{2 + (2x - 4t + 4)^2} \]

### Part (b): Finding y(x) at t = 1s

1. **Substitute \( t = 1 \) into the wave equation**:
Now we need to find \( y(x) \) at \( t = 1 \):
\[ y(x, 1) = \frac{10}{2 + (2x - 4(1) + 4)^2} \]

2. **Simplify the equation**:
Simplifying the expression:
\[ y(x, 1) = \frac{10}{2 + (2x - 4 + 4)^2} \]
This simplifies to:
\[ y(x, 1) = \frac{10}{2 + (2x)^2} \]

3. **Final result**:
Thus, the equation for \( y(x) \) at \( t = 1 \) second is:
\[ y(x, 1) = \frac{10}{2 + 4x^2} \]

### Summary of Answers:
- (a) The complete wave equation is:
\[ y(x, t) = \frac{10}{2 + (2x - 4t + 4)^2} \]
- (b) The equation at \( t = 1 \) second is:
\[ y(x, 1) = \frac{10}{2 + 4x^2} \]