Calculate the velocity of the car after accelerating for 30.5 m.

The car starts from rest, so its initial velocity ($u_1$) is $0 \text{ m/s}$.
The acceleration ($a_1$) is $5 \text{ m/s}^2$.
The distance traveled ($s_1$) is $30.5 \text{ m}$.
We use the kinematic equation: $v_1^2 = u_1^2 + 2a_1s_1$.

$$v_1^2 = (0 \text{ m/s})^2 + 2(5 \text{ m/s}^2)(30.5 \text{ m})$$ $$v_1^2 = 10 \times 30.5 \text{ m}^2/\text{s}^2$$ $$v_1^2 = 305 \text{ m}^2/\text{s}^2$$ $$v_1 = \sqrt{305} \text{ m/s}$$ $$v_1 \approx 17.464 \text{ m/s}$$

This velocity $v_1$ is the initial velocity for the braking phase.

For the braking phase, the initial velocity ($u_2$) is $v_1 = \sqrt{305} \text{ m/s}$.
The car stops, so the final velocity ($v_2$) is $0 \text{ m/s}$.
The time taken to stop ($t_2$) is $4.0 \text{ s}$.
We use the kinematic equation: $v_2 = u_2 + a_2t_2$.

$$0 \text{ m/s} = \sqrt{305} \text{ m/s} + a_2(4.0 \text{ s})$$ $$a_2(4.0 \text{ s}) = -\sqrt{305} \text{ m/s}$$ $$a_2 = -\frac{\sqrt{305}}{4.0} \text{ m/s}^2$$ $$a_2 \approx -\frac{17.464}{4.0} \text{ m/s}^2$$ $$a_2 \approx -4.366 \text{ m/s}^2$$

The acceleration during braking is approximately $-4.37 \text{ m/s}^2$.

We use the kinematic equation: $s_2 = \frac{u_2 + v_2}{2}t_2$.

$$s_2 = \frac{\sqrt{305} \text{ m/s} + 0 \text{ m/s}}{2}(4.0 \text{ s})$$ $$s_2 = \frac{\sqrt{305}}{2} \times 4.0 \text{ m}$$ $$s_2 = 2\sqrt{305} \text{ m}$$ $$s_2 \approx 2 \times 17.464 \text{ m}$$ $$s_2 \approx 34.928 \text{ m}$$

The distance traveled while braking is approximately $34.9 \text{ m}$.

The acceleration while braking is $\boxed{-4.37 \text{ m/s}^2}$.
The distance traveled while braking is $\boxed{34.9 \text{ m}}$.