Solve it in English

The initial speed is $u = 20 \text{ m s}^{-1}$.
The final speed is $v = 8 \text{ m s}^{-1}$.
The deceleration is $0.4 \text{ m s}^{-2}$, so the acceleration is $a = -0.4 \text{ m s}^{-2}$.

\begin{align*} 8 - 20 &= -0.4t \ -12 &= -0.4t \ t &= \frac{-12}{-0.4} \ t &= 30 s \end{align*}
The time for which the car is decelerating is $\boxed{\text{30 s}}$.

• Segment 1 (Constant speed $20 \text{ m s}^{-1}$ for $25 \text{ s}$): $$d_1 = \text{speed} \times \text{time} = 20 \text{ m s}^{-1} \times 25 \text{ s} = 500 \text{ m}$$
• Segment 2 (Deceleration from $20 \text{ m s}^{-1}$ to $8 \text{ m s}^{-1}$ over $30 \text{ s}$): $$d_2 = \left(\frac{u+v}{2}\right)t = \left(\frac{20 \text{ m s}^{-1} + 8 \text{ m s}^{-1}}{2}\right) \times 30 \text{ s} = \left(\frac{28}{2}\right) \times 30 \text{ s} = 14 \text{ m s}^{-1} \times 30 \text{ s} = 420 \text{ m}$$
• Segment 3 (Constant speed $8 \text{ m s}^{-1}$ for $60 \text{ s}$): $$d_3 = \text{speed} \times \text{time} = 8 \text{ m s}^{-1} \times 60 \text{ s} = 480 \text{ m}$$

The total distance is $1960 \text{ m}$.

In this segment, the car accelerates from $u = 8 \text{ m s}^{-1}$ to $v = 20 \text{ m s}^{-1}$ over a distance $d_4 = 560 \text{ m}$.
Using the equation $d = \left(\frac{u+v}{2}\right)t$:

Total time $T = t_1 + t_2 + t_3 + t_4$

The total time taken for the car to move from $A$ to $B$ is $\boxed{\text{155 s}}$.