On a two lane road , car (A) is travelling with a speed of `36 km h^(-1)`. Tho car ` 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 accelration of car (B) is required to avoid and accident.

Explanation

Velocity of a car A, `V_(A)=36 km//h=10 m//s`
Velocity of car B, `V_(B)=54 km//h=15 m//s`
Velocity of car C, `V_(C)=54 km//h=15 m//s`
Relative velocity of car B with respect to car A
`V_(BA)=V_(B)-V_(A)=15-10=5 m//s`
Relative velocity of car C with respect to car A
`V_(CA)=V_(C)+V_(A)=15+10=25 m//s`
At a certain instance, both cars B and C are at the same distance from car A i.e., `s=1km=1000m`
Time taken (t) by car C to cover `1000m` is
`t=(1000)/(25)=40s`
The acceleration rpduced by car B is
`1000=5xx40+(1)/(2)at^(2)a=(1600)/(1600)=1 m//s^(2)`