on a two lane road, car A is travelling with a speed of 36 km/hr. Two cars B and C approach car A in opposite directions with a speed of 54 km/hr each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to over take A before C does. What minimum acceleration of car B is required to avoid an accident?
(1) slightly greater than 1 m s–²
(2) slightly less than 1 m s–²
(3) Equal to 1 m s–2
(4) None of these
Answers
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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 = 1000/ 25 = 40
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 +1/2 at^2
1000= 5 × 40 = 1/2 ×a × (40)^2
a = 1600/1600 = 1 m/s^2
I hope, this will help you
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Explanation:
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 = 1000 / 25 = 40 s
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 + (1/2)at2
1000 = 5 × 40 + (1/2) × a × (40)2
a = 1600 / 1600 = 1 ms-2