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 ?
Answers
Answer:
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
Answer:
Speed of A=36km/hr=36×
18
5
m/s=10m/s
Speed of B= Speed of C=54×
18
5
m/s=15m/s
Relative speed of A w.r.t C=10+15=25m/s
Time taken by C to overtake A=
25
1000
=40sec.
Distance travelled by A all this time =10×40=400m
So, B have to cover distance of (1000m+400m)
i.e. 1400m to take over a A before C does. in 40sec.
Now putting in formula;-
s=ut+
2
1
at
2
1400=15×40+
2
1
×a×(40)
2
or, 800=a×800
a=1m/s
2
Ans
Explanation: