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. The minimum required acceleration of car B to avoid an accident is
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Velocity of car A,
Velocity of car B,
Velocity of car C,
Relative velocity of car B w.r.t. car A
Relative velocity of car C w.r.t. car A is
At a certain instant, both cars B and C are at the same distance form car A
i.e., AB = BC = 1 lm = 1000m
The taken by car C to cover 1 km to reach car a 
In order to avoid an accident, the car B accelerates such that it overtakes car A in less than 40s. Let the minimum required accelerations be a. Then,
, t = 40s, S = 1000 m, a = ?
As 
800a = 1000 – 200 = 800 or a = 1 
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