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