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 `54 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?

Answer

1

Explanation

The situation can be rought shown in the Fig.
Let `C` take time `t` to overtake `A`.
.
`d_(rel) =1000 m, v_(rel) =(10 +15) =25 m s^(-1)`
Here `t=(drel)/(v_(rel) =(1000)/(25) =40 s`
Let acceleration of `B` be a for overtaking . ltbtgt `d_(rel) =1000 m, v_(rel) =15 - 10 =5 m s^(-1)`
`d_(rel) =a` and `t=40 s`
Using `d_(rel) =u_(rel) t+(1)/(2) a_(rel) t^(2)`
`S_(2) =(1)/(2) at^(2) =(1)/(2) 2xx 2xx 5^(2) rArr a=1 m s^(-2)`.