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, vx = 36 km/h = 10 m/s
Velocity of car B, vy = 54 km/h = 15 m/s
Velocity of car C, vz = 54 km/h = 15 m/s
Relative velocity of car B with respect to car A,
vyx = vy – vx = 15 – 10 = 5 m/s
Relative velocity of car C with respect to car A,
vzx = vz – (– vz) = 15 + 10 = 25 m/s
At a certain dinstance, 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 = 1000/25 = 40s
Hence, to avoid an accident, car B must cover the same distance in a maximum of 40 s. From second equation of motion, minimum acceleration (a) produced by car B can be obtained as:
s = ut + ½ at2
1000 = 5 x 40 + ½ x a x 402
a = 1600 / 1600 = 1 m/s2
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Welcome to the NCERT Solutions for Class 11 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 8: On a two-lane road, car A is travelling with a speed of 36 km h–1. Two cars B and C approach c....
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