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Question
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?
Solution 1
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 at^2`
`1000 = 5 xx 40 +1/2 xx a xx (40)^2`
`a= 1600/1600 = 1 "m/s"^2`
Solution 2
Speed of car A, `v_A = 35 "km/h" = 36 xx 5/18 = 10 "m/s"`
Speed of car B, `v_B = 54 "km/h" = 54 xx 5/18 = 15 "m/s"`
Relative speed of car A w.r.t car `C = v_(AC) = (10+15) ms^(-1) = 25 ms^(-1)`
Relative speed of car B w.r.t car `A = v_(BA) = (15-10) ms^(-1) = 5 "m/s"`
Time taken by car C to cover distance AC,
`t = 1000/v_(AC) = 1000/25 = 40s`
if a is the acceleration, then
`S = ut + 1/2 at^2`
`=>1000 = 5 xx 40 +1/2a xx (40)^2`
`=> a = (1000-200)/800 = 1 "m/s"^2`
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