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

A

Explanation

The situation can be roughly shown in the Figure. Let C take time t to overtake A. `d_C//A=1000m `
`v_(C//A) =(10+15) =25 ms^(-1)`
` (##VMC_PHY_XI_WOR_BOK_01_C02_E03_043_S01.png" width="80%">
Here ` t=( d_(C//A))/(v_(C//A)) =(1000)/(25) =40s `
Let acceleration of B be a for overtaking
`d_(B//A) =1000 m ," "u_(B//A)= 15-10 =5 ms ^(-1) ," " a_(B//A) =a and t = 40 s `
Using ` d_(B//A) =u_(B//A) t +(1)/(2) a_(B//A)t^(2)`
` 1000 =(5xx 40 )+(1)/(2) a (40 ) ^(2) " " 1000 =200 +800 , a =1ms^(-2)`