On a two lane road, car `A` is travelling with a speed of `36 km h^(-1)`, Two cars `B and C` approach car `A` in opposite directions with a speed of `34 km h^(-1)`. At a certain instant, when the distance `AB` is equal to `AC`, both `1` km `B` decided to overtake `A` before `C` does. What minimum acceleration of car `B` is required to avoid and accident?.

Explanation

To solve the problem step-by-step, we will analyze the motion of the cars involved and apply the relevant equations of motion.

### Step 1: Convert Speeds to Meters per Second
First, we need to convert the speeds of the cars from kilometers per hour to meters per second.

- Speed of car A:
\[
V_A = 36 \text{ km/h} = 36 \times \frac{1000}{3600} = 10 \text{ m/s}
...