14.
On a two lane road, a car A is travelling with a speed of 36 kmhr-1 Two cars B and C approach
car A in opposite direction with a speed of 54 kmhrl each. At a certain instant, when the
distance AB equals AC, both being 1 km, B decides to overtake A before C does. What
maximum acceleration for B is required to avoid an accident?
a) 1
b) 2
C) 1.5
d) 0.5
Answers
Answer:
option a = 1
Explanation:
Car A is in the middle. Car B and C are on either side of car A. Let car B travel in the same direction as A.
Va = speed of car A = 36 kmph
Vb = - 54 kmph = Vc
Relative velocity of c wrt A : 54+36 = 90 kmph
distance between them = 1 km
time to cross = 1/90 hrs
Initial Relative velocity of car B wrt A : 54 - 36 = 18 kmph
max. Time to cross = 1/90 hrs
Distance : 1 km
s = u t + 1/2 a t²
1 = 18 / 90 + 1/2 * a /90²
a = 2*90 * 72 *1000/(3600*3600) m/s^2
a = 1 m/s^2 or 12,960 km/hour^2
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