To solve this problem, we need to integrate the velocity components to find the position and then calculate the total distance traveled in 1 second. Given the velocity components:
\[ v_x = -y, \quad v_y = a, \quad v_z = v_2 \]
At \( t = 0 \), the initial position is \((v_2, 0, 0)\).
1. **Integrate \( v_y \) to find \( y(t) \):**
\[ v_y = a \Rightarrow y(t) = at \]
2. **Integrate \( v_z \) to find \( z(t) \):**
\[ v_z = v_2 \Rightarrow z(t) = v_2 t \]
3. **Substitute \( y(t) \) into \( v_x \):**
\[ v_x = -y = -at \]
4. **Integrate \( v_x \) to find \( x(t) \):**
\[ v_x = -at \Rightarrow x(t) = \int -at \, dt = -\frac{a t^2}{2} + C \]
Using the initial condition \( x(0) = v_2 \):
\[ v_2 = C \Rightarrow x(t) = v_2 - \frac{a t^2}{2} \]
Now, we have the position functions:
\[ x(t) = v_2 - \frac{a t^2}{2}, \quad y(t) = at, \quad z(t) = v_2 t \]
5. **Calculate the distance traveled in 1 second:**
\[ dx = x(1) - x(0) = \left(v_2 - \frac{a (1)^2}{2}\right) - v_2 = -\frac{a}{2} \]
\[ dy = y(1) - y(0) = a \cdot 1 - 0 = a \]
\[ dz = z(1) - z(0) = v_2 \cdot 1 - 0 = v_2 \]
The total distance traveled is:
\[ d = \sqrt{(dx)^2 + (dy)^2 + (dz)^2} = \sqrt{\left(-\frac{a}{2}\right)^2 + a^2 + v_2^2} \]
\[ d = \sqrt{\frac{a^2}{4} + a^2 + v_2^2} \]
\[ d = \sqrt{\frac{5a^2}{4} + v_2^2} \]
\[ d = \sqrt{\frac{5a^2}{4} + v_2^2} \]
So, the distance traveled by the particle in 1 second is:
\[ \sqrt{\frac{5a^2}{4} + v_2^2} \, \text{meters} \]
