Calculate the distance traveled during the acceleration phase.

The car starts from rest ($u = 0 \text{ m s}^{-1}$) and accelerates uniformly to $v = 20 \text{ m s}^{-1}$ in $t = 5 \text{ s}$.
Using the kinematic equation $s = \frac{u+v}{2}t$:

$$s_1 = \frac{0 \text{ m s}^{-1} + 20 \text{ m s}^{-1}}{2} \times 5 \text{ s}$$ $$s_1 = \frac{20 \text{ m s}^{-1}}{2} \times 5 \text{ s}$$ $$s_1 = 10 \text{ m s}^{-1} \times 5 \text{ s} = 50 \text{ m}$$

The car travels at a constant velocity of $v = 20 \text{ m s}^{-1}$ for $t = 10 \text{ s}$.
Using the formula $s = v \times t$:

$$s_2 = 20 \text{ m s}^{-1} \times 10 \text{ s} = 200 \text{ m}$$

The car applies brakes, slowing from $u = 20 \text{ m s}^{-1}$ to a stop ($v = 0 \text{ m s}^{-1}$) in $t = 6 \text{ s}$.
Using the kinematic equation $s = \frac{u+v}{2}t$:

$$s_3 = \frac{20 \text{ m s}^{-1} + 0 \text{ m s}^{-1}}{2} \times 6 \text{ s}$$ $$s_3 = \frac{20 \text{ m s}^{-1}}{2} \times 6 \text{ s}$$ $$s_3 = 10 \text{ m s}^{-1} \times 6 \text{ s} = 60 \text{ m}$$

The total distance is the sum of the distances from all three phases.

$$D_{\text{total}} = s_1 + s_2 + s_3$$ $$D_{\text{total}} = 50 \text{ m} + 200 \text{ m} + 60 \text{ m} = 310 \text{ m}$$

The total distance traveled is $\text{310 m}$.