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)` each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?

Explanation

At the instant when car `B` decides to overtake car `A`, the velocities of cars are:
`v_(A) =36xx(5)/(18)=10 m s^(-1)`
`v_(B)=54xx(5)/(18)=15 m sw^(-1)`
and `v_(C)=-54xx(5)/(18)=-15 m s^(-1)`
.
velocity of car `B` relative to `A`,
`vec v_(BA)=vec v_(B) -vec v_(A) =15-10=5 m s^(-1)`
Velocity of car `C` relative to `A`,
`vec v_(CA)=vec v_(c) -vec v_(A) =15-10=25 m s^(-1)`
Time required by car `C` to just cross `A`
`=(1000)/((CA)=(1000)/(25)=40 S`
In order to avoid accident, car `B` must overtake `A` in this time.
So
`1000=v_(BA)t+(1)/(2)a_(BA)t^(2)`
`1000=x 40+(1)/(2) a_(BA)xx 40^(2)`
`a_(BA)=1 m s^(-2)`
Thus, the minimum acceleration that car `B` requires to avoid an.