- To find the constant speed  \( v_2 \) of a car that travels 650 meters in 130 seconds with an acceleration of 1.3 m/s², constant speed \( v_2 \), and deceleration of 1 m/s², we use the following steps:

1. **Acceleration Phase:**
   - Time to accelerate: \( t_1 = \frac{v_2}{1.3} \)
   - Distance covered: \( s_1 = \frac{v_2^2}{2 \cdot 1.3} = \frac{v_2^2}{2.6} \)

2. **Deceleration Phase:**
   - Time to decelerate: \( t_3 = \frac{v_2}{1} = v_2 \)
   - Distance covered: \( s_3 = \frac{v_2^2}{2 \cdot 1} = \frac{v_2^2}{2} \)

3. **Constant Speed Phase:**
   - Time at constant speed: \( t_2 = 130 - t_1 - t_3 = 130 - \frac{v_2}{1.3} - v_2 \)
   - Distance covered: \( s_2 = v_2 \cdot t_2 \)

4. **Total Distance and Time Equations:**
   - Total distance: \( \frac{v_2^2}{2.6} + v_2 \left(130 - \frac{v_2}{1.3} - v_2 \right) + \frac{v_2^2}{2} = 650 \)
   - Total time: \( \frac{v_2}{1.3} + v_2 + \left(130 - \frac{v_2}{1.3} - v_2 \right) = 130 \)

5. **Quadratic Equation:**
   - Simplifying the distance equation leads to the quadratic equation:
     \[
     0.884615v_2^2 - 130v_2 + 650 = 0
     \]
   - Solving this quadratic equation using the quadratic formula:
     \[
     v_2 = \frac{130 \pm \sqrt{130^2 - 4 \cdot 0.884615 \cdot 650}}{2 \cdot 0.884615}
     \]
   - After calculating the discriminant and solving, we find the feasible solution for \( v_2 \) is approximately 5.18 m/s.

Thus, the constant speed \( v_2 \) is \(\boxed{5.18}\) m/s.