On a two - lane road, car A is travelling with a speed of 36 kmph. Two cars B and C approach car A in opposite direction with a speed of 54 kmph 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

Velocity of car `A=36km h^(-1)=10m s^(-1)`
velocity of car `B=54km h^(-1)=15 m s^(-1)`
Velocity of car `C=54km h^(-1)=15m s^(-1)`
The relative velocity of B w.r.t. A=`(15-10)m s^(-1)`
`=5m s^(-1)`
The relative velocity of C w.r.t. `A=(15+10)m s^(-1)`
`=25m s^(-1)`
At an instant when both the cars B and C are at same distance of 1km from car A
then time taken by C to cover a distance of `1000m=1000//25=40`s
Hence the car B should travel the same distance in a time less than 40s . Therefore, the minimum acceleration produced by car B can be found by using
`s=ut+(1)/(2)at^(2)`
`1000=5xx40+(1)/(2)xxaxx40xx40`
`800=(1)/(2)a xx 40xx40`
`a=(800xx2)/(40xx40)=1 ms^(-2)`