For uniformly accelerated motion, the average velocity over a time interval is equal to the instantaneous velocity at the midpoint of that time interval.

Let $d$ be the distance of each interval, $t_1$ be the time for the first interval, and $t_2$ be the time for the second interval.

For the first interval, the car travels a distance $d$ in time $t_1$. The average velocity is $\bar{v}_1 = d/t_1$. This is the instantaneous velocity, $v_A$, at the time midpoint $t_A = t_1/2$.

$$v_A = \frac{d}{t_1}$$

For the second interval, the car travels a distance $d$ in time $t_2$. This interval starts at time $t_1$ and ends at $t_1+t_2$. The average velocity is $\bar{v}_2 = d/t_2$. This is the instantaneous velocity, $v_B$, at the time midpoint of this interval, $t_B = t_1 + t_2/2$.

$$v_B = \frac{d}{t_2}$$

The acceleration $a$ is constant and can be calculated from these two instantaneous velocities and the time elapsed between them.

$$a = \frac{\Delta v}{\Delta t} = \frac{v_B - v_A}{t_B - t_A}$$

The time difference is:

$$\Delta t = t_B - t_A = \left(t_1 + \frac{t_2}{2}\right) - \frac{t_1}{2} = \frac{t_1 + t_2}{2}$$

Substituting the expressions for $v_A$, $v_B$, and $\Delta t$:

$$a = \frac{d/t_2 - d/t_1}{(t_1 + t_2)/2}$$ $$a = \frac{2d\left(\frac{1}{t_2} - \frac{1}{t_1}\right)}{t_1 + t_2} = \frac{2d\left(\frac{t_1 - t_2}{t_1 t_2}\right)}{t_1 + t_2}$$

This simplifies to the final expression for acceleration:

$$a = \frac{2d(t_1 - t_2)}{t_1 t_2 (t_1 + t_2)}$$

Now, we substitute the given values: $d = 50$ m, $t_1 = 5$ s, and $t_2 = 4$ s.

$$a = \frac{2(50)(5 - 4)}{5 \cdot 4 (5 + 4)} = \frac{100(1)}{20(9)} = \frac{100}{180}$$ $$a = \frac{5}{9} \text{ m/s}^2$$