3.8 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. What minimum acceleration of car B is required to avoid an accident ?
Velocity of car A, vA = 36 km/h = 10 m/s
Velocity of car B, vB = 54 km/h = 15 m/s
Velocity of car C, vC = 54 km/h = 15 m/s
Relative velocity of car B with respect to car A,
vBA=vB−vA=15−10=5 m/s
Relative velocity of car C with respect to car A,
vCA=vC−(−vA)=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 = 1 km = 1000 m
Time taken (t) by car C to cover 1000 m =100025=40s
Hence, to avoid an accident, car B must cover the same distance in a maximum of 40 s. From the second equation of motion, minimum acceleration (a) produced by car B can be obtained as:
s=ut+12at2
1000=5×40+12×a×(40)2
⇒a=16001600=1m/s2
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